Logic problem (1 Viewer)

[This_Guy]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.

You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

How is this guy able to avoid picking the red card in the BR and RB piles EVERYTIME?

 

[FatMax]

You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

I'm just getting back into this, so I apologize if it has already been said, but what do the 50%ers claim is wrong with this logic?

 

[FatMax]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.

You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

How is this guy able to avoid picking the red card in the BR and RB piles EVERYTIME?

I don't think he is talking about specific cards, rather, he is talking about what cards are contained within the set.

 

[sartre]

I might have struck upon what some of the confusion is. If the moderator looks at both cards, I think the odds change depending upon what the moderator's motivation is:

1) If the moderator always says "there's at least one black card" whenever there's a black card, then there's a 66% chance the other card is red.

2) If the moderator always says "there's at least one red card" whenever there's a red card, then there's a 66% chance the other card is black.

3) If the moderator randomly chooses to say either "there's at least one black card" or "there's at least one red card" whenever the cards are different, the odds that the other card is different are 50%.

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%. That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

By the way, fatguy, I though we agreed that the moderator doesn't look at both the cards. He just (randomly) picks one or the other. But he doens't know which one he picked and neither do you. All anybody knows is "at least one is black."

If he only looks at one and says "at least one is black," that's the same thing as him saying "the card I picked is black."

Sure, but that isn't the way the problem is worded, unless the guy doesn't know the sex of one of his kids.

OK, FatMax. An impartial arbiter. You tell us. this is the wording we eventually decided upon for our bet. I will accept Maurile or Ivan. I'm still not convinced that you are confused. But I will take your word on whether my wording is rtarded:

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

I told you several times I'd take the bet. Any of the incarnations of your hypothetical, except the one I pointed out was flawed. In each one of them, you were wrong.As for taking Ivan's word, well that won't do us any good, since he showed he doesn't understand the initial problem (as stated) either. You might as well ask FatMax. And why bother Maurile with this? Let's just do the card version, for real.

You're right in that I'm not impartial. And neither is Ivan. But to say neither of us understand the initial problem is a little bit insulting.

I apologize. But I don't think you comprehend the problem stated in the OP. I think you and Ivan too quickly assumed this was no different than the classic version of the puzzle. It is different, and it is an interesting difference.

 

[rick6668]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed <_< Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

 

[FatMax]

I apologize. But I don't think you comprehend the problem stated in the OP. I think you and Ivan too quickly assumed this was no different than the classic version of the puzzle. It is different, and it is an interesting difference.

Believe me when I say I comprehend it. I know the points you are bringing up. And I'm not saying with 100% confidence that I'm right. Not at all.Give me a few.

 

[FatMax]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

What he is saying is that with an even distribution, at least one black card will be represented in 300 out of 400 piles. Do you agree or disagree with those numbers?

 

[dancer]

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.

Chose a pile that contained at least one black card, or chose a card from the pair and it was black? Two different things.

You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

In last sentence, what is "it"? A selected card that is black? Or a pair of cards that contains at least one black card? Again, two different things. In this example, you have 300 pairs, and yes 200 of them will have one black and one red and 100 will have 2 black. So 2/3 of the time one is red and one is black.

But if your question is as you asked before, you don't have all 300 piles. Because someone is randomly selecting a card and saying it is black. For the RB and BR piles, this will only happen in half of the runs of the scenario, not in every one. So you end up with only 200 observations, not 300. And 100 are BB, 50 RB, 50 BR.

 

[rick6668]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

What he is saying is that with an even distribution, at least one black card will be represented in 300 out of 400 piles. Do you agree or disagree with those numbers?

I agree with those numbers.By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK,

From the scenario that was posted, this is run 400 times.

200 times he will choose one and it will be black.

200 times he will choose one and it will be red.

He chose ONLY one from the pair and looked at it.

200 times it was red. How many of these are RR?

 

[bostonfred]

bostonfred, can I get your opinion on this?

In the OP hypothetical, we go out of our way to say "you don't know if it is the first born or the second born." Which is the same as saying "you don't know if it is the first card picked or the second card picked." Right?

Yeah, it sounds like you have all of the math right, but this quote is the point of what little debate remains. It's not just "first born or second born". If I know that one of them is left handed, and the other one is right handed, and you tell me the left handed one is a male, that's every bit as significant as saying the taller one, the older one, or the one with more pimples is a male. Because at that point, we go from three possible combinations of BB/BG/GB:more pimple boy/less pimple boy

more pimple boy/less pimple girl

more pimple girl/less pimple boy

to

more pimple boy/less pimple boy

more pimple boy/less pimple girl

In this case, their argument is that if the father randomly selects one of their children to tell you about, then we go from three combinations to two:

boy he mentioned/boy he didn't

boy he mentioned/girl he didn't

Their argument is that if we asked the father if he has a boy, then we know that one of his kids is a boy, but we don't have any criteria by which to narrow down from 3 to 2 combinations. But the moment the father chooses to tell us that he has a boy, he is selecting one of his children to mention, and thus giving us a selection criteria.

