Logic problem (1 Viewer)

Sweet J said:

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.

Click to expand...

sartre said:

Sweet J said:

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.

Click to expand...

I think I might have a way of explaining this. The answer to your first two questions is techincally 50%, not 67%, here's why: (PLEASE KEEP READING)When you say 'One of them is a B' you are assigning a property to one of the children.

When you say 'At least one of them is a B' you are assigning a property to the pair.

It is a subtle but significant difference in how the information must be processed.

A man has two children, Pat and Terry. At least one of them is a boy. What is the probability that one of them is a girl? The answer to this is 2/3

A man has two children, Pat and Terry. Pat is a Boy. What is the probability that the other child, Terry, is a girl? The answer to this is 1/2

Click to expand...

sartre said:

Sweet J said:

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.

Click to expand...

I think I might have a way of explaining this. The answer to your first two questions is techincally 50%, not 67%, here's why: (PLEASE KEEP READING)When you say 'One of them is a B' you are assigning a property to one of the children.

When you say 'At least one of them is a B' you are assigning a property to the pair.

It is a subtle but significant difference in how the information must be processed.

A man has two children, Pat and Terry. At least one of them is a boy. What is the probability that one of them is a girl? The answer to this is 2/3

A man has two children, Pat and Terry. Pat is a Boy. What is the probability that the other child, Terry, is a girl? The answer to this is 1/2

Click to expand...

Sweet J said:

sartre said:

Sweet J said:

bostonfred said:

Sweet J said:

bostonfred said:

Sweet J said:

Then we don't really disagree. If a man has two children, and one of them is a boy, the chance that the other is a girl is 66%. That is the all I am the others are saying. If you want to make this another "more interesting" question, that's your prerogative.

Click to expand...

That argument was settled pages ago. Kutta, flapjacks, and co are saying that the fact that the guy VOLUNTEERS the information that he has a kid changes it to 50/50.

Click to expand...

Kutta, you asked for my opinion of BF's description of your interpretation of the problem. I have no opinion (other than unsure). I have seen the proof of the problem as I "clarified" it, and I believe that proof. I have not seen a proof of "your" clarification of the problem, and I'd be happy to see some 2/3 people smarter than me weigh in on it (sinn fein? fatguy? bf?)

Click to expand...

50/50 is the correct answer if you start with a random selection process. If a random father of two randomly selects one of his two children to talk about, and not the other, then the selection looks like BB father talks about older boy

BB father talks about younger boy

GB father talks about girl

GB father talks about boy

BG father talks about boy

BG father talks about girl

GG father talks about older girl

GG father talks about younger girl

We can cross out the four scenarios where a father talks about a girl. That leaves us with four scenarios - two BB and two BG/GB. Thus, 50/50.

If the selection process is not random - for example, if you asked the father, Is one of your children a boy? - then the odds are 2/3, because the correct way to set up the trial is based on the number of fathers you could be talking to, not the number of children they could be talking about.

Click to expand...

This is flawed, fred. See my cards analysis. You can't double count. If the dad talks about a "boy" he is talking about a boy from one of four groups:1. BB

2. BG

3. GB

4. GG.

If he talks about a boy, he is talking about either a boy from group 1, a boy from group 2, or a boy from group 3. You can't say "he is either talking about Joe from group 1 or Frank from group 1," because we are talking about FAMILIES, not individuals.

There are only 4 options of families, not 8 options of families. By analyzing this as 8 separate options, you are repeating some choices that shouldn't be repeated. You need, instead, to analyze it as 2 different groups of 4. PLease see my cards example.

Click to expand...

Bostonfred's analysis is not flawed at all, until he gets to the argument that we can't assume the man is randomly choosing one of his kids to talk about. It may be above your paygrade, and I'd love to explain it to you again, but you've already acknowledged you won't hear it from anyone aside from MT or a few select others.

biggamer3 said:

A man tells you he has two children. He then starts talking about his son. He does not tell you whether the son is the oldest child or the youngest child. What is the probability that his other child is a girl?

Click to expand...

"I have two kids. My son Pat just got back from Iraq" So from this we know there are two kids, Pat and NotPat. We don't know nor need to know which is older. What is the likelhood that NotPat is a girl?

50%

Click to expand...

You keep bringing other examples besides what I am putting forward. The only thing I have consistantly said was that if you have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise. People seem to be disagreeing with me, but then they argue things that are not related to what I put forward.The only reason I want to hear from a "select" group, is that I have grown tired of you people avoiding the proof for this simple request. Frankly, I'm tired of arguing with people who don't get it.

Click to expand...

sartre said:

Sweet J said:

sartre said:

Sweet J said:

bostonfred said:

Sweet J said:

bostonfred said:

Sweet J said:

Then we don't really disagree. If a man has two children, and one of them is a boy, the chance that the other is a girl is 66%. That is the all I am the others are saying. If you want to make this another "more interesting" question, that's your prerogative.

Click to expand...

That argument was settled pages ago. Kutta, flapjacks, and co are saying that the fact that the guy VOLUNTEERS the information that he has a kid changes it to 50/50.

Click to expand...

Kutta, you asked for my opinion of BF's description of your interpretation of the problem. I have no opinion (other than unsure). I have seen the proof of the problem as I "clarified" it, and I believe that proof. I have not seen a proof of "your" clarification of the problem, and I'd be happy to see some 2/3 people smarter than me weigh in on it (sinn fein? fatguy? bf?)

Click to expand...

50/50 is the correct answer if you start with a random selection process. If a random father of two randomly selects one of his two children to talk about, and not the other, then the selection looks like BB father talks about older boy

BB father talks about younger boy

GB father talks about girl

GB father talks about boy

BG father talks about boy

BG father talks about girl

GG father talks about older girl

GG father talks about younger girl

We can cross out the four scenarios where a father talks about a girl. That leaves us with four scenarios - two BB and two BG/GB. Thus, 50/50.

If the selection process is not random - for example, if you asked the father, Is one of your children a boy? - then the odds are 2/3, because the correct way to set up the trial is based on the number of fathers you could be talking to, not the number of children they could be talking about.

Click to expand...

This is flawed, fred. See my cards analysis. You can't double count. If the dad talks about a "boy" he is talking about a boy from one of four groups:1. BB

2. BG

3. GB

4. GG.

If he talks about a boy, he is talking about either a boy from group 1, a boy from group 2, or a boy from group 3. You can't say "he is either talking about Joe from group 1 or Frank from group 1," because we are talking about FAMILIES, not individuals.

There are only 4 options of families, not 8 options of families. By analyzing this as 8 separate options, you are repeating some choices that shouldn't be repeated. You need, instead, to analyze it as 2 different groups of 4. PLease see my cards example.

Click to expand...

Bostonfred's analysis is not flawed at all, until he gets to the argument that we can't assume the man is randomly choosing one of his kids to talk about. It may be above your paygrade, and I'd love to explain it to you again, but you've already acknowledged you won't hear it from anyone aside from MT or a few select others.

biggamer3 said:

A man tells you he has two children. He then starts talking about his son. He does not tell you whether the son is the oldest child or the youngest child. What is the probability that his other child is a girl?

Click to expand...

