Logic problem (1 Viewer)

[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?

 

[Sweet J]

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."

 

[fatguyinalittlecoat]

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."

 

[Nigel]

Do you agree with this Sweet J?:

Shuffle 1000 cards and deal them face down into pairs on a giant table. Randomly select one of the pairs and flip over the top card to reveal it to be black. The fact that that card, when dealt, was black has no bearing on what the card after it was - right? Let's say there were still 500 cards to be dealt when that pair was laid down, with virtually equal red and black still in the deck. The chance that the second card dealt to complete that pair is red is absolutely 50%.

The difference in your scenario is that the way you have worded it, the answer is 50%. If the pairs are in a stack (top bottom). Than you have equal probabilities of Red top/Red bottom, Red/black, black/red, and black/black.If you take the TOP card and it is black, than the next card can ONLY be red or black, for 50%. The other way to analyze it is to say that there are four equally likely pairs: RR, RB, BR, RR. You randomly take a pair. You are told that at least one of the cards is black. The odd that the other card is red is 66%.

Then fatguy's distinction of the moderator looking at only one card is important. You're saying that the moderator “randomly chooses one card” and reveals it to be black.There are four pairs of cards, face down on the table, one card next to the other. The card on the left in each pair is Card One, the one on the right is Card Two. The four combinations:

1. BB

2. RR

3. RB

4. BR

If the moderator said out loud that he looked at Card One, and it is black, you could then eliminate pairs two and three, where Card One is red. In this case, with BR and BB remaining, it would still be 50/50 that Card Two was red.

Do the odds change just because the moderator is not specifically revealing whether it is Card One or Card Two that he looked at? No they don’t. While by not being in the know we can not eliminate two of the combinations, it does not change the fact that two of them are eliminated as soon as he reveals (even if not out loud) that “Card Two is black”, leaving us with either BB or RB.

 

[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.

 

[Mr. Pickles]

500 pages

 

[Sweet J]

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."

Right. People have been accusing me of changing the hypo (I agree that it is the same thing. But I just wnat to clarify.)

 

[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.

But you are saying that if he turns a black card over there is a 66% chance the other one is red. And if he turns a red one over there is a 66% chance the other one is black.Doesn't add up.

 

[This_Guy]

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 the moderate only looks at 1 card then the other card is 50/50 black red.If the moderator looks at both cards and tells you there is at least 1 black card then there is a 66% it is red.The first scenario is only asking for the probability of 1 outcome in 1 event (That 1 card happens to be 1 particular color). The second scenario is asking about the probability of 1 outcome in 2 events (There is a black card what is the probability that there is also a red card)If you only know about 1 card then the color of the other card is independent of the first card. Doesn't matter if you know what actual card it is. If you know information that could pertain to either card you still have a possibility that either card fits the outcome you are calculating the odds for.

 

[bostonfred]

Imagine that I had 120 fathers in the room, each with two children, and assume that all distributions will be perfectly normal (e.g. 30 BB, 30 BG, 30 GB, 30 GG; coin flips come up heads exactly half the time, etc.).

A) First, I ask them, how many of you have male children? How many hands would be raised? I think both of us agree that the answer would be 90. I think we also agree that 60 of them would have just one boy, while only 30 would have two.

B) Next, I ask them to flip a coin. If it's heads, tell me the gender of your older child. If it's tails, tell me the gender of your younger child. How many of them would answer "Male"? Do you agree that it would be 60?

Of those 60, what would the gender distribution be? I contend that all 30 BB would answer male, since it doesn't matter what side the coin came up. But only 15 BGs would answer male, as would 15 GBs. Do you agree with this?

That's the 50/50 we're talking about. What changed? The fact that we're flipping coins to determine which child to report, instead of simply reporting "I have at leaast one son". Do you see why A and B are different problems, that require us to look at different populations?

I'm just not following how this refutes the following:A man has two children. One of them is a (B or G). We have already picked. The odds that the other child is opposite is 66%. I don't see how the above refutes that.

A man has two children. One of them is a B. The odds that the other is a girl is 66%. We agree to that, right?