Of course, if the boy were not randomly selected - for example, you're at a football game, you talk to the guy next to you and ask him how many kids he has, and he says two, and that boy right there's my son - then we're still looking at 2/3.

Excellent post again -- except for the last part. If "you talk to the guy next to you and ask him how many kids he has, and he says two, and that boy right there's my son" then the chance his other child is a girl is still only 50%!If you don't believe me, let's use the example someone posted earlier of Pat and Terry.

You're sitting next to the guy at the football game, and he says he has two kids, Pat and Terry. He then points to the QB and says, "that's Pat". Now do you think that makes Terry 66% likely to be a girl?

Here's what you're missing from the example - the selection criteria used by the father are no longer random. In the case that the father simply volunteers the gender of one of his children, he could choose to volunteer the gender of a boy or a girl, and randomly chose to volunteer the gender of a child that was a boy. Given four fathers, each of whom has two children, we have eight possible random selections, and can eliminate four (he didn't name a girl first).

But in the case where we're watching an all-boy football game, the father is no longer randomly selecting a child to name. He is identifiying the player in the game that is his son. Since he could only choose a boy, the selection criteria are no longer random, and thus the signficant population is based not the selection of the child to name, but the family whose child is named, and as we've discussed ad nauseum, for every three families that have a male child, only one has two male children.

 

[FatMax]

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.

Chose a pile that contained at least one black card, or chose a card from the pair and it was black? Two different things.

You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

In last sentence, what is "it"? A selected card that is black? Or a pair of cards that contains at least one black card? Again, two different things. In this example, you have 300 pairs, and yes 200 of them will have one black and one red and 100 will have 2 black. So 2/3 of the time one is red and one is black.

But if your question is as you asked before, you don't have all 300 piles. Because someone is randomly selecting a card and saying it is black. For the RB and BR piles, this will only happen in half of the runs of the scenario, not in every one. So you end up with only 200 observations, not 300. And 100 are BB, 50 RB, 50 BR.

Good post. The bolded is the crux of the disagreement for those who understand the problem.* I assume that we're talking about option A. Other think B.*by understand the problem, I talking about those who realize that we're not trying to use known past events to predict the future ... and there are a few of those guys here on the 50/50 side.

 

[kutta]

You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

I'm just getting back into this, so I apologize if it has already been said, but what do the 50%ers claim is wrong with this logic?

The problem is that he is MORE LIKELY to have picked a black card from the BB pile than he is to have picked it from the RB pile. He is equally likely to have picked the black card from either the BB pile, or the set of the (RB, and BR) piles. So the odds are not 33% that it came from one of the piles. The odds are heavier in favor of him picking the black card from the BB pile, so the odds of the black card coming from each pile are BB - 50, RB - 25, BR - 25%, RR - 0%.Now if you said, "I have two cards and at least one is black. What is the chance the other one is red?" Then it would be 66% because there are black cards in three piles. Just as in this problem:BBBBBBBBBBRRRRRRRRBRRRRRRRRRRRRRRRRRRIf I randomly select a stack, then randomly select a card and end up with a black card here, the odds that I picked it out of the first stack are very high. But if you pick a random stack and tell me that stack has at least one black card in it, what are the odds the stack it came from also has a red card, that is a completely different problem. Those odds are 2/3.

 

[This_Guy]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.

You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

How is this guy able to avoid picking the red card in the BR and RB piles EVERYTIME?

I don't think he is talking about specific cards, rather, he is talking about what cards are contained within the set.

Unless he can somehow avoid picking the red cards he would only end up with a sample space of 200 cards - 100 BB, 50 RB and 50 BR. You will lose half of the BR and RB cards because you will chose the red card sometimes.

 

[Sweet J]

I told you several times I'd take the bet. Any of the incarnations of your hypothetical, except the one I pointed out was flawed. In each one of them, you were wrong.

I'm not ignoring you. There is a real possibility that I am wrong. And if I make a promise to pay out all over the place, I'm gonna make someone unhappy cause I will ultimately forget to pay someone, or my wife will see a bunch of PP transactions and wondering who I'm paying 10 bucks at a time. The "bet" is really symbolic anyway.

 

[FatMax]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

What he is saying is that with an even distribution, at least one black card will be represented in 300 out of 400 piles. Do you agree or disagree with those numbers?

I agree with those numbers.By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK,

From the scenario that was posted, this is run 400 times.

200 times he will choose one and it will be black.

200 times he will choose one and it will be red.

He chose ONLY one from the pair and looked at it.

200 times it was red. How many of these are RR?

Right, and I understand that. What I don't like is that the moderator choses one card without looking at the other. To tie the analogy back to kids, that is like saying that a parent only knows the sex of one of his children. To me (and I could be wrong), I think that the moderator holds either RB, BR, or BB and knows what both cards are. Regardless, he shows one black.Now, that scenario is admittedly slanted to the 66% side, but that is how I read the question.