"I have two kids. My son Pat just got back from Iraq" So from this we know there are two kids, Pat and NotPat. We don't know nor need to know which is older. What is the likelhood that NotPat is a girl?

50%

Click to expand...

You keep bringing other examples besides what I am putting forward. The only thing I have consistantly said was that if you have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise. People seem to be disagreeing with me, but then they argue things that are not related to what I put forward.The only reason I want to hear from a "select" group, is that I have grown tired of you people avoiding the proof for this simple request. Frankly, I'm tired of arguing with people who don't get it.

Click to expand...

I actually have explained this to you before. The frustration comes in when you refuse to read and consider posts because they disagree with you and are not from established posters like MT.Forget four stacks. Why limit the sample size to something so small? Let's imagine a hundred decks of cards shuffled together. Or better yet, an infinite stack.

When you turn over the first card of the stack, no matter whether it was top card or bottom card, younger card or older card, the card still face down has a 50-50 chance of matching the color of that first card you flipped. About 8 pages ago FatMax posted this example but got the answer wrong. I suggested you both actually try it, but neither of you did. If you try the experiment you might see.

I'm bringing up other parallels in hopes that you might see the light. If you look at the OP, can you tell me how it is different from either of these puzzles?

A man has two kids. He tells you the gender of one of them. What are the odds the unmentioned child is of opposite gender?

A man has two kids. His son Pat just go back from Iraq. What are the odds his other child is a girl?

Click to expand...

sartre said:

Sweet J said:

sartre said:

Sweet J said:

bostonfred said:

Sweet J said:

bostonfred said:

Sweet J said:

Then we don't really disagree. If a man has two children, and one of them is a boy, the chance that the other is a girl is 66%. That is the all I am the others are saying. If you want to make this another "more interesting" question, that's your prerogative.

Click to expand...

That argument was settled pages ago. Kutta, flapjacks, and co are saying that the fact that the guy VOLUNTEERS the information that he has a kid changes it to 50/50.

Click to expand...

Kutta, you asked for my opinion of BF's description of your interpretation of the problem. I have no opinion (other than unsure). I have seen the proof of the problem as I "clarified" it, and I believe that proof. I have not seen a proof of "your" clarification of the problem, and I'd be happy to see some 2/3 people smarter than me weigh in on it (sinn fein? fatguy? bf?)

Click to expand...

50/50 is the correct answer if you start with a random selection process. If a random father of two randomly selects one of his two children to talk about, and not the other, then the selection looks like BB father talks about older boy

BB father talks about younger boy

GB father talks about girl

GB father talks about boy

BG father talks about boy

BG father talks about girl

GG father talks about older girl

GG father talks about younger girl

We can cross out the four scenarios where a father talks about a girl. That leaves us with four scenarios - two BB and two BG/GB. Thus, 50/50.

If the selection process is not random - for example, if you asked the father, Is one of your children a boy? - then the odds are 2/3, because the correct way to set up the trial is based on the number of fathers you could be talking to, not the number of children they could be talking about.

Click to expand...

This is flawed, fred. See my cards analysis. You can't double count. If the dad talks about a "boy" he is talking about a boy from one of four groups:1. BB

2. BG

3. GB

4. GG.

If he talks about a boy, he is talking about either a boy from group 1, a boy from group 2, or a boy from group 3. You can't say "he is either talking about Joe from group 1 or Frank from group 1," because we are talking about FAMILIES, not individuals.

There are only 4 options of families, not 8 options of families. By analyzing this as 8 separate options, you are repeating some choices that shouldn't be repeated. You need, instead, to analyze it as 2 different groups of 4. PLease see my cards example.

Click to expand...

Bostonfred's analysis is not flawed at all, until he gets to the argument that we can't assume the man is randomly choosing one of his kids to talk about. It may be above your paygrade, and I'd love to explain it to you again, but you've already acknowledged you won't hear it from anyone aside from MT or a few select others.

biggamer3 said:

A man tells you he has two children. He then starts talking about his son. He does not tell you whether the son is the oldest child or the youngest child. What is the probability that his other child is a girl?

Click to expand...

"I have two kids. My son Pat just got back from Iraq" So from this we know there are two kids, Pat and NotPat. We don't know nor need to know which is older. What is the likelhood that NotPat is a girl?

50%

Click to expand...

You keep bringing other examples besides what I am putting forward. The only thing I have consistantly said was that if you have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise. People seem to be disagreeing with me, but then they argue things that are not related to what I put forward.The only reason I want to hear from a "select" group, is that I have grown tired of you people avoiding the proof for this simple request. Frankly, I'm tired of arguing with people who don't get it.

Click to expand...

I actually have explained this to you before. The frustration comes in when you refuse to read and consider posts because they disagree with you and are not from established posters like MT.Forget four stacks. Why limit the sample size to something so small? Let's imagine a hundred decks of cards shuffled together. Or better yet, an infinite stack.

When you turn over the first card of the stack, no matter whether it was top card or bottom card, younger card or older card, the card still face down has a 50-50 chance of matching the color of that first card you flipped. About 8 pages ago FatMax posted this example but got the answer wrong. I suggested you both actually try it, but neither of you did. If you try the experiment you might see.

I'm bringing up other parallels in hopes that you might see the light. If you look at the OP, can you tell me how it is different from either of these puzzles?

A man has two kids. He tells you the gender of one of them. What are the odds the unmentioned child is of opposite gender?

A man has two kids. His son Pat just go back from Iraq. What are the odds his other child is a girl?

Click to expand...

Sweet J said:

sartre said:

Sweet J said:

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.

Click to expand...

I think I might have a way of explaining this. The answer to your first two questions is techincally 50%, not 67%, here's why: (PLEASE KEEP READING)When you say 'One of them is a B' you are assigning a property to one of the children.

When you say 'At least one of them is a B' you are assigning a property to the pair.

It is a subtle but significant difference in how the information must be processed.

A man has two children, Pat and Terry. At least one of them is a boy. What is the probability that one of them is a girl? The answer to this is 2/3

A man has two children, Pat and Terry. Pat is a Boy. What is the probability that the other child, Terry, is a girl? The answer to this is 1/2

Click to expand...

Jesus Christ. Fine, the question is: A man has two children, at least one of them is (either a boy or a girl). What are the odds that the other child is the opposite of what we picked..As far as your second question. I'd have to think on it, because it's not clear to me whether it is equivalent to saying: "The oldest is a boy, what is the percentage for the next child." If it is equivalent to saying that, than yes the answer is 50%. It doesn't really matter, though. Because that is not the contention I am making. If we can agree on the bolded, I'd be happy to stop.

Click to expand...

sartre said:

You really should've stopped a long, long, time ago, because for many pages we have all agreed on the solution to the classic problem. The question in the OP does not fit that form. We've been discussing issues in your second paragraph.

And as for the rest of my posts, I do hope you will mull them over, because they are correct and may shed some light on the logic for you.

Click to expand...

You have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise

Click to expand...

Sweet J said:

sartre said:

Sweet J said:

sartre said:

Sweet J said:

bostonfred said:

Sweet J said:

bostonfred said:

Sweet J said:

Then we don't really disagree. If a man has two children, and one of them is a boy, the chance that the other is a girl is 66%. That is the all I am the others are saying. If you want to make this another "more interesting" question, that's your prerogative.