A man has two children. One of them is a G. The odds that the other is a B is 66%. We agree to that, right?

This is the only thing I am saying. Assumptions are: (1) you have a two child family (or you've picked two cards). (2) you pick one child/card. (3) the second card is 2/3 more likely to be the opposite.

Can you refute this. If so, I'd happily agree with you.

Let's break this down into two separate cases. (1) you have a two child family (or you've picked two cards).

(2) you pick one child/card.

(3) the second card is 50/50 to be the opposite.

(1) you have a two child family (or you've picked two cards).

(2) you look at both cards, and verify that at least one of the child/cards is red/boy/tails/whatever.

(3) the second card is 2/3 more likely to be the opposite.

 

[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.

 

[Sweet J]

Do you agree with this Sweet J?:

Shuffle 1000 cards and deal them face down into pairs on a giant table. Randomly select one of the pairs and flip over the top card to reveal it to be black. The fact that that card, when dealt, was black has no bearing on what the card after it was - right? Let's say there were still 500 cards to be dealt when that pair was laid down, with virtually equal red and black still in the deck. The chance that the second card dealt to complete that pair is red is absolutely 50%.

The difference in your scenario is that the way you have worded it, the answer is 50%. If the pairs are in a stack (top bottom). Than you have equal probabilities of Red top/Red bottom, Red/black, black/red, and black/black.If you take the TOP card and it is black, than the next card can ONLY be red or black, for 50%. The other way to analyze it is to say that there are four equally likely pairs: RR, RB, BR, RR. You randomly take a pair. You are told that at least one of the cards is black. The odd that the other card is red is 66%.

Then fatguy's distinction of the moderator looking at only one card is important. You're saying that the moderator “randomly chooses one card” and reveals it to be black.There are four pairs of cards, face down on the table, one card next to the other. The card on the left in each pair is Card One, the one on the right is Card Two. The four combinations:

1. BB

2. RR

3. RB

4. BR

If the moderator said out loud that he looked at Card One, and it is black, you could then eliminate pairs two and three, where Card One is red. In this case, with BR and BB remaining, it would still be 50/50 that Card Two was red.

Do the odds change just because the moderator is not specifically revealing whether it is Card One or Card Two that he looked at? No they don’t. While by not being in the know we can not eliminate two of the combinations, it does not change the fact that two of them are eliminated as soon as he reveals (even if not out loud) that “Card Two is black”, leaving us with either BB or RB.

I've been saying over and over and over that the moderator doesn't know whether he picked "card one" or "card two." HE NEVER SAYS "card one" is black." If he did so, our analysis would be different. Because of that, you can only eliminate RR from the analysis. He MIGHT have picked up "card one" and he might have" picked up "card two." He doesn't know (he only looked at one card). And you don't know. All you know is that one of the cards is Red.

 

[kutta]

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."

Right. People have been accusing me of changing the hypo (I agree that it is the same thing. But I just wnat to clarify.)

Sweet J, you have been changing it a bit. We started this whole thing saying that there was a random selection of stacks and a random selection of one card. Heck, about 6 pages ago I proposed the exact same thing with coins.Anything other than "a guy picks one stack of two cards (BB, RB, BR, RR) and then randomly selects one of the cards to look at." is changing the hypo. And in this situation, the odds of the other card being the opposite color is 50/50. If you agree with the statement above, we are done. If you do not agree, we have more work to do.

 

[fatguyinalittlecoat]

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."

Right. People have been accusing me of changing the hypo (I agree that it is the same thing. But I just wnat to clarify.)

You haven't always been precise in your language.

 

[bryhamm]

Some of you guys are loonies. It's 50%, period. If this was about flipping coins and the guy flipped one in front of you (tails) and mentioned he planned to flip another when he gets home, some of you are arguing he has a 2/3 likelihood of flipping a tails at home just because we know he can't flip 2 heads at this point. That makes zero sense. Like coin flips, the possible sex of an unknown child is independent of any other piece of known information, and is 50%.

that's not what I'm saying. I'm saying if a person has flipped two coins. You were told by another person that one of the coins was heads. The probability that the OHTER coin was tails is 2/3.For you other guys: My card example is no different. You people trying to parse words are making different hypos. My hypo has alway been based on this basic analysis.

