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Think you understand Monty Hall? Try the Tuesday boy problem.

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Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#81

Post by zerostar07 »

jfager wrote: post #73

When in doubt, use brute force: 1BT 2BM 1BM 2BT 1BT 2BT 1BT 2BT I think the issue people are having is that they want to distinguish between the two instances of '1BT 2BT', but it doesn't work like that. Let's say I numbered each of the possibilities written above 1 through 28. If I handed you a piece of paper with '1BT 2BT' written on it, could you tell me which number that corresponded to? No - it would be one of t…

Alternatively, suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14 items). We have a bucket with 2 of each of these combinations, as there are 2 kids (total 28 items). The problem states that we have taken one "BT" item out of the bucket (so 27 items left), and are asking how many "B*" items are left in the bucket (13).

The confusion arises because one can consider the same problem with replacement, putting back the BT item in the bucket before picking a second time. It's something that's not clear if u are not thinking of enumeration.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#82

Post by bermanoid »

reitzensteinm wrote: post #80
jfager wrote: post #75

Earlier quoted context omitted.

You're making the exact mistake the author is cautioning against, which is assuming the day doesn't matter. Write out all the possibilities (see my other comment in this thread), eliminate the dupe, and you get 13/27.

Did you read my post carefully? I do get 13/27, when presented with the information from a neutral third party - i.e. filter for all 2 child families with one son male/Tuesday, what is the chance the other is also male/Tuesday. But the fact that the father voluntarily offered up the information changes the probability distribution. We can assume he's selecting one of his children at random, and revealing their birthd…

We can assume he's selecting one of his children at random, and revealing their birthday and gender.

This is the entire point of the article, IMO: we have to make some assumption about how we selected this guy to talk to, and how he chose what to tell us. It's not pinned down by the statement of the problem, and what you might consider a natural assumption is not necessarily what other people might assume.

Which is why these problems tend to suck...

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#83

Post by orangecat »

user24 wrote: post #46
ColinWright wrote: post #9

Earlier quoted context omitted.

No, you shouldn't "also discount every other symmetrical pair", for exactly the same reason as there's a 1/36 chance of rolling double 6, but 2/36 chance of rolling a six and a one. It's all to do with labellings, and it's the most common source of error[1] in statistics. [1] By "error" I mean calculations that then don't agree with the experimental results.

What about this: I have two teenagers. One is a boy of 13. Do we encounter a similar situation with regard to the odds of the second teenager being a boy? edit: I'm thinking the odds are exactly the same, 13/27, by coincidence, as there are seven possible teen ages. So then, what about this: One is a boy named George. Or One wears a black shirt. Or One likes chocolate. Doesn't this mean that the more information we g…

Doesn't this mean that the more information we gain about the boy, the less likely it makes it that his sibling is a brother?

More likely, but only if that information being true was a precondition for knowing about the boy in the first place. Take the following scenarios, assuming Alice knows Bob has exactly two children.

Alice: Do you have a son? Bob: Yes Alice: Pick one of your sons, and tell me the day of the week he was born Bob: Sunday

Here the day of week provides no additional information because Bob will always have an answer (like in Monty Hall, where Monty will always reveal a losing door), so the probability that Bob has two boys is 1/3.

Alice: Do you have a son who was born on a Sunday? Bob: Yes

Here having a son isn't enough; he also has to satisfy a condition that occurs with only 1/7 probability. Bob is more likely to be able to answer yes if he has two sons and thus two chances to satisfy that condition.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#84

Post by Delmania »

ars wrote: post #22
nadam wrote: post #21

As the currently top voted comment does not get it, I try to intuitively explain the paradox. No, it is not ambiguity of language. It says formally: I have two children. There exists a child of mine who is (boy and born on tuesday). And yes, the probability of the other child being a boy is 13/27. To understand this, try a more extreme case: When a child is born we generate a random number: rnd(1billion) Now the man…

Why can't the other boy also be born on Tuesday?

If the older child is the boy born on Tuesday, you have 7 chances the younger child is a boy. If the younger child is a boy born on Tueday, you only have 6 chances the older child is a boy. This is because the 7th chance (older is a boy born on Tueday) was covered in the first scenario.

Summarized from: http://www.sciencenews.org/view/generic/id/60598/title/Math_...

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#85

Post by [deleted] »

nadam wrote: post #21

As the currently top voted comment does not get it, I try to intuitively explain the paradox. No, it is not ambiguity of language. It says formally: I have two children. There exists a child of mine who is (boy and born on tuesday). And yes, the probability of the other child being a boy is 13/27. To understand this, try a more extreme case: When a child is born we generate a random number: rnd(1billion) Now the man…

[deleted]

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#86

Post by reitzensteinm »

bermanoid wrote: post #82
reitzensteinm wrote: post #80

Earlier quoted context omitted.

Did you read my post carefully? I do get 13/27, when presented with the information from a neutral third party - i.e. filter for all 2 child families with one son male/Tuesday, what is the chance the other is also male/Tuesday. But the fact that the father voluntarily offered up the information changes the probability distribution. We can assume he's selecting one of his children at random, and revealing their birthd…

We can assume he's selecting one of his children at random, and revealing their birthday and gender. This is the entire point of the article, IMO: we have to make some assumption about how we selected this guy to talk to, and how he chose what to tell us. It's not pinned down by the statement of the problem, and what you might consider a natural assumption is not necessarily what other people might assume. Which is w…

I definitely agree - I hate problems like this, because the difficulty is caused by ambiguity of English, not the problem itself.

