Logic problem (1 Viewer)

[rick6668]

kutta is totally costing me $10 here.

You want to change your position? I could use some help.

I meant he was costing me $10 because I thought he was going to break through the cloud of bong smoke that apparently surrounds your brain. My position remains the same.

Ok. Note that I changed/clarified the hypo. Let me know if that is a problem.I am convinced that the BBBBR / RBBBB is a red herring because it changes the odds and confuses things. But I am not adept enough with this material to know how.For the life of me, I don't know how the cards question as I've laid out is any different than the question of "family has two children, at least one is a boy, what are the odds that the other is a girl" but you guys apparently think so.

Your card scenario would have to say, "I look at both cards and tell you at least one of them is Black". It doesn't.

 

[Sweet J]

Fred, I see how my hypo is unclear. Maybe that was our problem. When I say you "randomly" select two cards, that was intended to mean that you pick either the first or the second card, but you don't know which one. Maybe the better way to say it is

"You randomly select a pair from the 100 group. Then one of the two cards is randomly selected. You are told that it is black, though you don't know if it was the card on the left or the right. What are the odds that the other card is red?"

I can see how this was not clear the first time. Do you still want to bet?

Let's go with this premise.You randomly select a pair from the 100 group. Then one of the two cards is randomly selected. You are told that it is black.

STOP.

1/2 the time the card randomly selected will be black. This is the scenario we want to move forward in.

1/2 the time the card randomly selected will be red. These are tossed out.

Let's look at the scenarios that are tossed out. All of the pairs that are RR, half of the pairs that are RB (where you randomly selected the R) and half the pairs that are BR (where you randomly selected the R).

This means that what you are left with when you have selected black is all of the pairs that are BB and half of the pairs or RB and BR which weren't tossed out above.

You are not helping your side. You can only toss out those pairs that you KNOW your card is not a part of.You KNOW that the card is not red, so you know you can toss out the odds that this is a RR pairs.

You don't know if this is the card on the left or the right. All you know is that it is Red. It COULD be the card on the left, and it COULD be the card on the right. So in the "possible" pairs group, you KNOW that it is from either the BB group, the RB group, or the BR group. Because that is all the info you have, you can't make any other assumptions. 1/3 chance of coming from any of these groups.

 

[Sweet J]

kutta is totally costing me $10 here.

You want to change your position? I could use some help.

I meant he was costing me $10 because I thought he was going to break through the cloud of bong smoke that apparently surrounds your brain. My position remains the same.

Ok. Note that I changed/clarified the hypo. Let me know if that is a problem.I am convinced that the BBBBR / RBBBB is a red herring because it changes the odds and confuses things. But I am not adept enough with this material to know how.For the life of me, I don't know how the cards question as I've laid out is any different than the question of "family has two children, at least one is a boy, what are the odds that the other is a girl" but you guys apparently think so.

Your card scenario would have to say, "I look at both cards and tell you at least one of them is Black". It doesn't.

Can you please tell me what the difference between "at least one card is black" (as I clarified the hypo), and "I look at both cards and tell you that at least one of them is Black."

 

[kutta]

OK. Think it over and let me know.

I will. It may take me a little while, because I've got some other stuff to do.Take a look how I clarified my hypo. Does this change your mind, or are you still sure that it is 50%.

It does not change my mind.But you have two slightly different hypos.

Was you have a pair of cards. You are told that at least one of them (you don't know which) is Black. What are the odds that the other card is Red.

and

"You randomly select a pair from the 100 group. Then one of the two cards is randomly selected. You are told that it is black, though you don't know if it was the card on the left or the right. What are the odds that the other card is red?"

In the second one you have a random selection process. In the first one you may or may not. This is exactly what caused the problem in the OP. Us 50/50 folks felt the selection process was random in the OP, where the 2/3 ers did not. Most of the 2/3 ers agreed that if the selection process is random, the odds become 50/50.If you have the set of BB, RB, BR, RR, and are asked, "What percentage of the time that you have a stack with a red card in it is the other card black?" the answer is 2/3.But if I ask, "Pick a stack and then pick a card. What are the chances that the other card in the stack is the opposite suit?" the answer is 50/50.I have four stacks:BBBBBBBBBBBBBBBBBBBBBRRRRRRRRRRRRRRRRRRRBRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRForget for a second that it doesn't perfectly map. If I ask you "What percentage of the time that you have a stack with a red card in it is there also a card black?" the answer is 2/3.This DOES NOT mean that if I pick a stack, then pick a card and it is black, there is an equal chance that card came from one of the three stacks with a black card in it. It is more likely that I selected the all B stack.