Sweet J, I think the problem is your usage of the word "other". As soon as you say "other" you have identified a specific coin/card/etc and want to know the odds of the non-specific coin/card/etc. The answer to that is 50% (and I am one who believes the answer to the original question is 66%).I think it is better if your question at the end is "what are the odds that there is a red card in the pair". The answer to that is 66%.

 

[Nigel]

Do you agree with this Sweet J?:

Shuffle 1000 cards and deal them face down into pairs on a giant table. Randomly select one of the pairs and flip over the top card to reveal it to be black. The fact that that card, when dealt, was black has no bearing on what the card after it was - right? Let's say there were still 500 cards to be dealt when that pair was laid down, with virtually equal red and black still in the deck. The chance that the second card dealt to complete that pair is red is absolutely 50%.

The difference in your scenario is that the way you have worded it, the answer is 50%. If the pairs are in a stack (top bottom). Than you have equal probabilities of Red top/Red bottom, Red/black, black/red, and black/black.If you take the TOP card and it is black, than the next card can ONLY be red or black, for 50%. The other way to analyze it is to say that there are four equally likely pairs: RR, RB, BR, RR. You randomly take a pair. You are told that at least one of the cards is black. The odd that the other card is red is 66%.

Then fatguy's distinction of the moderator looking at only one card is important. You're saying that the moderator “randomly chooses one card” and reveals it to be black.There are four pairs of cards, face down on the table, one card next to the other. The card on the left in each pair is Card One, the one on the right is Card Two. The four combinations:

1. BB

2. RR

3. RB

4. BR

If the moderator said out loud that he looked at Card One, and it is black, you could then eliminate pairs two and three, where Card One is red. In this case, with BR and BB remaining, it would still be 50/50 that Card Two was red.

Do the odds change just because the moderator is not specifically revealing whether it is Card One or Card Two that he looked at? No they don’t. While by not being in the know we can not eliminate two of the combinations, it does not change the fact that two of them are eliminated as soon as he reveals (even if not out loud) that “Card Two is black”, leaving us with either BB or RB.

I've been saying over and over and over that the moderator doesn't know whether he picked "card one" or "card two." HE NEVER SAYS "card one" is black." If he did so, our analysis would be different. Because of that, you can only eliminate RR from the analysis. He MIGHT have picked up "card one" and he might have" picked up "card two." He doesn't know (he only looked at one card). And you don't know. All you know is that one of the cards is Red.

Fine, it doesn't matter if he says it or even if he knows it. The reality is that when he looks at one card, it is either Card One or Card Two, half of the pool of outcomes is off the board, and the two remaining possible outcomes contain one red and one black as the second card.

 

[kutta]

Imagine that I had 120 fathers in the room, each with two children, and assume that all distributions will be perfectly normal (e.g. 30 BB, 30 BG, 30 GB, 30 GG; coin flips come up heads exactly half the time, etc.).

A) First, I ask them, how many of you have male children? How many hands would be raised? I think both of us agree that the answer would be 90. I think we also agree that 60 of them would have just one boy, while only 30 would have two.

B) Next, I ask them to flip a coin. If it's heads, tell me the gender of your older child. If it's tails, tell me the gender of your younger child. How many of them would answer "Male"? Do you agree that it would be 60?

Of those 60, what would the gender distribution be? I contend that all 30 BB would answer male, since it doesn't matter what side the coin came up. But only 15 BGs would answer male, as would 15 GBs. Do you agree with this?

That's the 50/50 we're talking about. What changed? The fact that we're flipping coins to determine which child to report, instead of simply reporting "I have at leaast one son". Do you see why A and B are different problems, that require us to look at different populations?

I'm just not following how this refutes the following:A man has two children. One of them is a (B or G). We have already picked. The odds that the other child is opposite is 66%. I don't see how the above refutes that.

A man has two children. One of them is a B. The odds that the other is a girl is 66%. We agree to that, right?