Click to expand...

That argument was settled pages ago. Kutta, flapjacks, and co are saying that the fact that the guy VOLUNTEERS the information that he has a kid changes it to 50/50.

Click to expand...

Kutta, you asked for my opinion of BF's description of your interpretation of the problem. I have no opinion (other than unsure). I have seen the proof of the problem as I "clarified" it, and I believe that proof. I have not seen a proof of "your" clarification of the problem, and I'd be happy to see some 2/3 people smarter than me weigh in on it (sinn fein? fatguy? bf?)

Click to expand...

50/50 is the correct answer if you start with a random selection process. If a random father of two randomly selects one of his two children to talk about, and not the other, then the selection looks like BB father talks about older boy

BB father talks about younger boy

GB father talks about girl

GB father talks about boy

BG father talks about boy

BG father talks about girl

GG father talks about older girl

GG father talks about younger girl

We can cross out the four scenarios where a father talks about a girl. That leaves us with four scenarios - two BB and two BG/GB. Thus, 50/50.

If the selection process is not random - for example, if you asked the father, Is one of your children a boy? - then the odds are 2/3, because the correct way to set up the trial is based on the number of fathers you could be talking to, not the number of children they could be talking about.

Click to expand...

This is flawed, fred. See my cards analysis. You can't double count. If the dad talks about a "boy" he is talking about a boy from one of four groups:1. BB

2. BG

3. GB

4. GG.

If he talks about a boy, he is talking about either a boy from group 1, a boy from group 2, or a boy from group 3. You can't say "he is either talking about Joe from group 1 or Frank from group 1," because we are talking about FAMILIES, not individuals.

There are only 4 options of families, not 8 options of families. By analyzing this as 8 separate options, you are repeating some choices that shouldn't be repeated. You need, instead, to analyze it as 2 different groups of 4. PLease see my cards example.

Click to expand...

Bostonfred's analysis is not flawed at all, until he gets to the argument that we can't assume the man is randomly choosing one of his kids to talk about. It may be above your paygrade, and I'd love to explain it to you again, but you've already acknowledged you won't hear it from anyone aside from MT or a few select others.

biggamer3 said:

A man tells you he has two children. He then starts talking about his son. He does not tell you whether the son is the oldest child or the youngest child. What is the probability that his other child is a girl?

Click to expand...

"I have two kids. My son Pat just got back from Iraq" So from this we know there are two kids, Pat and NotPat. We don't know nor need to know which is older. What is the likelhood that NotPat is a girl?

50%

Click to expand...

You keep bringing other examples besides what I am putting forward. The only thing I have consistantly said was that if you have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise. People seem to be disagreeing with me, but then they argue things that are not related to what I put forward.The only reason I want to hear from a "select" group, is that I have grown tired of you people avoiding the proof for this simple request. Frankly, I'm tired of arguing with people who don't get it.

Click to expand...

I actually have explained this to you before. The frustration comes in when you refuse to read and consider posts because they disagree with you and are not from established posters like MT.Forget four stacks. Why limit the sample size to something so small? Let's imagine a hundred decks of cards shuffled together. Or better yet, an infinite stack.

When you turn over the first card of the stack, no matter whether it was top card or bottom card, younger card or older card, the card still face down has a 50-50 chance of matching the color of that first card you flipped. About 8 pages ago FatMax posted this example but got the answer wrong. I suggested you both actually try it, but neither of you did. If you try the experiment you might see.

I'm bringing up other parallels in hopes that you might see the light. If you look at the OP, can you tell me how it is different from either of these puzzles?

A man has two kids. He tells you the gender of one of them. What are the odds the unmentioned child is of opposite gender?

A man has two kids. His son Pat just go back from Iraq. What are the odds his other child is a girl?

Click to expand...

This might be our problem.We are not talking about "picking up the first card in a deck, and guessing the suit (Red or Black) of the next card in the deck." That would be 50% chance of being the opposite suit.

We are talking about "dealing two cards, and picking up one of those two cards, and guessing the color of the next." The answer would be 66% chance of being the opposite suit.

Click to expand...

Sweet J said:

sartre said:

You really should've stopped a long, long, time ago, because for many pages we have all agreed on the solution to the classic problem. The question in the OP does not fit that form. We've been discussing issues in your second paragraph.

And as for the rest of my posts, I do hope you will mull them over, because they are correct and may shed some light on the logic for you.

Click to expand...

You just disagreed with it above. You disagreed to this statement:

You have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise

Click to expand...

Or, if you don't like the "four stacks" you can say: You have a deck of cards, and you shuffle the deck. You deal yourself two cards. You look at one of them. What are the chances that the other card is the opposite? The answer is 66%.

If you disagree, can you show me how it is different from this: A man has two children, at least one of them is (either a boy or a girl). What are the odds that the other child is the opposite of what we picked.

Click to expand...

sartre said:

Sweet J said:

FatMax said:

FatMax said:

FlapJacks said:

FatMax said:

Do the card thing.

Here is an even easier one, if you like doing stuff face down. Make 26 piles of two cards. Turn over one card. The second card will have a 66% chance of being opposite.

Just write down on a paper. Doesn't matter if the first card is red or black. That way, you don't discard anything.

Click to expand...

BRRB

BB

RR

It looks like half of the sample space is opposite

Click to expand...

Yeah, do that 26 times. A sample size of four isn't going to get it.

Click to expand...

Bump.

Click to expand...

Yes, this is the best, simplest, example, of the "opposite" question.

Click to expand...

Clearly neither of you actually tried this, and are simply pretending to understand probabilities!

(For those without the desire to try the bolded experiment, the probability approaches 50%, not 66%, as the number of trials grows. It does not support their argument but rather is a counterexample)

Click to expand...

sartre said:

Sweet J said:

We are not talking about "picking up the first card in a deck, and guessing the suit (Red or Black) of the next card in the deck." That would be 50% chance of being the opposite suit.

We are talking about "dealing two cards, and picking up one of those two cards, and guessing the color of the next." The answer would be 66% chance of being the opposite suit.

Click to expand...

How do you figure the bolded? I find it equivalent to the statement before it. And in both cases, the answer is 50-50.Let's look at the bolded: Fresh shuffled deck, I deal you the two of clubs and another card face down. You're saying you can guess with 66% accuracy that the face down card is red? But we know there are 25 black cards unaccounted for and 26 red ones. So the probability of it being opposite is only slightly above 50% (and certainly not 2/3). It converges on 50% as we add decks.

Click to expand...

Sweet J said:

sartre said:

Sweet J said:

sartre said:

Sweet J said:

bostonfred said:

Sweet J said:

bostonfred said:

Sweet J said:

Then we don't really disagree. If a man has two children, and one of them is a boy, the chance that the other is a girl is 66%. That is the all I am the others are saying. If you want to make this another "more interesting" question, that's your prerogative.

Click to expand...

That argument was settled pages ago. Kutta, flapjacks, and co are saying that the fact that the guy VOLUNTEERS the information that he has a kid changes it to 50/50.

Click to expand...

Kutta, you asked for my opinion of BF's description of your interpretation of the problem. I have no opinion (other than unsure). I have seen the proof of the problem as I "clarified" it, and I believe that proof. I have not seen a proof of "your" clarification of the problem, and I'd be happy to see some 2/3 people smarter than me weigh in on it (sinn fein? fatguy? bf?)