 

[rick6668]

Fred, I see how my hypo is unclear. Maybe that was our problem. When I say you "randomly" select two cards, that was intended to mean that you pick either the first or the second card, but you don't know which one. Maybe the better way to say it is

"You randomly select a pair from the 100 group. Then one of the two cards is randomly selected. You are told that it is black, though you don't know if it was the card on the left or the right. What are the odds that the other card is red?"

I can see how this was not clear the first time. Do you still want to bet?

Let's go with this premise.You randomly select a pair from the 100 group. Then one of the two cards is randomly selected. You are told that it is black.

STOP.

1/2 the time the card randomly selected will be black. This is the scenario we want to move forward in.

1/2 the time the card randomly selected will be red. These are tossed out.

Let's look at the scenarios that are tossed out. All of the pairs that are RR, half of the pairs that are RB (where you randomly selected the R) and half the pairs that are BR (where you randomly selected the R).

This means that what you are left with when you have selected black is all of the pairs that are BB and half of the pairs or RB and BR which weren't tossed out above.

You are not helping your side. You can only toss out those pairs that you KNOW your card is not a part of.You KNOW that the card is not red, so you know you can toss out the odds that this is a RR pairs.

You don't know if this is the card on the left or the right. All you know is that it is Red. It COULD be the card on the left, and it COULD be the card on the right. So in the "possible" pairs group, you KNOW that it is from either the BB group, the RB group, or the BR group. Because that is all the info you have, you can't make any other assumptions. 1/3 chance of coming from any of these groups.

Are you saying that I will never randomly pick a red card from a Red/Black, Black/Red stack? If that is possible, I must eliminate those as well.

 

[Sweet J]

You have four stacks on the table BB, BR, RB, and RR (i.e., equal distribution). You pick a card (either the first or the second card). You look at it. It is black. Now, you give me a dollar every time the other card is Red. I give you a dollar every time the other card is black. Shuffle, repeat (although every time we end up with equal distribution). And if you chose a Red Card, every time the other card is black, I get a dollar, and every time the other card is Red, you get a dollar.

I'm going to make a lot of money.

How are you missing this?

I just hope I'm still in the thread when you finally figure out you're wrong.

How about this. Let's PM Maurile this quesition. If he says you are right, I will PP you $10. If I am right, you can either PP me $10, or you can friend me on facebook, so I can oogle your wife.

There are 100 pairs, each with equal distribution.

25 RR pairs.

25 RB pairs.

25 BR pairs.

25 BB pairs.

You randomly select a pair from the 100 group. You then randomly select one of the two cards. You look at your card. It is black. What are the odds that the other card is red.

I say the answer is 66%. You say it's 50%.

Fred, I see how my hypo is unclear. Maybe that was our problem. When I say you "randomly" select two cards, that was intended to mean that you pick either the first or the second card, but you don't know which one. Maybe the better way to say it is "You randomly select a pair from the 100 group. Then one of the two cards is randomly selected. You are told that it is black, though you don't know if it was the card on the left or the right. What are the odds that the other card is red?"

I can see how this was not clear the first time. Do you still want to bet?

bump, so that our bet is on the current page. fatguy has said he has read my hypo as clarified and he still wants to bet. Fine.Now, someone just piped in to say that my hypo, as clarified above, cannot be interpreted as saying: "I look at both cards and tell you at least one of them is Black." But if you read the hypo above, it says: "Then one of the two cards is randomly selected. You are told that it is black."

How is that any different than "I tell you that at least one of them is black."

 

[kutta]

You have four stacks on the table BB, BR, RB, and RR (i.e., equal distribution). You pick a card (either the first or the second card). You look at it. It is black. Now, you give me a dollar every time the other card is Red. I give you a dollar every time the other card is black. Shuffle, repeat (although every time we end up with equal distribution). And if you chose a Red Card, every time the other card is black, I get a dollar, and every time the other card is Red, you get a dollar.

I'm going to make a lot of money.

How are you missing this?

I just hope I'm still in the thread when you finally figure out you're wrong.

How about this. Let's PM Maurile this quesition. If he says you are right, I will PP you $10. If I am right, you can either PP me $10, or you can friend me on facebook, so I can oogle your wife.

There are 100 pairs, each with equal distribution.

25 RR pairs.

25 RB pairs.

25 BR pairs.

25 BB pairs.

You randomly select a pair from the 100 group. You then randomly select one of the two cards. You look at your card. It is black. What are the odds that the other card is red.

I say the answer is 66%. You say it's 50%.

Fred, I see how my hypo is unclear. Maybe that was our problem. When I say you "randomly" select two cards, that was intended to mean that you pick either the first or the second card, but you don't know which one. Maybe the better way to say it is "You randomly select a pair from the 100 group. Then one of the two cards is randomly selected. You are told that it is black, though you don't know if it was the card on the left or the right. What are the odds that the other card is red?"