A man has two children. One of them is a G. The odds that the other is a B is 66%. We agree to that, right?

This is the only thing I am saying. Assumptions are: (1) you have a two child family (or you've picked two cards). (2) you pick one child/card. (3) the second card is 2/3 more likely to be the opposite.

Can you refute this. If so, I'd happily agree with you.

Let's break this down into two separate cases. (1) you have a two child family (or you've picked two cards).

(2) you pick one child/card.

(3) the second card is 50/50 to be the opposite.

(1) you have a two child family (or you've picked two cards).

(2) you look at both cards, and verify that at least one of the child/cards is red/boy/tails/whatever.

(3) the second card is 2/3 more likely to be the opposite.

You gotta take over here. The rest of us are getting worn down.

 

[bostonfred]

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 the moderator only looks at one card, how can he know that at least one is black? Does this moderator have magical powers that helps him to pick black cards?

 

[Sweet J]

Imagine that I had 120 fathers in the room, each with two children, and assume that all distributions will be perfectly normal (e.g. 30 BB, 30 BG, 30 GB, 30 GG; coin flips come up heads exactly half the time, etc.).

A) First, I ask them, how many of you have male children? How many hands would be raised? I think both of us agree that the answer would be 90. I think we also agree that 60 of them would have just one boy, while only 30 would have two.

B) Next, I ask them to flip a coin. If it's heads, tell me the gender of your older child. If it's tails, tell me the gender of your younger child. How many of them would answer "Male"? Do you agree that it would be 60?

Of those 60, what would the gender distribution be? I contend that all 30 BB would answer male, since it doesn't matter what side the coin came up. But only 15 BGs would answer male, as would 15 GBs. Do you agree with this?

That's the 50/50 we're talking about. What changed? The fact that we're flipping coins to determine which child to report, instead of simply reporting "I have at leaast one son". Do you see why A and B are different problems, that require us to look at different populations?

I'm just not following how this refutes the following:A man has two children. One of them is a (B or G). We have already picked. The odds that the other child is opposite is 66%. I don't see how the above refutes that.

A man has two children. One of them is a B. The odds that the other is a girl is 66%. We agree to that, right?

A man has two children. One of them is a G. The odds that the other is a B is 66%. We agree to that, right?

This is the only thing I am saying. Assumptions are: (1) you have a two child family (or you've picked two cards). (2) you pick one child/card. (3) the second card is 2/3 more likely to be the opposite.

Can you refute this. If so, I'd happily agree with you.

Let's break this down into two separate cases. (1) you have a two child family (or you've picked two cards).

(2) you pick one child/card.

(3) the second card is 50/50 to be the opposite.

(1) you have a two child family (or you've picked two cards).

(2) you look at both cards, and verify that at least one of the child/cards is red/boy/tails/whatever.

(3) the second card is 2/3 more likely to be the opposite.

I think I get our disconnect: I agree that the first set of propositions is 50/50. If you say "you have picked one child/card; what is the next child/card." Easy 50/50.

I agree with your second group of propositions. I guess I just don't think you need to have looked at both cards.

I can't tell if we are arguing semantics or not, however. I guess my argument is that "looking at both cards, and verifying that at least one of the child/cards is red/boy whatever" is equivalent to saying "you picked a child/card from the pair, but don't know which one you picked. It could be the older/younger, it could be the top/bottom. All you know is that at least one of the child/card is the color/sex that you've chosen."

You insist that a person needs to look at BOTH cards. I say that you only need to verify that AT LEAST ONE of the cards is black/boy/ whatever. Is that the only difference? Do you still think you need to look at both cards? Why does that make a difference?

 

[Sweet J]

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 the moderator only looks at one card, how can he know that at least one is black? Does this moderator have magical powers that helps him to pick black cards?

How can you possibly be asking this quesiton? When he looks at it, and he sees that it is black, he knows that at least one is black. What he doesn't know is whether the top/bottom or left/right is black. If neither card is black, he would have said "at least one card is Red," and there would be a 66% chance that the other card is black.