Click to expand...

50/50 is the correct answer if you start with a random selection process. If a random father of two randomly selects one of his two children to talk about, and not the other, then the selection looks like BB father talks about older boy

BB father talks about younger boy

GB father talks about girl

GB father talks about boy

BG father talks about boy

BG father talks about girl

GG father talks about older girl

GG father talks about younger girl

We can cross out the four scenarios where a father talks about a girl. That leaves us with four scenarios - two BB and two BG/GB. Thus, 50/50.

If the selection process is not random - for example, if you asked the father, Is one of your children a boy? - then the odds are 2/3, because the correct way to set up the trial is based on the number of fathers you could be talking to, not the number of children they could be talking about.

Click to expand...

This is flawed, fred. See my cards analysis. You can't double count. If the dad talks about a "boy" he is talking about a boy from one of four groups:1. BB

2. BG

3. GB

4. GG.

If he talks about a boy, he is talking about either a boy from group 1, a boy from group 2, or a boy from group 3. You can't say "he is either talking about Joe from group 1 or Frank from group 1," because we are talking about FAMILIES, not individuals.

There are only 4 options of families, not 8 options of families. By analyzing this as 8 separate options, you are repeating some choices that shouldn't be repeated. You need, instead, to analyze it as 2 different groups of 4. PLease see my cards example.

Click to expand...

Bostonfred's analysis is not flawed at all, until he gets to the argument that we can't assume the man is randomly choosing one of his kids to talk about. It may be above your paygrade, and I'd love to explain it to you again, but you've already acknowledged you won't hear it from anyone aside from MT or a few select others.

biggamer3 said:

A man tells you he has two children. He then starts talking about his son. He does not tell you whether the son is the oldest child or the youngest child. What is the probability that his other child is a girl?

Click to expand...

"I have two kids. My son Pat just got back from Iraq" So from this we know there are two kids, Pat and NotPat. We don't know nor need to know which is older. What is the likelhood that NotPat is a girl?

50%

Click to expand...

You keep bringing other examples besides what I am putting forward. The only thing I have consistantly said was that if you have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise. People seem to be disagreeing with me, but then they argue things that are not related to what I put forward.The only reason I want to hear from a "select" group, is that I have grown tired of you people avoiding the proof for this simple request. Frankly, I'm tired of arguing with people who don't get it.

Click to expand...

I actually have explained this to you before. The frustration comes in when you refuse to read and consider posts because they disagree with you and are not from established posters like MT.Forget four stacks. Why limit the sample size to something so small? Let's imagine a hundred decks of cards shuffled together. Or better yet, an infinite stack.

When you turn over the first card of the stack, no matter whether it was top card or bottom card, younger card or older card, the card still face down has a 50-50 chance of matching the color of that first card you flipped. About 8 pages ago FatMax posted this example but got the answer wrong. I suggested you both actually try it, but neither of you did. If you try the experiment you might see.

I'm bringing up other parallels in hopes that you might see the light. If you look at the OP, can you tell me how it is different from either of these puzzles?

A man has two kids. He tells you the gender of one of them. What are the odds the unmentioned child is of opposite gender?

A man has two kids. His son Pat just go back from Iraq. What are the odds his other child is a girl?

Click to expand...

This might be our problem.We are not talking about "picking up the first card in a deck, and guessing the suit (Red or Black) of the next card in the deck." That would be 50% chance of being the opposite suit.

We are talking about "dealing two cards, and picking up one of those two cards, and guessing the color of the next." The answer would be 66% chance of being the opposite suit.

Click to expand...

sartre said:

Sweet J said:

sartre said:

You really should've stopped a long, long, time ago, because for many pages we have all agreed on the solution to the classic problem. The question in the OP does not fit that form. We've been discussing issues in your second paragraph.

And as for the rest of my posts, I do hope you will mull them over, because they are correct and may shed some light on the logic for you.

Click to expand...

You just disagreed with it above. You disagreed to this statement:

You have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise

Click to expand...

Or, if you don't like the "four stacks" you can say: You have a deck of cards, and you shuffle the deck. You deal yourself two cards. You look at one of them. What are the chances that the other card is the opposite? The answer is 66%.

If you disagree, can you show me how it is different from this: A man has two children, at least one of them is (either a boy or a girl). What are the odds that the other child is the opposite of what we picked.

Click to expand...

Sweet J, if you try this experiment you will see the answer is most certainly not 66%. I don't know what else to tell you other than your solution is 100% wrong. We covered this several pages ago:

Click to expand...

bryhamm said:

Some of you in the 66% camp who are coming up with the scenarios are botching it. Sweet J, the above example is incorrect because you are keeping all combinations in play doing it the bolded way.

If you have:

R-R

R-B

B-R

B-B

If you pick from the first stack and get red, there is a 50-50 chance the other card is R or B. If you pick from the 2nd stack and get B, you still have the same 50-50 shot that the first stack is R.

The 66% comes in to play when you say "if both stacks are B we are ignoring the outcome" (i.e. the same as saying "there is at least 1 R card").

Click to expand...

Sweet J said:

sartre said:

Sweet J said:

sartre said:

You really should've stopped a long, long, time ago, because for many pages we have all agreed on the solution to the classic problem. The question in the OP does not fit that form. We've been discussing issues in your second paragraph.

And as for the rest of my posts, I do hope you will mull them over, because they are correct and may shed some light on the logic for you.

Click to expand...

You just disagreed with it above. You disagreed to this statement:

You have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise

Click to expand...

Or, if you don't like the "four stacks" you can say: You have a deck of cards, and you shuffle the deck. You deal yourself two cards. You look at one of them. What are the chances that the other card is the opposite? The answer is 66%.

If you disagree, can you show me how it is different from this: A man has two children, at least one of them is (either a boy or a girl). What are the odds that the other child is the opposite of what we picked.

Click to expand...

Sweet J, if you try this experiment you will see the answer is most certainly not 66%. I don't know what else to tell you other than your solution is 100% wrong. We covered this several pages ago:

Click to expand...

Look, you must be fishing. You disagreed with FatMax's proposition: Here is an even easier one, if you like doing stuff face down. Make 26 piles of two cards. Turn over one card. The second card will have a 66% chance of being opposite. Tell me which statement you disagree with:

1. With a 100 card deck, you make 50 pairs.

2. With 50 pairs, assuming equal distribution, you have the following pairs: 25 pairs of RR, 25 pairs of RB, 25 pairs of BR, 25 pairs of BB.

3. You pick up one of the cards from one of the pairs. Let's say you picked up a Red card.

4. Before you pick up the next card, you want to guess the odds regarding what color it will be.

5. The possible pairs that that card could come from is as follows: It is either (a) from the 25 RR pairs, (b) from the 25 RB pairs, or © from the 25 BR pairs.

6. You have a 25/75 chance that the other card is Red.

7. You have a 50/75 chance that the other card is Black.

Which one do you disagree with.

Click to expand...

Sweet J said:

bryhamm said:

Some of you in the 66% camp who are coming up with the scenarios are botching it. Sweet J, the above example is incorrect because you are keeping all combinations in play doing it the bolded way.