I can see how this was not clear the first time. Do you still want to bet?

bump, so that our bet is on the current page. fatguy has said he has read my hypo as clarified and he still wants to bet. Fine.Now, someone just piped in to say that my hypo, as clarified above, cannot be interpreted as saying: "I look at both cards and tell you at least one of them is Black." But if you read the hypo above, it says: "Then one of the two cards is randomly selected. You are told that it is black."

How is that any different than "I tell you that at least one of them is black."

Because in the first case you will ALWAYS say there is a black card in the cases of RB, BR, and BB. In the second case you will NOT always say one of the cards is black.

 

[Sweet J]

If you have the set of BB, RB, BR, RR, and are asked, "What percentage of the time that you have a stack with a red card in it is the other card black?" the answer is 2/3.

This is what I put forward. This is what I always put forward. If you have the set of: BB, RB, BR, RR. If you pick a stack randomly. If you are told that one card is a certain suit (either red or black). What percentage of the time do you have a stack containing a color that is different than the card that you were orignially told.that is the hypo as I clarified. It was that simple. I'm glad we agree.

 

[rick6668]

kutta is totally costing me $10 here.

You want to change your position? I could use some help.

I meant he was costing me $10 because I thought he was going to break through the cloud of bong smoke that apparently surrounds your brain. My position remains the same.

Ok. Note that I changed/clarified the hypo. Let me know if that is a problem.I am convinced that the BBBBR / RBBBB is a red herring because it changes the odds and confuses things. But I am not adept enough with this material to know how.For the life of me, I don't know how the cards question as I've laid out is any different than the question of "family has two children, at least one is a boy, what are the odds that the other is a girl" but you guys apparently think so.

Your card scenario would have to say, "I look at both cards and tell you at least one of them is Black". It doesn't.

Can you please tell me what the difference between "at least one card is black" (as I clarified the hypo), and "I look at both cards and tell you that at least one of them is Black."

Your HYPO is NOT "at least one card is black" it says you pick one randomly and it is black. There is a difference.When you pick one randomly and it is black, your sample has been reduced by 50%.When you say "at least one card is black" or "I look at both cards and tell you that at least one of them is Black.", your sample has been reduced by 25%.

 

[kutta]

If you have the set of BB, RB, BR, RR, and are asked, "What percentage of the time that you have a stack with a red card in it is the other card black?" the answer is 2/3.

This is what I put forward. This is what I always put forward. If you have the set of: BB, RB, BR, RR. If you pick a stack randomly. If you are told that one card is a certain suit (either red or black). What percentage of the time do you have a stack containing a color that is different than the card that you were orignially told.that is the hypo as I clarified. It was that simple. I'm glad we agree.

It is not the hypo you put forward. "you are told" is different from "you randomly select."We agreed on the "you are told" back on page 3.

 

[Sweet J]

But if I ask, "Pick a stack and then pick a card. What are the chances that the other card in the stack is the opposite suit?" the answer is 50/50.

I agree. I have always agreed. Because doing this connotates order. If you have the stacks RR, RB, BR, and BB. And you pick the left card and it's red, Than the potentail stacks are RR or RB. 50/50. But that wasn't my hypo. In my hypo, your stack chosen was random. And the card chosen for you was random, and you don't know if it came from the left or right (or top or bottom, or wherever).

 

[sartre]

I'm beginning to think Sweet J is being purposely obtuse (i.e., fishing).

A deck of cards and 5-10 minutes would give him the answer. He's rather keep posting flawed conjecture and altering hypos though.

This conversation was much more intriguing when it was bostonfred and a couple of the other more rational 2/3er's covering their position.

 

[Sweet J]

kutta is totally costing me $10 here.

You want to change your position? I could use some help.

I meant he was costing me $10 because I thought he was going to break through the cloud of bong smoke that apparently surrounds your brain. My position remains the same.

Ok. Note that I changed/clarified the hypo. Let me know if that is a problem.I am convinced that the BBBBR / RBBBB is a red herring because it changes the odds and confuses things. But I am not adept enough with this material to know how.For the life of me, I don't know how the cards question as I've laid out is any different than the question of "family has two children, at least one is a boy, what are the odds that the other is a girl" but you guys apparently think so.

Your card scenario would have to say, "I look at both cards and tell you at least one of them is Black". It doesn't.

Can you please tell me what the difference between "at least one card is black" (as I clarified the hypo), and "I look at both cards and tell you that at least one of them is Black."

Your HYPO is NOT "at least one card is black" it says you pick one randomly and it is black. There is a difference.When you pick one randomly and it is black, your sample has been reduced by 50%.When you say "at least one card is black" or "I look at both cards and tell you that at least one of them is Black.", your sample has been reduced by 25%.