 

[Sweet J]

"Joe has two kids, at least one is a boy. what are the odds on the other."

flawed

What is flawed here? This is the classic 66% problem.

 

[Sweet J]

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.

Chose a pile that contained at least one black card, or chose a card from the pair and it was black? Two different things.

He choses a pile that contained at least one black card 300 times.

 

[kutta]

Right, and I understand that. What I don't like is that the moderator choses one card without looking at the other. To tie the analogy back to kids, that is like saying that a parent only knows the sex of one of his children. To me (and I could be wrong), I think that the moderator holds either RB, BR, or BB and knows what both cards are. Regardless, he shows one black.Now, that scenario is admittedly slanted to the 66% side, but that is how I read the question.

We are no longer talking about kids. That just confuses the issue. Let's stick with cards.We agree if the moderator knows both cards and simply says "at least one is black," then there is a 2/3 chance the other is red.If the moderator only looks at one card and calls it out, there is a 50/50 chance the other is the other color.AND if the moderator knows both cards but somehow randomly chooses to reveal only one of them, there is a 50/50 chance the other is red.We can argue all day if the OP was the 1st case or the 3rd case. But we are now arguing about cards.

 

[Sweet J]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

I am not changing it. I am saying he is picking "piles," not "cards." He make a selection of piles 400 times. Of those 400 times he selects a pile, 1/4 of the time it will come from the BB pile (100 piles), 1/4 BR(100), 1/4 RB(100), and 1/4 RR (100).Now he is told that a black is in his pile. So he knows he has one of three piles: it is either one of the 100 BB, 100 BR, or 100 RB. He has a roughly equivalent chance of getting any one. 1/3 each. That means he only has a 1/3 chance of getting another Black, and 2/3 chance of Red.

 

[Sweet J]

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.

Chose a pile that contained at least one black card, or chose a card from the pair and it was black? Two different things.

You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

In last sentence, what is "it"? A selected card that is black? Or a pair of cards that contains at least one black card? Again, two different things. In this example, you have 300 pairs, and yes 200 of them will have one black and one red and 100 will have 2 black. So 2/3 of the time one is red and one is black.

But if your question is as you asked before, you don't have all 300 piles. Because someone is randomly selecting a card and saying it is black. For the RB and BR piles, this will only happen in half of the runs of the scenario, not in every one. So you end up with only 200 observations, not 300. And 100 are BB, 50 RB, 50 BR.

Good post. The bolded is the crux of the disagreement for those who understand the problem.* I assume that we're talking about option A. Other think B.*by understand the problem, I talking about those who realize that we're not trying to use known past events to predict the future ... and there are a few of those guys here on the 50/50 side.

Yes, option A.

 

[kutta]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

I am not changing it. I am saying he is picking "piles," not "cards." He make a selection of piles 400 times. Of those 400 times he selects a pile, 1/4 of the time it will come from the BB pile (100 piles), 1/4 BR(100), 1/4 RB(100), and 1/4 RR (100).Now he is told that a black is in his pile. So he knows he has one of three piles: it is either one of the 100 BB, 100 BR, or 100 RB. He has a roughly equivalent chance of getting any one. 1/3 each. That means he only has a 1/3 chance of getting another Black, and 2/3 chance of Red.

Again with the wording. "Now he is told" is different from "he randomly selects a card." For the life of me I can't see why you don't see the difference.

 

[FatMax]

Right, and I understand that. What I don't like is that the moderator choses one card without looking at the other. To tie the analogy back to kids, that is like saying that a parent only knows the sex of one of his children. To me (and I could be wrong), I think that the moderator holds either RB, BR, or BB and knows what both cards are. Regardless, he shows one black.Now, that scenario is admittedly slanted to the 66% side, but that is how I read the question.

We are no longer talking about kids. That just confuses the issue. Let's stick with cards.We agree if the moderator knows both cards and simply says "at least one is black," then there is a 2/3 chance the other is red.If the moderator only looks at one card and calls it out, there is a 50/50 chance the other is the other color.AND if the moderator knows both cards but somehow randomly chooses to reveal only one of them, there is a 50/50 chance the other is red.We can argue all day if the OP was the 1st case or the 3rd case. But we are now arguing about cards.

OK, no kids. The moderator has to know both cards. The second case doesn't apply to this problem. And I agree with the third, assuming that if the random first draw shows red, it is discarded.So I guess it come down to if you believe he randomly picks the card.

 

[Sweet J]

"I have two cards and at least one is black. What is the chance the other one is red?"

That is all I've been saying. And we agree. I'm not sure what there is to fight about.

 

[sartre]

"Joe has two kids, at least one is a boy. what are the odds on the other."

flawed

What is flawed here? This is the classic 66% problem.

Anyone else care to show what the flaw is? Sweet J might take it more seriously if it came from someone in the 2/3 camp.