If you have:

R-R

R-B

B-R

B-B

If you pick from the first stack and get red, there is a 50-50 chance the other card is R or B. If you pick from the 2nd stack and get B, you still have the same 50-50 shot that the first stack is R.

The 66% comes in to play when you say "if both stacks are B we are ignoring the outcome" (i.e. the same as saying "there is at least 1 R card").

Click to expand...

You are wrong. If you pick up the first stack and it is red, than you know it came from a stack that looks like one of the following:RR

RB

BR.

Because you know it can't have come from the BB stack.

Each of those stacks has equal probabilities of occuring. For example, if there were 100 stacks to begin with, there are 25 stacks of each. If one of the cards is Red, there is a 66% chance that the other card is Black. It is that simple.

Click to expand...

sartre said:

Sweet J said:

sartre said:

Sweet J said:

sartre said:

You really should've stopped a long, long, time ago, because for many pages we have all agreed on the solution to the classic problem. The question in the OP does not fit that form. We've been discussing issues in your second paragraph.

And as for the rest of my posts, I do hope you will mull them over, because they are correct and may shed some light on the logic for you.

Click to expand...

You just disagreed with it above. You disagreed to this statement:

You have four equally distributed stacks of two, and you pick a card/child/coin from one, what are the odds that the other one is the opposite? I have shown that it is 66%. Please do the math if you feel otherwise

Click to expand...

Or, if you don't like the "four stacks" you can say: You have a deck of cards, and you shuffle the deck. You deal yourself two cards. You look at one of them. What are the chances that the other card is the opposite? The answer is 66%.

If you disagree, can you show me how it is different from this: A man has two children, at least one of them is (either a boy or a girl). What are the odds that the other child is the opposite of what we picked.

Click to expand...

Sweet J, if you try this experiment you will see the answer is most certainly not 66%. I don't know what else to tell you other than your solution is 100% wrong. We covered this several pages ago:

Click to expand...

Look, you must be fishing. You disagreed with FatMax's proposition: Here is an even easier one, if you like doing stuff face down. Make 26 piles of two cards. Turn over one card. The second card will have a 66% chance of being opposite. Tell me which statement you disagree with:

1. With a 100 200 card deck, you make 50 100 pairs.

2. With 50 pairs, assuming equal distribution, you have the following pairs: 25 pairs of RR, 25 pairs of RB, 25 pairs of BR, 25 pairs of BB.

3. You pick up one of the cards from one of the pairs. Let's say you picked up a Red card.

4. Before you pick up the next card, you want to guess the odds regarding what color it will be.

5. The possible pairs that that card could come from is as follows: It is either (a) from the 25 RR pairs, (b) from the 25 RB pairs, or © from the 25 BR pairs.

6. You have a 25/75 chance that the other card is Red.

7. You have a 50/75 chance that the other card is Black.

Which one do you disagree with.

Click to expand...

#5 is where you make your error. I really wish you would just try this and see for yourself. If not, at least take the word of bryhamm or bostonfred (who explained this to you several pages ago).

Click to expand...

kutta said:

Sweet J said:

bryhamm said:

Some of you in the 66% camp who are coming up with the scenarios are botching it. Sweet J, the above example is incorrect because you are keeping all combinations in play doing it the bolded way.

If you have:

R-R

R-B

B-R

B-B

If you pick from the first stack and get red, there is a 50-50 chance the other card is R or B. If you pick from the 2nd stack and get B, you still have the same 50-50 shot that the first stack is R.

The 66% comes in to play when you say "if both stacks are B we are ignoring the outcome" (i.e. the same as saying "there is at least 1 R card").

Click to expand...

You are wrong. If you pick up the first stack and it is red, than you know it came from a stack that looks like one of the following:RR

RB

BR.

Because you know it can't have come from the BB stack.

Each of those stacks has equal probabilities of occuring. For example, if there were 100 stacks to begin with, there are 25 stacks of each. If one of the cards is Red, there is a 66% chance that the other card is Black. It is that simple.

Click to expand...

But you do not have an equal % chance of drawing a red card from each stack. So if you did draw a red card, IT MORE LIKELY CAME FROM the RR stack.I tried to make this point earlier, but you didn't address. I think that is the part you are missing.

Each stack has a red card in it, so if the question was, "does your stack have a red card?" the answer is yes. If the question is, "if your stack has a red card, what is the chance the other one is black", the answer is 2/3. But if you randomly flip one of those cards over and look at it and it is red, the odds of the other being black is 50/50 because you could have possibly flipped over a black card. In the example where you are asked if a card is red, you do not have the possibility of flipping over a black card.

Click to expand...

kutta said:

But you do not have an equal % chance of drawing a red card from each stack. So if you did draw a red card, IT MORE LIKELY CAME FROM the RR stack.I tried to make this point earlier, but you didn't address. I think that is the part you are missing.

Click to expand...

Sweet J said:

Bostonfred agrees with me. I found my mistake. But it was a careless error.So, with my correction, please explain to me how, if you have 100 stacks, and you picked a red card, it did not come from one of the 25 pairs of RR, 25 pairs of RB, or 25 pairs of BR. I am eagerly awaiting.

Click to expand...

sartre said:

Sweet J said:

Bostonfred agrees with me. I found my mistake. But it was a careless error.So, with my correction, please explain to me how, if you have 100 stacks, and you picked a red card, it did not come from one of the 25 pairs of RR, 25 pairs of RB, or 25 pairs of BR. I am eagerly awaiting.

Click to expand...

bostonfred disagreed with you regarding FatMax's proposition. FatMax posed it to support his argument, but came to the wrong answer. Had he (and you) come to the correct answer, you would see how it supports the 50% position to the OP. What made it even funnier is that neither of you guys actually tried the experiment, but asked other to do so.TRY THE EXPERIMENT. Then come back after the light bulb moment.

Click to expand...

Sweet J said:

bryhamm said:

Some of you in the 66% camp who are coming up with the scenarios are botching it. Sweet J, the above example is incorrect because you are keeping all combinations in play doing it the bolded way.

If you have:

R-R

R-B

B-R

B-B

If you pick from the first stack and get red, there is a 50-50 chance the other card is R or B. If you pick from the 2nd stack and get B, you still have the same 50-50 shot that the first stack is R.

The 66% comes in to play when you say "if both stacks are B we are ignoring the outcome" (i.e. the same as saying "there is at least 1 R card").

Click to expand...

You are wrong. If you pick up the first stack and it is red, than you know it came from a stack that looks like one of the following:RR

RB

BR.

Because you know it can't have come from the BB stack.

Each of those stacks has equal probabilities of occuring. For example, if there were 100 stacks to begin with, there are 25 stacks of each. If one of the cards is Red, there is a 66% chance that the other card is Black. It is that simple.

Click to expand...

sartre said:

Sweet J said:

FatMax said:

FatMax said:

FlapJacks said:

FatMax said:

Do the card thing.

Here is an even easier one, if you like doing stuff face down. Make 26 piles of two cards. Turn over one card. The second card will have a 66% chance of being opposite.

Just write down on a paper. Doesn't matter if the first card is red or black. That way, you don't discard anything.

Click to expand...

BRRB

BB

RR

It looks like half of the sample space is opposite

Click to expand...