When I wrote: "you pick one card randomly at it is black" i meant that you don't know whether it came from the left or the right (i.e., the same as "you are told what it is"), only that you have one of the two. That is why I clarified my hypo. fatguy looked at my clarified hypo and still wanted to bet

 

[kutta]

But if I ask, "Pick a stack and then pick a card. What are the chances that the other card in the stack is the opposite suit?" the answer is 50/50.

I agree. I have always agreed. Because doing this connotates order. If you have the stacks RR, RB, BR, and BB. And you pick the left card and it's red, Than the potentail stacks are RR or RB. 50/50. But that wasn't my hypo. In my hypo, your stack chosen was random. And the card chosen for you was random, and you don't know if it came from the left or right (or top or bottom, or wherever).

I am starting to think you are fishing here now too. What the heck difference does it make if it came from the left or right?If you pick a stack at random, then pick a card at random from that stack, the odds are 50/50 the other card will be the opposite color. I can't tell if you agree with that or not.

 

[Sweet J]

Recap: Let's see if we can all agree:

There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.

I say it is 66%. Agree or Disagree?

If you agree, and you've always agreed, then I apologize for being unclear. I tried to clarify it the minute it was apparent that some people thought I was indicating that you knew whether the card was the left or right.

 

[fatguyinalittlecoat]

Your HYPO is NOT "at least one card is black" it says you pick one randomly and it is black. There is a difference.When you pick one randomly and it is black, your sample has been reduced by 50%.When you say "at least one card is black" or "I look at both cards and tell you that at least one of them is Black.", your sample has been reduced by 25%.

When I wrote: "you pick one card randomly at it is black" i meant that you don't know whether it came from the left or the right (i.e., the same as "you are told what it is"), only that you have one of the two. That is why I clarified my hypo. fatguy looked at my clarified hypo and still wanted to bet

There are two different hypos:1) I look at one of the two cards and say "it's black." In this case, there's a 50% chance the other is red.2) I look at both of the two cards and say "at least one of these is black." In this case, there's a 66% chance the other is red.It isn't always clear which hypo you're talking about.

 

[sartre]

kutta is totally costing me $10 here.

You want to change your position? I could use some help.

I meant he was costing me $10 because I thought he was going to break through the cloud of bong smoke that apparently surrounds your brain. My position remains the same.

Ok. Note that I changed/clarified the hypo. Let me know if that is a problem.I am convinced that the BBBBR / RBBBB is a red herring because it changes the odds and confuses things. But I am not adept enough with this material to know how.For the life of me, I don't know how the cards question as I've laid out is any different than the question of "family has two children, at least one is a boy, what are the odds that the other is a girl" but you guys apparently think so.

Your card scenario would have to say, "I look at both cards and tell you at least one of them is Black". It doesn't.

Can you please tell me what the difference between "at least one card is black" (as I clarified the hypo), and "I look at both cards and tell you that at least one of them is Black."

Your HYPO is NOT "at least one card is black" it says you pick one randomly and it is black. There is a difference.When you pick one randomly and it is black, your sample has been reduced by 50%.When you say "at least one card is black" or "I look at both cards and tell you that at least one of them is Black.", your sample has been reduced by 25%.

When I wrote: "you pick one card randomly at it is black" i meant that you don't know whether it came from the left or the right (i.e., the same as "you are told what it is"), only that you have one of the two. That is why I clarified my hypo. fatguy looked at my clarified hypo and still wanted to bet

I'd still like to take the bet too. Even though you keep adjusting your hypo, you haven't adjusted it enough to the point it gives you the answer you so desperately want.

 

[fatguyinalittlecoat]

There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.

Did the person look at both cards before saying this or just one?

 

[Sweet J]

But if I ask, "Pick a stack and then pick a card. What are the chances that the other card in the stack is the opposite suit?" the answer is 50/50.

I agree. I have always agreed. Because doing this connotates order. If you have the stacks RR, RB, BR, and BB. And you pick the left card and it's red, Than the potentail stacks are RR or RB. 50/50. But that wasn't my hypo. In my hypo, your stack chosen was random. And the card chosen for you was random, and you don't know if it came from the left or right (or top or bottom, or wherever).

I am starting to think you are fishing here now too. What the heck difference does it make if it came from the left or right?If you pick a stack at random, then pick a card at random from that stack, the odds are 50/50 the other card will be the opposite color.

I can't tell if you agree with that or not.

You keep saying that the term "pick a card at random from the stack" is important. I've said "I will try and clarify." So I've said (continually) fine, call it "you are told that the card is one color. What are the odds that the remaining color matches the color you are told."

 

[kutta]

But if I ask, "Pick a stack and then pick a card. What are the chances that the other card in the stack is the opposite suit?" the answer is 50/50.