 

[Sweet J]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

I am not changing it. I am saying he is picking "piles," not "cards." He make a selection of piles 400 times. Of those 400 times he selects a pile, 1/4 of the time it will come from the BB pile (100 piles), 1/4 BR(100), 1/4 RB(100), and 1/4 RR (100).Now he is told that a black is in his pile. So he knows he has one of three piles: it is either one of the 100 BB, 100 BR, or 100 RB. He has a roughly equivalent chance of getting any one. 1/3 each. That means he only has a 1/3 chance of getting another Black, and 2/3 chance of Red.

Again with the wording. "Now he is told" is different from "he randomly selects a card." For the life of me I can't see why you don't see the difference.

I clarified a long, long, long, time ago. There is a deck of cards, with equal probabilities of being RR, RB, BR, and BB. He is told that one of the cards is (either red or black). What are the odds that the other card is different? It is 66%.This was the bet that I had with fatguy. I'll bump it.

 

[Sweet J]

"Joe has two kids, at least one is a boy. what are the odds on the other."

flawed

What is flawed here? This is the classic 66% problem.

Anyone else care to show what the flaw is? Sweet J might take it more seriously if it came from someone in the 2/3 camp.

Just tell me. Joe has two kids. At least one is a boy. That is the same thing as saying "he is told that at least one is a boy."

 

[FatMax]

"Joe has two kids, at least one is a boy. what are the odds on the other."

flawed

What is flawed here? This is the classic 66% problem.

Anyone else care to show what the flaw is? Sweet J might take it more seriously if it came from someone in the 2/3 camp.

The punctuation?

 

[rick6668]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

I am not changing it. I am saying he is picking "piles," not "cards." He make a selection of piles 400 times. Of those 400 times he selects a pile, 1/4 of the time it will come from the BB pile (100 piles), 1/4 BR(100), 1/4 RB(100), and 1/4 RR (100).Now he is told that a black is in his pile. So he knows he has one of three piles: it is either one of the 100 BB, 100 BR, or 100 RB. He has a roughly equivalent chance of getting any one. 1/3 each. That means he only has a 1/3 chance of getting another Black, and 2/3 chance of Red.

He is not told black is in his pile, he is told that a card was chosen from his pile and it was black. There is a difference, until you see it you will continue to apply flawed logic.In your scenario,

So he looks at one card from the pile selected and it is black 300 out of 400 times?

 

[Sweet J]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

 

[Sweet J]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

I am not changing it. I am saying he is picking "piles," not "cards." He make a selection of piles 400 times. Of those 400 times he selects a pile, 1/4 of the time it will come from the BB pile (100 piles), 1/4 BR(100), 1/4 RB(100), and 1/4 RR (100).Now he is told that a black is in his pile. So he knows he has one of three piles: it is either one of the 100 BB, 100 BR, or 100 RB. He has a roughly equivalent chance of getting any one. 1/3 each. That means he only has a 1/3 chance of getting another Black, and 2/3 chance of Red.

He is not told black is in his pile, he is told that a card was chosen from his pile and it was black. There is a difference, until you see it you will continue to apply flawed logic.In your scenario,

So he looks at one card from the pile selected and it is black 300 out of 400 times?

No. Of the 300 times that he is told that a black card is in his pile, it came from either the BB pile (100 times), the BR pile (100 times), or the RB pile (100 times).

 

[FatMax]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

For you to be right, I think he has to look at both cards.

 

[Sweet J]

I'm trying to take out the "motivation" angle. By saying YOU pick the "pair" (equal chances of BB BR RB RR), and the pair to the moderator, and the moderator randomly choses one of the cards without looking at the other. I think if he choses one, and it is BLACK, then the chances that the other card being red is 66%.

That's what I've been trying to get across, anyway. God, I hope that is what I have been getting across.

Fine. Stick with one scenario and stop changing it. This is it.Let's do the bolded part 100 times.

Would you agree that 50 times he will choose a black card and 50 times he will choose a red card?

I'm not changing. I have to keep clarifying because people are mincing words. But yes, I agree that 50 times he will choose black, and 50 times he will choose a red.

OK. So of those 50/100 that he chose black, how many are BB, BR and RB respectively?

Can we change it to say that he did it 400 times to make the math easier? If so, That meant he ended up choosing a black card 200 times and a red card 200 time. And that also means he chose BB 100 times, he chose BR 100 times, he chose RB 100 times, and he chose RR 100 times. Of course, the times that he picked a red, he didn't say "the card is black." He said "the card is red."Oops on second thought, it looks like I didn't answer your question. Of the times that he chose black, 1/3 of the time it was BB, 1/3 of the time it was BR, and 1/3 of the time it was RB. So of 300 times he picked a black card, 100 of the times it will have come from the BB pile, 100 from the BR pile, and 100 from the RB pile.

OK.. hang on..So he picked a black card 300 times?

We have 400 times.

200 times he selects red

200 times he selects black. - you agreed to this.