Yeah, do that 26 times. A sample size of four isn't going to get it.

Click to expand...

Bump.

Click to expand...

Yes, this is the best, simplest, example, of the "opposite" question.

Click to expand...

Clearly neither of you actually tried this, and are simply pretending to understand probabilities!

(For those without the desire to try the bolded experiment, the probability approaches 50%, not 66%, as the number of trials grows. It does not support their argument but rather is a counterexample)

Click to expand...

bryhamm said:

Sweet J said:

bryhamm said:

Some of you in the 66% camp who are coming up with the scenarios are botching it. Sweet J, the above example is incorrect because you are keeping all combinations in play doing it the bolded way.

If you have:

R-R

R-B

B-R

B-B

If you pick from the first stack and get red, there is a 50-50 chance the other card is R or B. If you pick from the 2nd stack and get B, you still have the same 50-50 shot that the first stack is R.

The 66% comes in to play when you say "if both stacks are B we are ignoring the outcome" (i.e. the same as saying "there is at least 1 R card").

Click to expand...

You are wrong. If you pick up the first one stack and it has a red card, than you know it came from a stack that looks like one of the following:RR

RB

BR.

Because you know it can't have come from the BB stack.

Each of those stacks has equal probabilities of occuring. For example, if there were 100 stacks to begin with, there are 25 stacks of each. If one of the cards is Red, there is a 66% chance that the other card is Black. It is that simple.

Click to expand...

Uhhh, no. Look at the 2 bolded and red section. If you pick a card from the first stack and it is red, you CANNOT have B-R ... the first card is already red.

Click to expand...

Sweet J said:

sartre said:

Sweet J said:

Bostonfred agrees with me. I found my mistake. But it was a careless error.So, with my correction, please explain to me how, if you have 100 stacks, and you picked a red card, it did not come from one of the 25 pairs of RR, 25 pairs of RB, or 25 pairs of BR. I am eagerly awaiting.

Click to expand...

bostonfred disagreed with you regarding FatMax's proposition. FatMax posed it to support his argument, but came to the wrong answer. Had he (and you) come to the correct answer, you would see how it supports the 50% position to the OP. What made it even funnier is that neither of you guys actually tried the experiment, but asked other to do so.TRY THE EXPERIMENT. Then come back after the light bulb moment.

Click to expand...

This is insane. 100 stacks. Equal distribution. Four types of stacks. RB BR BB RR. You have a 25% chance of picking any one stack. You agree to this much, right?

Click to expand...

Sweet J said:

sartre said:

Sweet J said:

FatMax said:

FatMax said:

FlapJacks said:

FatMax said:

Do the card thing.

Here is an even easier one, if you like doing stuff face down. Make 26 piles of two cards. Turn over one card. The second card will have a 66% chance of being opposite.

Just write down on a paper. Doesn't matter if the first card is red or black. That way, you don't discard anything.

Click to expand...

BRRB

BB

RR

It looks like half of the sample space is opposite

Click to expand...

Yeah, do that 26 times. A sample size of four isn't going to get it.

Click to expand...

Bump.

Click to expand...

Yes, this is the best, simplest, example, of the "opposite" question.

Click to expand...

Clearly neither of you actually tried this, and are simply pretending to understand probabilities!

(For those without the desire to try the bolded experiment, the probability approaches 50%, not 66%, as the number of trials grows. It does not support their argument but rather is a counterexample)

Click to expand...

You do understand, right, that the question isn't: "what are the odds that the stacks are the same color?" The quesiton, rather is: "you've picked a card, what are the odds that the other card is opposite?"We are answering the second question, which is absolutely 66%.

Click to expand...

sartre said:

Sweet J said:

sartre said:

Sweet J said:

Bostonfred agrees with me. I found my mistake. But it was a careless error.So, with my correction, please explain to me how, if you have 100 stacks, and you picked a red card, it did not come from one of the 25 pairs of RR, 25 pairs of RB, or 25 pairs of BR. I am eagerly awaiting.

Click to expand...

bostonfred disagreed with you regarding FatMax's proposition. FatMax posed it to support his argument, but came to the wrong answer. Had he (and you) come to the correct answer, you would see how it supports the 50% position to the OP. What made it even funnier is that neither of you guys actually tried the experiment, but asked other to do so.TRY THE EXPERIMENT. Then come back after the light bulb moment.

Click to expand...

This is insane. 100 stacks. Equal distribution. Four types of stacks. RB BR BB RR. You have a 25% chance of picking any one stack. You agree to this much, right?

Click to expand...

So you're not going to try it. OK, well some people you just can't reach.

Click to expand...

sartre said:

Sweet J said:

sartre said:

Sweet J said:

FatMax said:

FatMax said:

FlapJacks said:

FatMax said:

Do the card thing.

Here is an even easier one, if you like doing stuff face down. Make 26 piles of two cards. Turn over one card. The second card will have a 66% chance of being opposite.

Just write down on a paper. Doesn't matter if the first card is red or black. That way, you don't discard anything.

Click to expand...

BRRB

BB

RR

It looks like half of the sample space is opposite

Click to expand...

Yeah, do that 26 times. A sample size of four isn't going to get it.

Click to expand...

Bump.

Click to expand...

Yes, this is the best, simplest, example, of the "opposite" question.

Click to expand...

Clearly neither of you actually tried this, and are simply pretending to understand probabilities!

(For those without the desire to try the bolded experiment, the probability approaches 50%, not 66%, as the number of trials grows. It does not support their argument but rather is a counterexample)

Click to expand...

You do understand, right, that the question isn't: "what are the odds that the stacks are the same color?" The quesiton, rather is: "you've picked a card, what are the odds that the other card is opposite?"We are answering the second question, which is absolutely 66%.

Click to expand...

1. I understand the question2. You understand the question

3. I understand the answer (50%)

4. :(

Click to expand...

Sweet J said:

kutta said:

Sweet J said:

bryhamm said:

Some of you in the 66% camp who are coming up with the scenarios are botching it. Sweet J, the above example is incorrect because you are keeping all combinations in play doing it the bolded way.

If you have:

R-R

R-B

B-R

B-B

If you pick from the first stack and get red, there is a 50-50 chance the other card is R or B. If you pick from the 2nd stack and get B, you still have the same 50-50 shot that the first stack is R.

The 66% comes in to play when you say "if both stacks are B we are ignoring the outcome" (i.e. the same as saying "there is at least 1 R card").

Click to expand...

You are wrong. If you pick up the first stack and it is red, than you know it came from a stack that looks like one of the following:RR

RB

BR.

Because you know it can't have come from the BB stack.

Each of those stacks has equal probabilities of occuring. For example, if there were 100 stacks to begin with, there are 25 stacks of each. If one of the cards is Red, there is a 66% chance that the other card is Black. It is that simple.

Click to expand...

But you do not have an equal % chance of drawing a red card from each stack. So if you did draw a red card, IT MORE LIKELY CAME FROM the RR stack.I tried to make this point earlier, but you didn't address. I think that is the part you are missing.