But even given other assumptions as to why the father selects the child he does - sort by date, males first etc, the author's answer of 13/27 is still almost certainly wrong - he should have just taken the father out of the equation completely.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#87

Post by NickPollard »

reitzensteinm wrote: post #58

I've been thinking about this for an hour, and I'm now convinced that the author is wrong. The fact that we found out about one of the children from the father means that all probabilities are not equal, even though they're treated here like they are. The difference is between the information being offered, and determined independantly. I'll do this with the boy/girl problem, for simplicities sake. If we ask a man if…

The issue here is with assumptions - you have made a different set of assumptions from the author, and hence are getting a different result.

A lot of people here are having similar issues, by misreading exactly what the initial proposition means.

Your reasoning above relies on the 'likeliness' of a man giving you the information, which is something that is not meant to be a part of the problem. Although it is phrased as a man 'telling' you something, that statement is really a metaphor for 'you determine the following piece of information, 100% truthfully'.

In particular, your explanation assigns agency to the man - that if he does in fact have a son, he may or may not choose to reveal the truth 'I have a son'. However it makes no allowance for the man lying - so you are assuming if he answers it will be truthfully, but you are allowing him the lie of omission.

Whilst there is nothing specific that rules out your interpretation, it is not what is intended. Read it instead as:

----

There exists a man, A.

A has exactly two children.

The statement 'A has at least one Son, B' is true

What is the chance that the statement 'The Non-B child of A, is a son' is true?

----

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#88

Post by danparsonson »

ars wrote: post #53
ajanuary wrote: post #48

Earlier quoted context omitted.

When considering the case that the older son was the one born on a Tuesday, that gives 14/28 possibilities. One of those 14 is the case that both were born on Tuesday. When considering the case that the younger son was the one born on a Tuesday, that gives 14/28 possibilities. One of those 14 is the case that both were born on a Tuesday. But woops, we've already covered the case that both were born on a Tuesday in ou…

No, you do not remove the duplication. If you have two kids there are 4 possibilities, not 3: BG, BG, BB, GG BG seems to be the same as GB, except that it's not. And it's not in this case either.

I think I've spotted where your misunderstanding is.

BG is only not the same as GB if there is some other information available - which was born first, what their names are, hair colour, etc., because then you'd be saying something like

Boy born first, Girl born second

Girl born first, Boy born second

and those are two distinct possibilities. The point is that they are only distinct if you have this extra information, which we don't. We have no way of differentiating between the two children except for gender, and therefore BG and GB both just say 'one male child and one female child'. You might assume that the order of the two letters specifies the order in which the children were born, but that's a false assumption because it's not stated anywhere.

If you include the order, there are four possibilities: a) Boy first, Boy second, b) Boy first, Girl second, c) Girl first, Boy second, d) Girl first, Girl second. If you don't, there are three possibilities: a) two boys, b) two girls, c) one boy and one girl. The unspecified information about order is the subtle but important difference.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#89

Post by jfager »

zerostar07 wrote: post #81
jfager wrote: post #73

When in doubt, use brute force: 1BT 2BM 1BM 2BT 1BT 2BT 1BT 2BT I think the issue people are having is that they want to distinguish between the two instances of '1BT 2BT', but it doesn't work like that. Let's say I numbered each of the possibilities written above 1 through 28. If I handed you a piece of paper with '1BT 2BT' written on it, could you tell me which number that corresponded to? No - it would be one of t…

Alternatively, suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14 items). We have a bucket with 2 of each of these combinations, as there are 2 kids (total 28 items). The problem states that we have taken one "BT" item out of the bucket (so 27 items left), and are asking how many "B*" items are left in the bucket (13). The confusion arises because one can consider the same problem with repla…

Having 2 each of BM, GM, etc. is implicitly labeling 1BM, 2BM, 1GM, 2GM, etc. If you sampled with replacement, you have to admit the possibility that you draw the same BT twice. It's a stretch to suggest that the statement is ambiguous in this way, as it would imply that the father could have the same child twice.

Re: Think you understand Monty Hall? Try the Tuesday boy problem.

#90

Post by DougBTX »

zerostar07 wrote: post #81
jfager wrote: post #73

When in doubt, use brute force: 1BT 2BM 1BM 2BT 1BT 2BT 1BT 2BT I think the issue people are having is that they want to distinguish between the two instances of '1BT 2BT', but it doesn't work like that. Let's say I numbered each of the possibilities written above 1 through 28. If I handed you a piece of paper with '1BT 2BT' written on it, could you tell me which number that corresponded to? No - it would be one of t…

Alternatively, suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14 items). We have a bucket with 2 of each of these combinations, as there are 2 kids (total 28 items). The problem states that we have taken one "BT" item out of the bucket (so 27 items left), and are asking how many "B*" items are left in the bucket (13). The confusion arises because one can consider the same problem with repla…

What would the question have to be to make the following the answer?

Suppose we label the combinations of child-day BM, GM, BT, GT ... BU, GU (14 items). We have two buckets which both contain a copy of each combination (total 28, 14 in each bucket). The problem states that we have taken one "BT" item out of one bucket, and are asking how many "B" items are left in the other bucket (7 out of 14).

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