I was trying to make the math easier, but it appearently didn't work. He picked 400 separate sets of cards. In those 400 picks, he chose a pile that contained at least one black card 300 times. Of those 300 times he chose a pile containing at least one black card, 100 times it came from the BB pile, 100 times it came from the RB pile, and 100 times it came from a BR pile.You can't start the analysis with "he picks a black card" because that ends up specifying WHICH of the two pairs he picked. The whole idea is that you don't know. So you start with "he picked a pile." Do you agree that if he picks a pair 400 times, that 100 times it will be BB, 100 it will be BR, and 100 it will be RB? Then you just do the math. One out of three times it will have come from the BB pile, and 2/3 from the RB BR pile.

You keep changing it, so this is confirmed Only 200 times out of 400 he selectes black (50%), this you agreed to but now are saying 300.

By only looking at one card he recudes the scenarios by 200.

By looking at both cards he reduces the scenarios by 100.

I am not changing it. I am saying he is picking "piles," not "cards." He make a selection of piles 400 times. Of those 400 times he selects a pile, 1/4 of the time it will come from the BB pile (100 piles), 1/4 BR(100), 1/4 RB(100), and 1/4 RR (100).Now he is told that a black is in his pile. So he knows he has one of three piles: it is either one of the 100 BB, 100 BR, or 100 RB. He has a roughly equivalent chance of getting any one. 1/3 each. That means he only has a 1/3 chance of getting another Black, and 2/3 chance of Red.

Again with the wording. "Now he is told" is different from "he randomly selects a card." For the life of me I can't see why you don't see the difference.

Kind of. As I stated before, when I said "he randomly selects a card" my intention was to signify that he doesn't know if it is the first or the second card. It clarifies things to just say "he is told that the card is black" with is the same as saying "he is told that one of the cards is black." (because, by definition, if the card chosen was black, then "one of the cards is black.")

 

[rick6668]

He is not told black is in his pile, he is told that a card was chosen from his pile and it was black. There is a difference, until you see it you will continue to apply flawed logic.

In your scenario,

So he looks at one card from the pile selected and it is black 300 out of 400 times?

No. Of the 300 times that he is told that a black card is in his pile, it came from either the BB pile (100 times), the BR pile (100 times), or the RB pile (100 times).

Fail

 

[Sweet J]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

For you to be right, I think he has to look at both cards.

Why, and how is it different than saying "Joe has two kids. One of Joe's kids is a boy, what are the odds that the other is a girl?" We agree that this is 66%. But that situation doesn't require "looking" at the sex of the two kids.Same thing. "you've been dealt two cards. At least one of the cards is black. What are the odds that the other card is red? (my hypo, above) doesn't require you look at both the cards.

 

[Sweet J]

He is not told black is in his pile, he is told that a card was chosen from his pile and it was black. There is a difference, until you see it you will continue to apply flawed logic.

In your scenario,

So he looks at one card from the pile selected and it is black 300 out of 400 times?

No. Of the 300 times that he is told that a black card is in his pile, it came from either the BB pile (100 times), the BR pile (100 times), or the RB pile (100 times).

Fail

show me.You agree that if a guy deals at a pair of cards (from an infinite deck) 400 times: 100 times his pair will be BB, 100 will be BR, 100 will be RB, and 100 will be RR. Do you disagree?

 

[FatMax]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

For you to be right, I think he has to look at both cards.

Why, and how is it different than saying "Joe has two kids. One of Joe's kids is a boy, what are the odds that the other is a girl?" We agree that this is 66%. But that situation doesn't require "looking" at the sex of the two kids.Same thing. "you've been dealt two cards. At least one of the cards is black. What are the odds that the other card is red? (my hypo, above) doesn't require you look at both the cards.

If the moderator doesn't look at both cards, he could have a group containing black and not be aware of it.

 

[kutta]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

For you to be right, I think he has to look at both cards.

Why, and how is it different than saying "Joe has two kids. One of Joe's kids is a boy, what are the odds that the other is a girl?" We agree that this is 66%. But that situation doesn't require "looking" at the sex of the two kids.Same thing. "you've been dealt two cards. At least one of the cards is black. What are the odds that the other card is red? (my hypo, above) doesn't require you look at both the cards.

You keep saying you are not changing the hypo, but you are.

You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card.

is different from:

you've been dealt two cards. At least one of the cards is black. What are the odds that the other card is red?

In your first example, let's say the moderator picked the RB stack and flips over the R. He will NOT tell you "at least one card is black." He can't.In your second example, WITH THE EXACT SAME STACK, if someone knows the suit of BOTH CARDS, he will always say, "at least one card is black."

Do you see the difference? Please take some time and think about this and only this. This is all that matters. Nothing else. No kids, no four stacks, no who said what and who picked what. Just these two scenarios.

 

[Sweet J]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

For you to be right, I think he has to look at both cards.

Why, and how is it different than saying "Joe has two kids. One of Joe's kids is a boy, what are the odds that the other is a girl?" We agree that this is 66%. But that situation doesn't require "looking" at the sex of the two kids.Same thing. "you've been dealt two cards. At least one of the cards is black. What are the odds that the other card is red? (my hypo, above) doesn't require you look at both the cards.