Each stack has a red card in it, so if the question was, "does your stack have a red card?" the answer is yes. If the question is, "if your stack has a red card, what is the chance the other one is black", the answer is 2/3. But if you randomly flip one of those cards over and look at it and it is red, the odds of the other being black is 50/50 because you could have possibly flipped over a black card. In the example where you are asked if a card is red, you do not have the possibility of flipping over a black card.

Click to expand...

No, you are wrong. You are chosing from 4 stacks of pairs, you are not choosing from "8 cards." 4 stacks, each with 25% chance of getting chosen. After you choose a stack, you randomly look at one card. The odds that the other card is the opposite color is 66%.

Click to expand...

Sweet J said:

sartre said:

Sweet J said:

sartre said:

Sweet J said:

FatMax said:

FatMax said:

FlapJacks said:

FatMax said:

Do the card thing.

Here is an even easier one, if you like doing stuff face down. Make 26 piles of two cards. Turn over one card. The second card will have a 66% chance of being opposite.

Just write down on a paper. Doesn't matter if the first card is red or black. That way, you don't discard anything.

Click to expand...

BRRB

BB

RR

It looks like half of the sample space is opposite

Click to expand...

Yeah, do that 26 times. A sample size of four isn't going to get it.

Click to expand...

Bump.

Click to expand...

Yes, this is the best, simplest, example, of the "opposite" question.

Click to expand...

Clearly neither of you actually tried this, and are simply pretending to understand probabilities!

(For those without the desire to try the bolded experiment, the probability approaches 50%, not 66%, as the number of trials grows. It does not support their argument but rather is a counterexample)

Click to expand...

You do understand, right, that the question isn't: "what are the odds that the stacks are the same color?" The quesiton, rather is: "you've picked a card, what are the odds that the other card is opposite?"We are answering the second question, which is absolutely 66%.

Click to expand...

1. I understand the question2. You understand the question

3. I understand the answer (50%)

4. :(

Click to expand...

Let's do this together. I have layed out 100 pairs of cards on my desk. Equal distribution. That means 25 pairs of RR, 25 pairs of RB, 25 pairs of BR, 25 pairs of BB. Ready? I'm going to pick a card pair.

Now, tell me, is there an equal possibility that I pick each category? 25% chance of picking RR? 25% chance of picking BR? 25% chance of picking RB? 25% chance of picking BB?

Click to expand...

kutta said:

I am willing to make this bet with you all day long. I don't know how to explain it anymore. The simplest way is this:

BB

RB

BR

RR

Forget everything else about the problem. Pick one stack. You have either picked BB, RB, BR, or RR.

If you picked BB or RR, when you flip one card, the other one will be the same. (Totally agree)

If you picked RB or BR, when you flip one card, the other one will be different. (Totally agree)

Does that make sense?

Click to expand...

Sweet J said:

Let's go with your example. You picked up BB or RR. Now, flip over a card. It is B. Don't you see that we can't include the RR anymore? What are the odds that the other card is also B? That card came from one of three "types" of stacks. Either a BB, a RB, or a BR.

Yes, there are 50% of the stacks that are "the same." but after you've separated them, and picked one card, that is where you can't include BOTH RR and BB.

Click to expand...

kutta said:

Sweet J said:

sartre said:

Sweet J said:

sartre said:

Sweet J said:

FatMax said:

FatMax said:

FlapJacks said:

FatMax said:

Do the card thing.

Here is an even easier one, if you like doing stuff face down. Make 26 piles of two cards. Turn over one card. The second card will have a 66% chance of being opposite.

Just write down on a paper. Doesn't matter if the first card is red or black. That way, you don't discard anything.

Click to expand...

BRRB

BB

RR

It looks like half of the sample space is opposite

Click to expand...

Yeah, do that 26 times. A sample size of four isn't going to get it.

Click to expand...

Bump.

Click to expand...

Yes, this is the best, simplest, example, of the "opposite" question.

Click to expand...

Clearly neither of you actually tried this, and are simply pretending to understand probabilities!

(For those without the desire to try the bolded experiment, the probability approaches 50%, not 66%, as the number of trials grows. It does not support their argument but rather is a counterexample)

Click to expand...

You do understand, right, that the question isn't: "what are the odds that the stacks are the same color?" The quesiton, rather is: "you've picked a card, what are the odds that the other card is opposite?"We are answering the second question, which is absolutely 66%.

Click to expand...

1. I understand the question2. You understand the question

3. I understand the answer (50%)

4. :(

Click to expand...

Let's do this together. I have layed out 100 pairs of cards on my desk. Equal distribution. That means 25 pairs of RR, 25 pairs of RB, 25 pairs of BR, 25 pairs of BB. Ready? I'm going to pick a card pair.

Now, tell me, is there an equal possibility that I pick each category? 25% chance of picking RR? 25% chance of picking BR? 25% chance of picking RB? 25% chance of picking BB?

Click to expand...

I really hope you see that you are going down a losing road here. You really need to open up and see that this problem is different from what you think it is.I will play this game with you. Yes, equal probability that you picked one of the four stacks.

Next move.

Click to expand...

fatguyinalittlecoat said:

Sweet J said:

Let's go with your example. You picked up BB or RR. Now, flip over a card. It is B. Don't you see that we can't include the RR anymore? What are the odds that the other card is also B? That card came from one of three "types" of stacks. Either a BB, a RB, or a BR.

Yes, there are 50% of the stacks that are "the same." but after you've separated them, and picked one card, that is where you can't include BOTH RR and BB.

Click to expand...

If the first card is a "B", wouldn't you also eliminate the "RB" stack as a possibility?

Click to expand...

Sweet J said:

kutta said:

I am willing to make this bet with you all day long. I don't know how to explain it anymore. The simplest way is this:

BB

RB

BR

RR

Forget everything else about the problem. Pick one stack. You have either picked BB, RB, BR, or RR.

If you picked BB or RR, when you flip one card, the other one will be the same. (Totally agree)

If you picked RB or BR, when you flip one card, the other one will be different. (Totally agree)

Does that make sense?

Click to expand...

Kutta, you are SOOOO close to seeing itYou are confusing "what are the odds that the other one will be the same before you flip" with "you've picked a card, what are the odds that the other matches."

Let's go with your example. You picked up BB or RR. Now, flip over a card. It is B. Don't you see that we can't include the RR anymore? What are the odds that the other card is also B? That card came from one of three "types" of stacks. Either a BB, a RB, or a BR.

Yes, there are 50% of the stacks that are "the same." but after you've separated them, and picked one card, that is where you can't include BOTH RR and BB.

Can someone smarter than me explain this better?

Click to expand...

fatguyinalittlecoat said:

Sweet J said:

Let's go with your example. You picked up BB or RR. Now, flip over a card. It is B. Don't you see that we can't include the RR anymore? What are the odds that the other card is also B? That card came from one of three "types" of stacks. Either a BB, a RB, or a BR.

Yes, there are 50% of the stacks that are "the same." but after you've separated them, and picked one card, that is where you can't include BOTH RR and BB.

Click to expand...

If the first card is a "B", wouldn't you also eliminate the "RB" stack as a possibility?

Click to expand...

Sweet J said:

fatguyinalittlecoat said:

Sweet J said:

Let's go with your example. You picked up BB or RR. Now, flip over a card. It is B. Don't you see that we can't include the RR anymore? What are the odds that the other card is also B? That card came from one of three "types" of stacks. Either a BB, a RB, or a BR.