I agree. I have always agreed. Because doing this connotates order. If you have the stacks RR, RB, BR, and BB. And you pick the left card and it's red, Than the potentail stacks are RR or RB. 50/50. But that wasn't my hypo. In my hypo, your stack chosen was random. And the card chosen for you was random, and you don't know if it came from the left or right (or top or bottom, or wherever).

I am starting to think you are fishing here now too. What the heck difference does it make if it came from the left or right?If you pick a stack at random, then pick a card at random from that stack, the odds are 50/50 the other card will be the opposite color.

I can't tell if you agree with that or not.

You keep saying that the term "pick a card at random from the stack" is important. I've said "I will try and clarify." So I've said (continually) fine, call it "you are told that the card is one color. What are the odds that the remaining color matches the color you are told."

Saying "you are told" is not the same as picking a card at random.If I look at both cards and tell you one of them is black, I will do that 100% of the time with the BB stack, 100% of the time with the RB stack, and 100% of the time with the BR stack.

If I randomly flip one card over and tell you it is black, I will do that 100% of the time with the BB stack, 50% of the time with the RB stack, and 50% of the time with the BR stack.

They are different scenarios.

 

[Sweet J]

There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.

Did the person look at both cards before saying this or just one?

It doesn't matter. You picked a pair. You handed the pair to some moderater. He looks at one (you don't know which one). And tells you "at least one of the cards is black." What are the odds that the other card is red.Now, I don't know why you would quibble over the term "at least one of the cards" because it is apparent when you look at one and it is black, that at least one of them is black.And regardless of whether he has looked at the cards, he has told you "at least one of them is black." that is all the information you have.

 

[fatguyinalittlecoat]

There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.

Did the person look at both cards before saying this or just one?

It doesn't matter. You picked a pair. You handed the pair to some moderater. He looks at one (you don't know which one). And tells you "at least one of the cards is black." What are the odds that the other card is red.Now, I don't know why you would quibble over the term "at least one of the cards" because it is apparent when you look at one and it is black, that at least one of them is black.And regardless of whether he has looked at the cards, he has told you "at least one of them is black." that is all the information you have.

I'm just trying to get you to clarify your hypo. Did the moderator only look at one card?

 

[kutta]

There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.

Did the person look at both cards before saying this or just one?

It doesn't matter. You picked a pair. You handed the pair to some moderater. He looks at one (you don't know which one). And tells you "at least one of the cards is black." What are the odds that the other card is red.Now, I don't know why you would quibble over the term "at least one of the cards" because it is apparent when you look at one and it is black, that at least one of them is black.And regardless of whether he has looked at the cards, he has told you "at least one of them is black." that is all the information you have.

50/50.If he looks at both and tells you at least one is black, it's 2/3.

 

[Sweet J]

But if I ask, "Pick a stack and then pick a card. What are the chances that the other card in the stack is the opposite suit?" the answer is 50/50.

I agree. I have always agreed. Because doing this connotates order. If you have the stacks RR, RB, BR, and BB. And you pick the left card and it's red, Than the potentail stacks are RR or RB. 50/50. But that wasn't my hypo. In my hypo, your stack chosen was random. And the card chosen for you was random, and you don't know if it came from the left or right (or top or bottom, or wherever).

I am starting to think you are fishing here now too. What the heck difference does it make if it came from the left or right?If you pick a stack at random, then pick a card at random from that stack, the odds are 50/50 the other card will be the opposite color.

I can't tell if you agree with that or not.

You keep saying that the term "pick a card at random from the stack" is important. I've said "I will try and clarify." So I've said (continually) fine, call it "you are told that the card is one color. What are the odds that the remaining color matches the color you are told."

Saying "you are told" is not the same as picking a card at random.If I look at both cards and tell you one of them is black, I will do that 100% of the time with the BB stack, 100% of the time with the RB stack, and 100% of the time with the BR stack.

If I randomly flip one card over and tell you it is black, I will do that 100% of the time with the BB stack, 50% of the time with the RB stack, and 50% of the time with the BR stack.

They are different scenarios.

My use of the word random" was meant to signify that you don't know which of the two was looked at. I tried to clarify that it means "the information you have is that at least one of the cards is black."I don't think it matters whether he looked at the cards or not. You picked a pair. You are told that one of the cards is black. Or you are told that one of the cards is red. In either case, the remaining card will be different from the card you are told 66% of the time.

 

[fatguyinalittlecoat]

I don't think it matters whether he looked at the cards or not.

It does.

 

[kutta]

But if I ask, "Pick a stack and then pick a card. What are the chances that the other card in the stack is the opposite suit?" the answer is 50/50.

I agree. I have always agreed. Because doing this connotates order. If you have the stacks RR, RB, BR, and BB. And you pick the left card and it's red, Than the potentail stacks are RR or RB. 50/50. But that wasn't my hypo. In my hypo, your stack chosen was random. And the card chosen for you was random, and you don't know if it came from the left or right (or top or bottom, or wherever).