If the moderator doesn't look at both cards, he could have a group containing black and not be aware of it.

If his group doesn't include a black card, he says, "at least one of the cards is red." Then we have the following potential sets: RR, RB, BR. 66% chance that the set is not the same color.

 

[dgreen]

Really soon one of you is going to say the right words to convince the other side they are wrong. I can feel it. It's about to happen.

 

[FatMax]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

For you to be right, I think he has to look at both cards.

Why, and how is it different than saying "Joe has two kids. One of Joe's kids is a boy, what are the odds that the other is a girl?" We agree that this is 66%. But that situation doesn't require "looking" at the sex of the two kids.Same thing. "you've been dealt two cards. At least one of the cards is black. What are the odds that the other card is red? (my hypo, above) doesn't require you look at both the cards.

If the moderator doesn't look at both cards, he could have a group containing black and not be aware of it.

If his group doesn't include a black card, he says, "at least one of the cards is red." Then we have the following potential sets: RR, RB, BR. 66% chance that the set is not the same color.

Ah. That does fit within your hypo above.

 

[kutta]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

For you to be right, I think he has to look at both cards.

Why, and how is it different than saying "Joe has two kids. One of Joe's kids is a boy, what are the odds that the other is a girl?" We agree that this is 66%. But that situation doesn't require "looking" at the sex of the two kids.Same thing. "you've been dealt two cards. At least one of the cards is black. What are the odds that the other card is red? (my hypo, above) doesn't require you look at both the cards.

If the moderator doesn't look at both cards, he could have a group containing black and not be aware of it.

If his group doesn't include a black card, he says, "at least one of the cards is red." Then we have the following potential sets: RR, RB, BR. 66% chance that the set is not the same color.

Ah. That does fit within your hypo above.

NO! Fat Max, I thought we had you.Let's just simplify.

BB

RB

BR

RR

Pick a stack. Pick a card.

If you picked BB or RR, the 2nd card will be the same as the first. If you picked RB or BR the 2nd card will be different from the first.

It is soooo simple. There is nothing else to it. Don't try to fit it into the kid problem.

BUT, if I phrased it this way: If you pick a stack that has a red card in it, what are the odds it also contains a black card, THEN the odds are 2/3. Because the RB and BR stacks are always included, when they are only included 1/2 the time with a random selection.

 

[FatMax]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

For you to be right, I think he has to look at both cards.

Why, and how is it different than saying "Joe has two kids. One of Joe's kids is a boy, what are the odds that the other is a girl?" We agree that this is 66%. But that situation doesn't require "looking" at the sex of the two kids.Same thing. "you've been dealt two cards. At least one of the cards is black. What are the odds that the other card is red? (my hypo, above) doesn't require you look at both the cards.

If the moderator doesn't look at both cards, he could have a group containing black and not be aware of it.

If his group doesn't include a black card, he says, "at least one of the cards is red." Then we have the following potential sets: RR, RB, BR. 66% chance that the set is not the same color.

Ah. That does fit within your hypo above.

NO! Fat Max, I thought we had you.Let's just simplify.

BB

RB

BR

RR

Pick a stack. Pick a card.

If you picked BB or RR, the 2nd card will be the same as the first. If you picked RB or BR the 2nd card will be different from the first.

It is soooo simple. There is nothing else to it. Don't try to fit it into the kid problem.

BUT, if I phrased it this way: If you pick a stack that has a red card in it, what are the odds it also contains a black card, THEN the odds are 2/3. Because the RB and BR stacks are always included, when they are only included 1/2 the time with a random selection.

Wait a second. I said it fit. I didn't say it was right.

 

[Sweet J]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

"You are told the suit of the card." It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%. Is anybody going to say I changed the hypo?

For you to be right, I think he has to look at both cards.

Why, and how is it different than saying "Joe has two kids. One of Joe's kids is a boy, what are the odds that the other is a girl?" We agree that this is 66%. But that situation doesn't require "looking" at the sex of the two kids.Same thing. "you've been dealt two cards. At least one of the cards is black. What are the odds that the other card is red? (my hypo, above) doesn't require you look at both the cards.

If the moderator doesn't look at both cards, he could have a group containing black and not be aware of it.

If his group doesn't include a black card, he says, "at least one of the cards is red." Then we have the following potential sets: RR, RB, BR. 66% chance that the set is not the same color.

Ah. That does fit within your hypo above.

NO! Fat Max, I thought we had you.Let's just simplify.

BB

RB

BR

RR

Pick a stack. Pick a card.

If you picked BB or RR, the 2nd card will be the same as the first. If you picked RB or BR the 2nd card will be different from the first.

BUT, if I phrased it this way: If you pick a stack that has a red card in it, what are the odds it also contains a black card, THEN the odds are 2/3. Because the RB and BR stacks are always included, when they are only included 1/2 the time with a random selection.