 

[rick6668]

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 the moderator only looks at one card, how can he know that at least one is black? Does this moderator have magical powers that helps him to pick black cards?

How can you possibly be asking this quesiton? When he looks at it, and he sees that it is black, he knows that at least one is black. What he doesn't know is whether the top/bottom or left/right is black. If neither card is black, he would have said "at least one card is Red," and there would be a 66% chance that the other card is black.

How can he know neither card is black when he only looks at one?Nice fishing trip.

 

[Sweet J]

Some of you guys are loonies. It's 50%, period. If this was about flipping coins and the guy flipped one in front of you (tails) and mentioned he planned to flip another when he gets home, some of you are arguing he has a 2/3 likelihood of flipping a tails at home just because we know he can't flip 2 heads at this point. That makes zero sense. Like coin flips, the possible sex of an unknown child is independent of any other piece of known information, and is 50%.

that's not what I'm saying. I'm saying if a person has flipped two coins. You were told by another person that one of the coins was heads. The probability that the OHTER coin was tails is 2/3.For you other guys: My card example is no different. You people trying to parse words are making different hypos. My hypo has alway been based on this basic analysis.

Sweet J, I think the problem is your usage of the word "other". As soon as you say "other" you have identified a specific coin/card/etc and want to know the odds of the non-specific coin/card/etc. The answer to that is 50% (and I am one who believes the answer to the original question is 66%).I think it is better if your question at the end is "what are the odds that there is a red card in the pair". The answer to that is 66%.

Don't give me that Nancy mincing "oh, I didn't know you meant that" explaination. Say you were wrong and move on. If you know that one card is Black, and that is all you know, then "the odds that there is a red card in the pair" is the same thing as saying "what is the other card." If any of you try to argue otherwise, I call bullshit on that nancy bs.

 

[Chaka]

I got it...I got it...I got it...

...I ain't got it.

 

[Sweet J]

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."

Right. People have been accusing me of changing the hypo (I agree that it is the same thing. But I just wnat to clarify.)

You haven't always been precise in your language.

I've admitted that. That's why i have been trying very hard to clarify. And every time I clarify, I ask you, "are you ok with this clarification?"

 

[Nugget]

I'm going to go with 50%.

 

[Loan Sharks]

Man I have been playing Roulette all wrong

If Red comes up on the first spin I should bet a ton on Red the second spin.

 

[fatguyinalittlecoat]

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."

Right. People have been accusing me of changing the hypo (I agree that it is the same thing. But I just wnat to clarify.)

You haven't always been precise in your language.

I've admitted that. That's why i have been trying very hard to clarify. And every time I clarify, I ask you, "are you ok with this clarification?"

Right, I'm just saying that's what the accusations are about. People think you're arguing different things because, well, sometimes it seems like you are.

 

[Sweet J]

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 the moderator only looks at one card, how can he know that at least one is black? Does this moderator have magical powers that helps him to pick black cards?

How can you possibly be asking this quesiton? When he looks at it, and he sees that it is black, he knows that at least one is black. What he doesn't know is whether the top/bottom or left/right is black. If neither card is black, he would have said "at least one card is Red," and there would be a 66% chance that the other card is black.

How can he know neither card is black when he only looks at one?Nice fishing trip.

That's not what I am saying. I am saying that one of the two cards is selected, but you don't know which of the two. Nobody knows. All you know is that you were told the card. You were told it was black. So this is the information you have: A par of cards, each with equal chances of being R/B. You are told one is black.Hence, "at least one of the cards is black." The odds of the other being red is 66%.

 

[FatMax]

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.

 

[TheAristocrat]

I can't believe that this is still going.