I am starting to think you are fishing here now too. What the heck difference does it make if it came from the left or right?If you pick a stack at random, then pick a card at random from that stack, the odds are 50/50 the other card will be the opposite color.

I can't tell if you agree with that or not.

You keep saying that the term "pick a card at random from the stack" is important. I've said "I will try and clarify." So I've said (continually) fine, call it "you are told that the card is one color. What are the odds that the remaining color matches the color you are told."

Saying "you are told" is not the same as picking a card at random.If I look at both cards and tell you one of them is black, I will do that 100% of the time with the BB stack, 100% of the time with the RB stack, and 100% of the time with the BR stack.

If I randomly flip one card over and tell you it is black, I will do that 100% of the time with the BB stack, 50% of the time with the RB stack, and 50% of the time with the BR stack.

They are different scenarios.

My use of the word random" was meant to signify that you don't know which of the two was looked at. I tried to clarify that it means "the information you have is that at least one of the cards is black."I don't think it matters whether he looked at the cards or not. You picked a pair. You are told that one of the cards is black. Or you are told that one of the cards is red. In either case, the remaining card will be different from the card you are told 66% of the time.

I can't believe that you can't see the difference here.Can you see the difference between someone looking at BOTH cards and telling you one of them is black, and someone just randomly selecting one card and telling you it is black? If not, read my post that you quoted and tell me what you disagree with.

 

[Sweet J]

There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.

Did the person look at both cards before saying this or just one?

It doesn't matter. You picked a pair. You handed the pair to some moderater. He looks at one (you don't know which one). And tells you "at least one of the cards is black." What are the odds that the other card is red.Now, I don't know why you would quibble over the term "at least one of the cards" because it is apparent when you look at one and it is black, that at least one of them is black.And regardless of whether he has looked at the cards, he has told you "at least one of them is black." that is all the information you have.

I'm just trying to get you to clarify your hypo. Did the moderator only look at one card?

You don't know. All you know is that you picked, randomly, one stack from the four of the stacks: RR, RB, BR, RR. You are told "one of the cards is black."I'm trying to be true to my hypy. My hypo, as I've tried to state, is that "you are told that one of the cards (you don't know which) is a certain color, what are the odds that the remaining card is a different color?"

 

[sartre]

Recap: Let's see if we can all agree:There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.I say it is 66%. Agree or Disagree?If you agree, and you've always agreed, then I apologize for being unclear. I tried to clarify it the minute it was apparent that some people thought I was indicating that you knew whether the card was the left or right.

This lastest adjustment makes the problem flawed and unsolvable. 1) "There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. "We're fine so far2) "Now, you are told that at least one of the cards is either red or black."OK, still fine, we've assigned a property TO THE PAIR of cards. We have made no assignments to either either of the individual cards.3) "What are the odds that the remaining card is a different suit."The question is invalid!. How can we talk about the 'other card' when we haven't identified an individual of the pair yet?If you change "at least one of" in your latest hypo to "one of" (in statement 2) then the question is valid. And the answer to it is 50%

 

[Nigel]

Hi, I'm new here, and this thread is much too long to read through. I'm sure it's alredy been said, but how does learning the result of one coin flip alter the odds for results of a second?

It doesn't. The answer is 50%

 

[Sweet J]

Your HYPO is NOT "at least one card is black" it says you pick one randomly and it is black. There is a difference.When you pick one randomly and it is black, your sample has been reduced by 50%.When you say "at least one card is black" or "I look at both cards and tell you that at least one of them is Black.", your sample has been reduced by 25%.

When I wrote: "you pick one card randomly at it is black" i meant that you don't know whether it came from the left or the right (i.e., the same as "you are told what it is"), only that you have one of the two. That is why I clarified my hypo. fatguy looked at my clarified hypo and still wanted to bet

There are two different hypos:1) I look at one of the two cards and say "it's black." In this case, there's a 50% chance the other is red.2) I look at both of the two cards and say "at least one of these is black." In this case, there's a 66% chance the other is red.It isn't always clear which hypo you're talking about.

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?

 

[sartre]

There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.

Did the person look at both cards before saying this or just one?

It doesn't matter. You picked a pair. You handed the pair to some moderater. He looks at one (you don't know which one). And tells you "at least one of the cards is black." What are the odds that the other card is red.Now, I don't know why you would quibble over the term "at least one of the cards" because it is apparent when you look at one and it is black, that at least one of them is black.And regardless of whether he has looked at the cards, he has told you "at least one of them is black." that is all the information you have.

I'm just trying to get you to clarify your hypo. Did the moderator only look at one card?

You don't know. All you know is that you picked, randomly, one stack from the four of the stacks: RR, RB, BR, RR. You are told "one of the cards is black."I'm trying to be true to my hypy. My hypo, as I've tried to state, is that "you are told that one of the cards (you don't know which) is a certain color, what are the odds that the remaining card is a different color?"