Yes, there are 50% of the stacks that are "the same." but after you've separated them, and picked one card, that is where you can't include BOTH RR and BB.

Click to expand...

If the first card is a "B", wouldn't you also eliminate the "RB" stack as a possibility?

Click to expand...

This is where there is confusion. You picked one card randomly from the pair. It could have been the first card, or it could have been the second card. This is why this "card" problem simulates the "not knowing the older/younger child problem. You didn't pick the "first" card, you picked "one of the two cards." Both RB and BR are in play.

Click to expand...

kutta said:

Sweet J said:

kutta said:

I am willing to make this bet with you all day long. I don't know how to explain it anymore. The simplest way is this:

BB

RB

BR

RR

Forget everything else about the problem. Pick one stack. You have either picked BB, RB, BR, or RR.

If you picked BB or RR, when you flip one card, the other one will be the same. (Totally agree)

If you picked RB or BR, when you flip one card, the other one will be different. (Totally agree)

Does that make sense?

Click to expand...

Kutta, you are SOOOO close to seeing itYou are confusing "what are the odds that the other one will be the same before you flip" with "you've picked a card, what are the odds that the other matches."

Let's go with your example. You picked up BB or RR. Now, flip over a card. It is B. Don't you see that we can't include the RR anymore? What are the odds that the other card is also B? That card came from one of three "types" of stacks. Either a BB, a RB, or a BR.

Yes, there are 50% of the stacks that are "the same." but after you've separated them, and picked one card, that is where you can't include BOTH RR and BB.

Can someone smarter than me explain this better?

Click to expand...

NO! If you flipped a B, you can't say it came from the RB stack. You are counting RB and BR even when you flip over a B first. If you flip a B, your choice of stacks are:BB

BR

That's it! How can you say that you flip a B and you can still be talking about the RB stack?

Click to expand...

Sweet J said:

So, we have equal chances that the card came from the BB pile, the RB pile, or the BR pile. Agree?

Click to expand...

Sweet J said:

kutta said:

Sweet J said:

Let's do this together. I have layed out 100 pairs of cards on my desk. Equal distribution. That means 25 pairs of RR, 25 pairs of RB, 25 pairs of BR, 25 pairs of BB.

Ready? I'm going to pick a card pair.

Now, tell me, is there an equal possibility that I pick each category? 25% chance of picking RR? 25% chance of picking BR? 25% chance of picking RB? 25% chance of picking BB?

Click to expand...

I really hope you see that you are going down a losing road here. You really need to open up and see that this problem is different from what you think it is.I will play this game with you. Yes, equal probability that you picked one of the four stacks.

Next move.

Click to expand...

Ok, I pick a card. It is black.Now, there are still 100 pairs on the desk. We don't know if it came from the BB pile, the BR pile, or the RB pile. Are the odds that it came from the BB pile, the BR pile, and the RB pile equal? I.e., is there an equal probability that it came from the BB, the BR, or the RB?

Click to expand...

fatguyinalittlecoat said:

Sweet J said:

fatguyinalittlecoat said:

Sweet J said:

Let's go with your example. You picked up BB or RR. Now, flip over a card. It is B. Don't you see that we can't include the RR anymore? What are the odds that the other card is also B? That card came from one of three "types" of stacks. Either a BB, a RB, or a BR.

Yes, there are 50% of the stacks that are "the same." but after you've separated them, and picked one card, that is where you can't include BOTH RR and BB.

Click to expand...

If the first card is a "B", wouldn't you also eliminate the "RB" stack as a possibility?

Click to expand...

This is where there is confusion. You picked one card randomly from the pair. It could have been the first card, or it could have been the second card. This is why this "card" problem simulates the "not knowing the older/younger child problem. You didn't pick the "first" card, you picked "one of the two cards." Both RB and BR are in play.

Click to expand...

Well, this is dumb, but if you insist on this explanation, then you still have a problem. The "B" card is twice as likely to have come from a "BB" stack than an "RB" or "BR" stack.

Click to expand...

Sweet J said:

fatguyinalittlecoat said:

Sweet J said:

fatguyinalittlecoat said:

Sweet J said:

Let's go with your example. You picked up BB or RR. Now, flip over a card. It is B. Don't you see that we can't include the RR anymore? What are the odds that the other card is also B? That card came from one of three "types" of stacks. Either a BB, a RB, or a BR.

Yes, there are 50% of the stacks that are "the same." but after you've separated them, and picked one card, that is where you can't include BOTH RR and BB.

Click to expand...

If the first card is a "B", wouldn't you also eliminate the "RB" stack as a possibility?

Click to expand...

This is where there is confusion. You picked one card randomly from the pair. It could have been the first card, or it could have been the second card. This is why this "card" problem simulates the "not knowing the older/younger child problem. You didn't pick the "first" card, you picked "one of the two cards." Both RB and BR are in play.

Click to expand...

Well, this is dumb, but if you insist on this explanation, then you still have a problem. The "B" card is twice as likely to have come from a "BB" stack than an "RB" or "BR" stack.

Click to expand...

No it's not. There are four stacks, with four equal probabilities. BB, BR, RB, RR. You are picking "stacks" NOT "cards." You FIRST (randomly) pick a stack, and THEN you (randomly) pick a card. So where are you disagreeing?Are you disagreeing that you have a 1 in 4 chance of picking BB, BR, RB, RR if you chose one of the four "stacks"? I would think not!

Once you pick a "stack," now choose a card. If it is black, there is a 33% chance it came from the BB stack, a 33% chance it came from the BR stack, and a 33% chance it came from a RB stack. Do you disagree?

Click to expand...

Sweet J said:

Once you pick a "stack," now choose a card. If it is black, there is a 33% chance it came from the BB stack, a 33% chance it came from the BR stack, and a 33% chance it came from a RB stack. Do you disagree?

Click to expand...

fatguyinalittlecoat said:

Sweet J said:

fatguyinalittlecoat said:

Sweet J said:

Let's go with your example. You picked up BB or RR. Now, flip over a card. It is B. Don't you see that we can't include the RR anymore? What are the odds that the other card is also B? That card came from one of three "types" of stacks. Either a BB, a RB, or a BR.

Yes, there are 50% of the stacks that are "the same." but after you've separated them, and picked one card, that is where you can't include BOTH RR and BB.

Click to expand...

If the first card is a "B", wouldn't you also eliminate the "RB" stack as a possibility?

Click to expand...

This is where there is confusion. You picked one card randomly from the pair. It could have been the first card, or it could have been the second card. This is why this "card" problem simulates the "not knowing the older/younger child problem. You didn't pick the "first" card, you picked "one of the two cards." Both RB and BR are in play.

Click to expand...

Well, this is dumb, but if you insist on this explanation, then you still have a problem. The "B" card is twice as likely to have come from a "BB" stack than an "RB" or "BR" stack.

Click to expand...

fatguyinalittlecoat said:

Sweet J said:

Once you pick a "stack," now choose a card. If it is black, there is a 33% chance it came from the BB stack, a 33% chance it came from the BR stack, and a 33% chance it came from a RB stack. Do you disagree?

Click to expand...

Yes I disagree.

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Sweet J said:

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?

Click to expand...