Sweet J 160

kutta 106

FatMax 78

FlapJacks 67

sartre 48

bryhamm 41

mon 39

Cavalier 38

bostonfred 32

fatguyinalittlecoat 31

videoguy505 28

Dave Baker 25

adonis 25

IvanKaramazov 24

Short Corner 22

rick6668 17

dgreen 15

Random 15

Sinn Fein 13

biggamer3 12

DaVinci 12

Big Dumb Ape 12

parrot 11

BoltThrower 11

GregR 10

Chaka 10

bshipper 9

Girl A+ 9

CBusAlex 9

TheAristocrat 8

moby fan 7

Nigel 5

pantagrapher 5

sirelfman 5

guncrib 5

munga30 4

Nugget 4

Big Petey 4

guru_007 4

JBPH 4

GoSensGo 3

dancer 3

JAA 3

The Commish 3

warpedone 3

The Noid 3

D_House 2

Dante Hicks 2

Football Girl 2

Harris 2

perry147 2

Moe Green 2

Jiggyonthehut 2

Juxtatarot 2

rotohuff 2

This_Guy 2

PsychoMan 2

safariplanet 2

Gigantomachia 2

INRIhab 2

rock753 2

The Iguana 2

Dexter 2

SlimShady 2

TidesofWar 2

wildbill 2

Chuck Norris 2

chet 2

Mr. Yuk 2

Mr. Pickles 2

da_budman 1

BigJim® 1

Das Boot 1

Loan Sharks 1

Sulla 1

Jack Burton 1

Socrates11 1

Arid Filch 1

Greco 1

Topes 1

JetsWillWin 1

3MTA3 1

mlavwilson 1

Uncle Humuna 1

Popinski 1

Ashy Larry 1

Verbal Kint 1

Memphis Foundry 1

MikeMan 1

Franknbeans 1

top dog 1

Notorious T.R.E. 1

Men-in-Cleats 1

Dragons 1

BroadwayG 1

thesurfshop19 1

Cyclones 1

Linus Scrimmage 1

WetDream 1

ETA: I can't believe that Timschochet hasn't weighed in on this issue either. Weird.

 

[Sweet J]

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."

Right. People have been accusing me of changing the hypo (I agree that it is the same thing. But I just wnat to clarify.)

You haven't always been precise in your language.

I've admitted that. That's why i have been trying very hard to clarify. And every time I clarify, I ask you, "are you ok with this clarification?"

Right, I'm just saying that's what the accusations are about. People think you're arguing different things because, well, sometimes it seems like you are.

Sure. There is a difference between "changing the hypo because you realize that your hypo as first presented gives you the answer you don't want," and "changing the hypo because you didn't think about the wording good enough." I'll admit to the latter, but I'm touchy about the former. That's all.

 

[FatMax]

Man I have been playing Roulette all wrongIf Red comes up on the first spin I should bet a ton on Red the second spin.

See, this is the stuff that is annoying to me. BigJim said asserted much the same. The problem isn't talking about probabilities of future events.

 

[Nigel]

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 the moderator only looks at one card, how can he know that at least one is black? Does this moderator have magical powers that helps him to pick black cards?

How can you possibly be asking this quesiton? When he looks at it, and he sees that it is black, he knows that at least one is black. What he doesn't know is whether the top/bottom or left/right is black. If neither card is black, he would have said "at least one card is Red," and there would be a 66% chance that the other card is black.

How can he know neither card is black when he only looks at one?Nice fishing trip.

That's not what I am saying. I am saying that one of the two cards is selected, but you don't know which of the two. Nobody knows. All you know is that you were told the card. You were told it was black. So this is the information you have: A par of cards, each with equal chances of being R/B. You are told one is black.Hence, "at least one of the cards is black." The odds of the other being red is 66%.

Even if you can't say for sure which two it is, in reality two of your combinations are eliminated as soon as you learn what the revealed card is. Agree or disagree?

 

[Sweet J]

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%.

 

[kutta]

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 the moderator only looks at one card, how can he know that at least one is black? Does this moderator have magical powers that helps him to pick black cards?

How can you possibly be asking this quesiton? When he looks at it, and he sees that it is black, he knows that at least one is black. What he doesn't know is whether the top/bottom or left/right is black. If neither card is black, he would have said "at least one card is Red," and there would be a 66% chance that the other card is black.

How can he know neither card is black when he only looks at one?Nice fishing trip.