And I'll still take this bet. Do you remember the hypo I proposed back on, oh, page 10 maybe? The one you responded with the Mencken quote? Look that one up and you will wee why I'll take this bet any day, for any amount.Are we on for $10?

 

[kutta]

There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.

Did the person look at both cards before saying this or just one?

It doesn't matter. You picked a pair. You handed the pair to some moderater. He looks at one (you don't know which one). And tells you "at least one of the cards is black." What are the odds that the other card is red.Now, I don't know why you would quibble over the term "at least one of the cards" because it is apparent when you look at one and it is black, that at least one of them is black.And regardless of whether he has looked at the cards, he has told you "at least one of them is black." that is all the information you have.

I'm just trying to get you to clarify your hypo. Did the moderator only look at one card?

You don't know. All you know is that you picked, randomly, one stack from the four of the stacks: RR, RB, BR, RR. You are told "one of the cards is black."I'm trying to be true to my hypy. My hypo, as I've tried to state, is that "you are told that one of the cards (you don't know which) is a certain color, what are the odds that the remaining card is a different color?"

We have been talking about a random choice of stacks and a random choice of cards. Let's stick to that. We will start to obfuscate the issue if we quibble over what was said and what wasn't said. Your hypo is unclear because HOW the information is obtained is very important. Let's stick to concrete stuff that we can prove with stats.You still think that if I have the four options, take a random stack, then take a random card, the odds of the other card in the stack being the opposite suit is 2/3. This is the issue we need to address.

 

[fatguyinalittlecoat]

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%.

 

[Sweet J]

Recap: Let's see if we can all agree:There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.I say it is 66%. Agree or Disagree?If you agree, and you've always agreed, then I apologize for being unclear. I tried to clarify it the minute it was apparent that some people thought I was indicating that you knew whether the card was the left or right.

This lastest adjustment makes the problem flawed and unsolvable. 1) "There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. "We're fine so far2) "Now, you are told that at least one of the cards is either red or black."OK, still fine, we've assigned a property TO THE PAIR of cards. We have made no assignments to either either of the individual cards.3) "What are the odds that the remaining card is a different suit."The question is invalid!. How can we talk about the 'other card' when we haven't identified an individual of the pair yet?

Step 2: Let's go ahead and pick a color. Black. You've done step one, picked a pair. You've done step two: You are told that one of the two cards is black.Step 3: What are the odds that the other card is red? I say its 66%. The reason you think it is "invalid" is because you keep trying to identify if the black card is the "first in the pair" or the "second in the pair." The reason this is such a neat trick is that because you don't know, it is 66%. If you knew which individual of the pair you were dealing with, it would be 50%.

 

[kutta]

Your HYPO is NOT "at least one card is black" it says you pick one randomly and it is black. There is a difference.When you pick one randomly and it is black, your sample has been reduced by 50%.When you say "at least one card is black" or "I look at both cards and tell you that at least one of them is Black.", your sample has been reduced by 25%.

When I wrote: "you pick one card randomly at it is black" i meant that you don't know whether it came from the left or the right (i.e., the same as "you are told what it is"), only that you have one of the two. That is why I clarified my hypo. fatguy looked at my clarified hypo and still wanted to bet

There are two different hypos:1) I look at one of the two cards and say "it's black." In this case, there's a 50% chance the other is red.2) I look at both of the two cards and say "at least one of these is black." In this case, there's a 66% chance the other is red.It isn't always clear which hypo you're talking about.

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?

I am fine with this hypo.50/50

 

[sartre]

Recap: Let's see if we can all agree:There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. Now, you are told that at least one of the cards is either red or black. What are the odds that the remaining card is a different suit.I say it is 66%. Agree or Disagree?If you agree, and you've always agreed, then I apologize for being unclear. I tried to clarify it the minute it was apparent that some people thought I was indicating that you knew whether the card was the left or right.

This lastest adjustment makes the problem flawed and unsolvable. 1) "There is a perfectly distributed sample of "pairs" of cards. RR, RB, BR, BB. A pair is randomly picked. "We're fine so far2) "Now, you are told that at least one of the cards is either red or black."OK, still fine, we've assigned a property TO THE PAIR of cards. We have made no assignments to either either of the individual cards.3) "What are the odds that the remaining card is a different suit."The question is invalid!. How can we talk about the 'other card' when we haven't identified an individual of the pair yet?

Step 2: Let's go ahead and pick a color. Black. You've done step one, picked a pair. You've done step two: You are told that one of the two cards is black.Step 3: What are the odds that the other card is red? I say its 66%. The reason you think it is "invalid" is because you keep trying to identify if the black card is the "first in the pair" or the "second in the pair." The reason this is such a neat trick is that because you don't know, it is 66%. If you knew which individual of the pair you were dealing with, it would be 50%.