Except I have said, over and over and over again: "The quesiton is NOT whether the cards match. The quesiton is once you identify one of the cards, what are the odds that the non-identified card matches the identifies card."You have the correct analysis in your second statement.

 

[Sweet J]

BUT, if I phrased it this way: If you pick a stack that has a red card in it, what are the odds it also contains a black card, THEN the odds are 2/3. Because the RB and BR stacks are always included, when they are only included 1/2 the time with a random selection.

That's what I am saying my hypo is. I am NOT (and I never was) saying "what are the odds that the cards match." I am saying you pick a stack that has a (fill in the black card), what are the odds that the other card matches. HUGE difference.

 

[Sweet J]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Are you ok with this hypo?

Well, I'm happy with its clarity. But not with your answer. If he only looks at one card, the odds of the other card being red are 50%.

Once again: The above is the hypo. You are told the suit of the card. It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%.

Kutta, sarte, fatmax, fatguy. Is someone disagreeing?

 

[bostonfred]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Once again: The above is the hypo. You are told the suit of the card. It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%.

Kutta, sarte, fatmax, fatguy. Is someone disagreeing?

If what you are saying were true, then you should go play blackjack. Although the cards are not revealed to you all at once, the cards are already in order on the top of the deck, so the example is identical to the one you've given. The dealer then reveals the results of the first trial. The odds of winning a hand are roughly 50/50. If you sat down and played two hands of blackjack, you could bet small on the first hand. If you won the first one, you would simply leave. If you lost the first one, then your odds of winning the second one are 2/3, so you would increase your bet. Over the long run, you could never lose.

But this is clearly not the case, or all those pretty buildings in Vegas wouldn't be there. Something's wrong. So I'm raising the ante. You're not just disagreeing with everybody else in this thread, you're disagreeing with everyone in Vegas, and everyone who has ever gambled. If you believe you've found some secret edge, though, then I'm sure that Dodds would like to know.

 

[bostonfred]

bostonfred, can I get your opinion on this?

In the OP hypothetical, we go out of our way to say "you don't know if it is the first born or the second born." Which is the same as saying "you don't know if it is the first card picked or the second card picked." Right?

Yeah, it sounds like you have all of the math right, but this quote is the point of what little debate remains. It's not just "first born or second born". If I know that one of them is left handed, and the other one is right handed, and you tell me the left handed one is a male, that's every bit as significant as saying the taller one, the older one, or the one with more pimples is a male. Because at that point, we go from three possible combinations of BB/BG/GB:more pimple boy/less pimple boy

more pimple boy/less pimple girl

more pimple girl/less pimple boy

to

more pimple boy/less pimple boy

more pimple boy/less pimple girl

In this case, their argument is that if the father randomly selects one of their children to tell you about, then we go from three combinations to two:

boy he mentioned/boy he didn't

boy he mentioned/girl he didn't

Their argument is that if we asked the father if he has a boy, then we know that one of his kids is a boy, but we don't have any criteria by which to narrow down from 3 to 2 combinations. But the moment the father chooses to tell us that he has a boy, he is selecting one of his children to mention, and thus giving us a selection criteria.

Of course, if the boy were not randomly selected - for example, you're at a football game, you talk to the guy next to you and ask him how many kids he has, and he says two, and that boy right there's my son - then we're still looking at 2/3.

Excellent post again -- except for the last part. If "you talk to the guy next to you and ask him how many kids he has, and he says two, and that boy right there's my son" then the chance his other child is a girl is still only 50%!If you don't believe me, let's use the example someone posted earlier of Pat and Terry.

You're sitting next to the guy at the football game, and he says he has two kids, Pat and Terry. He then points to the QB and says, "that's Pat". Now do you think that makes Terry 66% likely to be a girl?

Here's what you're missing from the example - the selection criteria used by the father are no longer random. In the case that the father simply volunteers the gender of one of his children, he could choose to volunteer the gender of a boy or a girl, and randomly chose to volunteer the gender of a child that was a boy. Given four fathers, each of whom has two children, we have eight possible random selections, and can eliminate four (he didn't name a girl first).

But in the case where we're watching an all-boy football game, the father is no longer randomly selecting a child to name. He is identifiying the player in the game that is his son. Since he could only choose a boy, the selection criteria are no longer random, and thus the signficant population is based not the selection of the child to name, but the family whose child is named, and as we've discussed ad nauseum, for every three families that have a male child, only one has two male children.

bump for sartre

 

[FatMax]

Fred's analogy doesn't make sense in this context. At least to me.

 

[Sweet J]

Ok, I will go out on a limb and say: You pick a pair, and hand the pair to the moderator. The moderator ramdomly picks either the left or the right, but you did not see him pick the card. You just know he picked one. He looks at JUST ONE CARD, and tells you the suit of that card. Let's say it's black. He tells you "at least one of the cards is black." The odds that the other card is red is 66%.

So, no, he didn't look at both cards.