That's not what I am saying. I am saying that one of the two cards is selected, but you don't know which of the two. Nobody knows. All you know is that you were told the card. You were told it was black. So this is the information you have: A par of cards, each with equal chances of being R/B. You are told one is black.Hence, "at least one of the cards is black." The odds of the other being red is 66%.

If one card is selected from the pair (you don't care who selected it, who told you what, what card it was, blah blah blah - someone somewhere picked a card and told you what it was) and you are told it is black, the odds are 50/50 the other one is red.

If someone looks at both cards and tells you at least one is black, the odds are 2/3 the other one is red.

So if a random guy comes up to you and says, "I made these piles and picked a stack, and at least one card in my stack is black, what are the odds the other is red," you can't answer that question accurately without additional information because we could be talking about the first or second situation above.

 

[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.

Unless this is answered this is

 

[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?

 

[Sweet J]

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 the moderator only looks at one card, how can he know that at least one is black? Does this moderator have magical powers that helps him to pick black cards?

How can you possibly be asking this quesiton? When he looks at it, and he sees that it is black, he knows that at least one is black. What he doesn't know is whether the top/bottom or left/right is black. If neither card is black, he would have said "at least one card is Red," and there would be a 66% chance that the other card is black.

How can he know neither card is black when he only looks at one?Nice fishing trip.

That's not what I am saying. I am saying that one of the two cards is selected, but you don't know which of the two. Nobody knows. All you know is that you were told the card. You were told it was black. So this is the information you have: A par of cards, each with equal chances of being R/B. You are told one is black.Hence, "at least one of the cards is black." The odds of the other being red is 66%.

Even if you can't say for sure which two it is, in reality two of your combinations are eliminated as soon as you learn what the revealed card is. Agree or disagree?

Disagree. Look, here are the two possibilities (in terms of "first card picked/second card picked): BB, BR, RB, RR.You are ONLY told that ONE of the cards is black. You don't know which card is black (the first card picked or the second card picked). Hence, BB, BR, and RB are still on the table as possible combinations. This is the same as saying: "Joe has two kids, at least one is a boy. what are the odds on the other." You don't know if the the boy is older or younger. So you the possible combinations are BB, BG, and GB.

 

[kutta]

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.

We had a discussion a few pages back about this. If the man randomly chose to reveal the sex of one of his kids, then we do, IMO, have this situation because he could have mentioned a girl. But we have reached some sort of tentative agreement on that. We decided the question was sufficiently vague, and that we should deal in absolutes like cards. There is no argument (except for Sweet J) about the math using cards, and that is what we are all trying to get to.

 

[FatMax]

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'm not sure. Let me think about that. Something seems off, but I can't put my finger on it.

 

[Sweet J]

If one card is selected from the pair (you don't care who selected it, who told you what, what card it was, blah blah blah - someone somewhere picked a card and told you what it was) and you are told it is black, the odds are 50/50 the other one is red.If someone looks at both cards and tells you at least one is black, the odds are 2/3 the other one is red.

In my opinion, those two hypotheticals give equivalent odds. The odds are 2/3.

 

[kutta]

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 the moderator only looks at one card, how can he know that at least one is black? Does this moderator have magical powers that helps him to pick black cards?

How can you possibly be asking this quesiton? When he looks at it, and he sees that it is black, he knows that at least one is black. What he doesn't know is whether the top/bottom or left/right is black. If neither card is black, he would have said "at least one card is Red," and there would be a 66% chance that the other card is black.

How can he know neither card is black when he only looks at one?Nice fishing trip.

That's not what I am saying. I am saying that one of the two cards is selected, but you don't know which of the two. Nobody knows. All you know is that you were told the card. You were told it was black. So this is the information you have: A par of cards, each with equal chances of being R/B. You are told one is black.Hence, "at least one of the cards is black." The odds of the other being red is 66%.

Even if you can't say for sure which two it is, in reality two of your combinations are eliminated as soon as you learn what the revealed card is. Agree or disagree?