You have no idea how dense you are being. Or, you are fishing.either way, I'd still like an easy $10. Are we on?

 

[kutta]

Sweet J. I think you are just refusing to see how a random selection is different from looking at both cards and being told. Here is what I posted earlier:

Saying "you are told" is not the same as picking a card at random.If I look at both cards and tell you one of them is black, I will do that 100% of the time with the BB stack, 100% of the time with the RB stack, and 100% of the time with the BR stack.If I randomly flip one card over and tell you it is black, I will do that 100% of the time with the BB stack, 50% of the time with the RB stack, and 50% of the time with the BR stack.They are different scenarios.

Do you see the difference there? You are trying to equate everything to the first scenario above, when the second scenario is what we are talking about with a random selection of one card.

 

[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%.

 

[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%.

Ok, where does the logic break down?1. four pairs. RR, RB, BR, BB.2. One pair is picked randomly.3. One card is picked randomly. You do not see with one.4. You are told the suit. Let's say the suit is BLACK.5. The card with the black suit came from the following three pairs: RB, BR, BB.You are saying that the BB was most likely to be picked, right? I have somethign to say about it. Gimme a sec.

 

[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%.

Ok, where does the logic break down?1. four pairs. RR, RB, BR, BB.

2. One pair is picked randomly.

3. One card is picked randomly. You do not see with one.

4. You are told the suit. Let's say the suit is BLACK.

5. The card with the black suit came from the following three pairs: RB, BR, BB.

You are saying that the BB was most likely to be picked, right? I have somethign to say about it. Gimme a sec.

Using your logic, where the result of the first card being revealed as black affects the odds of the second being either red or black, wouldn't the odds of it being red be just 40%?The pool of possible cards in your scenario, with one having been revealed as black, is:

BB BR RB

If you take the B out of any of those pairs, among the remaining two and a half pairs there are two Rs and three Bs. Why is there now a 66% chance of it being red?

No, I am not saying that the result of the first card being reveiled affects "the odds of the second card." I am saying that once you are told that the card is Black, you now know that there are twice as many piles with a red and a black card than there are piles with black and a black card. So you are twice as likely to have chosen the card from that bunch. Hence, 66/33.

 

[Nigel]

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%.

Ok, where does the logic break down?1. four pairs. RR, RB, BR, BB.

2. One pair is picked randomly.

3. One card is picked randomly. You do not see with one.

4. You are told the suit. Let's say the suit is BLACK.

5. The card with the black suit came from the following three pairs: RB, BR, BB.

You are saying that the BB was most likely to be picked, right? I have somethign to say about it. Gimme a sec.

Using your logic, where the result of the first card being revealed as black affects the odds of the second being either red or black, wouldn't the odds of it being red be just 40%?The pool of possible cards in your scenario, with one having been revealed as black, is:

BB BR RB

If you take the B out of any of those pairs, among the remaining two and a half pairs there are two Rs and three Bs. Why is there now a 66% chance of it being red?

No, I am not saying that the result of the first card being reveiled affects "the odds of the second card." I am saying that once you are told that the card is Black, you now know that there are twice as many piles with a red and a black card than there are piles with black and a black card. So you are twice as likely to have chosen the card from that bunch. Hence, 66/33.

Yeah, nevermind. I realized the flaw in that and deleted the post.

 

[Chaka]

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

 

[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%.

Ok, where does the logic break down?1. four pairs. RR, RB, BR, BB.

2. One pair is picked randomly.

3. One card is picked randomly. You do not see with one.

4. You are told the suit. Let's say the suit is BLACK.

5. The card with the black suit came from the following three pairs: RB, BR, BB.

You are saying that the BB was most likely to be picked, right? I have somethign to say about it. Gimme a sec.

Using your logic, where the result of the first card being revealed as black affects the odds of the second being either red or black, wouldn't the odds of it being red be just 40%?The pool of possible cards in your scenario, with one having been revealed as black, is:

BB BR RB

If you take the B out of any of those pairs, among the remaining two and a half pairs there are two Rs and three Bs. Why is there now a 66% chance of it being red?

No, I am not saying that the result of the first card being reveiled affects "the odds of the second card." I am saying that once you are told that the card is Black, you now know that there are twice as many piles with a red and a black card than there are piles with black and a black card. So you are twice as likely to have chosen the card from that bunch. Hence, 66/33.

And what they're (now) saying is that you have a better chance to pick a black at the start, since RR piles have been discarded. I'm not sure that it true though, since you can't discard RR piles until after the pick.

 

[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.

 

[BigJim®]

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%.

 

[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.

 

[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%.

 

[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?

 

[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%.

 

[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.