Once again: The above is the hypo. You are told the suit of the card. It is black. What are the odds that the other card is red? I say 66%. fred and satre and kutta have said 50%.

Kutta, sarte, fatmax, fatguy. Is someone disagreeing?

If what you are saying were true, then you should go play blackjack. Although the cards are not revealed to you all at once, the cards are already in order on the top of the deck, so the example is identical to the one you've given. The dealer then reveals the results of the first trial. The odds of winning a hand are roughly 50/50. If you sat down and played two hands of blackjack, you could bet small on the first hand. If you won the first one, you would simply leave. If you lost the first one, then your odds of winning the second one are 2/3, so you would increase your bet. Over the long run, you could never lose.

But this is clearly not the case, or all those pretty buildings in Vegas wouldn't be there. Something's wrong. So I'm raising the ante. You're not just disagreeing with everybody else in this thread, you're disagreeing with everyone in Vegas, and everyone who has ever gambled. If you believe you've found some secret edge, though, then I'm sure that Dodds would like to know.

The problem is that we are dealing with pre arrainged sets of two, NOT with "the first card is black, what is the second?" We are dealing with pre arranged pairs. All with equal probability. So, if you have four pairs in front of you (RR, RB, BR, and RR). And I say "pick one of those four." You now have a distinct pair of two cards. That pair is one of the two combinations of RR, RB, BR, or RR. We are agreed so far, right?

Now, if I said "I'm going to look at one of the cards from one of the pairs of cards. And I tell you that in your pair, you have at least one Black card." OK?

Now, you know all this information. Random pick of one of four pairs. You are told that one of the cards in one of the pairs is black.

If I say to you: I'll give you $1 if the other (non-picked) card is black (i,e., "they are the same"), and you give me $1 if the card is red.

Would we break even over the long haul, or would I make money?

Fred, fatmax, fatguy, kutta, satre, please answer.

That, in a nutshell, is my hypo.

 

[sartre]

bostonfred, can I get your opinion on this?

In the OP hypothetical, we go out of our way to say "you don't know if it is the first born or the second born." Which is the same as saying "you don't know if it is the first card picked or the second card picked." Right?

Yeah, it sounds like you have all of the math right, but this quote is the point of what little debate remains. It's not just "first born or second born". If I know that one of them is left handed, and the other one is right handed, and you tell me the left handed one is a male, that's every bit as significant as saying the taller one, the older one, or the one with more pimples is a male. Because at that point, we go from three possible combinations of BB/BG/GB:more pimple boy/less pimple boy

more pimple boy/less pimple girl

more pimple girl/less pimple boy

to

more pimple boy/less pimple boy

more pimple boy/less pimple girl

In this case, their argument is that if the father randomly selects one of their children to tell you about, then we go from three combinations to two:

boy he mentioned/boy he didn't

boy he mentioned/girl he didn't

Their argument is that if we asked the father if he has a boy, then we know that one of his kids is a boy, but we don't have any criteria by which to narrow down from 3 to 2 combinations. But the moment the father chooses to tell us that he has a boy, he is selecting one of his children to mention, and thus giving us a selection criteria.

Of course, if the boy were not randomly selected - for example, you're at a football game, you talk to the guy next to you and ask him how many kids he has, and he says two, and that boy right there's my son - then we're still looking at 2/3.

Excellent post again -- except for the last part. If "you talk to the guy next to you and ask him how many kids he has, and he says two, and that boy right there's my son" then the chance his other child is a girl is still only 50%!If you don't believe me, let's use the example someone posted earlier of Pat and Terry.

You're sitting next to the guy at the football game, and he says he has two kids, Pat and Terry. He then points to the QB and says, "that's Pat". Now do you think that makes Terry 66% likely to be a girl?

Here's what you're missing from the example - the selection criteria used by the father are no longer random. In the case that the father simply volunteers the gender of one of his children, he could choose to volunteer the gender of a boy or a girl, and randomly chose to volunteer the gender of a child that was a boy. Given four fathers, each of whom has two children, we have eight possible random selections, and can eliminate four (he didn't name a girl first).

But in the case where we're watching an all-boy football game, the father is no longer randomly selecting a child to name. He is identifiying the player in the game that is his son. Since he could only choose a boy, the selection criteria are no longer random, and thus the signficant population is based not the selection of the child to name, but the family whose child is named, and as we've discussed ad nauseum, for every three families that have a male child, only one has two male children.

bump for sartre

Sorry, I've actually been getting a little bit of work done this afternoon in between checking in here.I guess I see your angle, but don't see it's applicability to the OP. Your angle is that if the family is randomly selected, then the answer to the OP is 50% (this much we agree on, and my contention is that we can go no further than this with the info posted in the OP). But further:

you suggest that we can't stop there, because this father might be restricted to a set in some instances, such as men watching their sons in a football game. All guys in that set are guaranteed to have at least one son.

If this is what you are saying, I can respond, but if I'm completely misunderstanding you, could you try another example?