Disagree. Look, here are the two possibilities (in terms of "first card picked/second card picked): BB, BR, RB, RR.You are ONLY told that ONE of the cards is black. You don't know which card is black (the first card picked or the second card picked). Hence, BB, BR, and RB are still on the table as possible combinations. This is the same as saying: "Joe has two kids, at least one is a boy. what are the odds on the other." You don't know if the the boy is older or younger. So you the possible combinations are BB, BG, and GB.

But again, you are MORE LIKELY to pick a black card from the BB group than you are from the RB or BR group.You never answered that question, and it really is at the heart of this whole thing.

 

[kutta]

If one card is selected from the pair (you don't care who selected it, who told you what, what card it was, blah blah blah - someone somewhere picked a card and told you what it was) and you are told it is black, the odds are 50/50 the other one is red.If someone looks at both cards and tells you at least one is black, the odds are 2/3 the other one is red.

In my opinion, those two hypotheticals give equivalent odds. The odds are 2/3.

I am sorry, but your opinion is wrong.

 

[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.

 

[Nigel]

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 the moderator only looks at one card, how can he know that at least one is black? Does this moderator have magical powers that helps him to pick black cards?

How can you possibly be asking this quesiton? When he looks at it, and he sees that it is black, he knows that at least one is black. What he doesn't know is whether the top/bottom or left/right is black. If neither card is black, he would have said "at least one card is Red," and there would be a 66% chance that the other card is black.

How can he know neither card is black when he only looks at one?Nice fishing trip.

That's not what I am saying. I am saying that one of the two cards is selected, but you don't know which of the two. Nobody knows. All you know is that you were told the card. You were told it was black. So this is the information you have: A par of cards, each with equal chances of being R/B. You are told one is black.Hence, "at least one of the cards is black." The odds of the other being red is 66%.

Even if you can't say for sure which two it is, in reality two of your combinations are eliminated as soon as you learn what the revealed card is. Agree or disagree?

Disagree. Look, here are the two possibilities (in terms of "first card picked/second card picked): BB, BR, RB, RR.You are ONLY told that ONE of the cards is black. You don't know which card is black (the first card picked or the second card picked). Hence, BB, BR, and RB are still on the table as possible combinations. This is the same as saying: "Joe has two kids, at least one is a boy. what are the odds on the other." You don't know if the the boy is older or younger. So you the possible combinations are BB, BG, and GB.

Okay, then if it was Card One, then BB and BR remainIf it was Card Two, then BB and RB remain.

The total pool of possible unknown cards (in red type above) is four, and two of them are Red.

It doesn't matter if you know which of the two is unknown - the odds don't change.

eta: as long as the guy says "I looked at one of the cards, and it is black" then what I say here holds true. Wheter it was Card One or Card Two is irelevant. Again though, the guy can only have looked at one of the cards, not both.

 

[JBPH]

My attempt to speak in a language FBGs understand (please work with the analogy):

You are all-in on the Final Table. In this crazy game, the dealer only has 4 cards left. They are all Kings, one of each suit. After the flop, you determine you will win if a red and black king shows up. You will lose if both black cards or both white cards comes up on the turn/river.

Percent chance ESPN is showing on their screen that you will win: 50%

Now, Phil Helmuth is ranting and at one point scares the dealer so much that he jumps, revealing a red card, which is immediately killed and removed from play. The hand continues. There is now 1 red card and 2 black cards in the dealer's deck for the flop/river. (Remember, you will win if a red/black or black/red combination shows up on the turn/river).

ESPN showing your winning chance now at 66%. (RB wins, BR wins, BB loses - this is equiv. to man mentioning he has a son, and therefore eliminating the daughter/daughter possibility.)

The dealer shows a black card on the turn. Now, you must have a red to win on the river. There are two cards left in the dealer's hand. One red, one black.

ESPN now shows your winning chance is 50%. (This is equiv. to man telling you his first son is a boy.)

That's all this puzzle is.

 

[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.

 

[sartre]

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

flawed

 

[This_Guy]

But again, you are MORE LIKELY to pick a black card from the BB group than you are from the RB or BR group.

You never answered that question, and it really is at the heart of this whole thing.

This is the key to the entire debate. Understand this and you can arrive at the solution.

 

[FatMax]

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.