Tricky Probability Problem: Chance of Two Sons, One Born on Tuesday?

http://wmbriggs.com/blog/?p=2553 ^ | William M. Briggs

[mattstat]

I forgot to say about the twist: ignore leap years.

[dangus]

Still exactly 1/3.

[pelican001]

“No, they’re no different possibilities. You don’t know birth order.”

But that’s the point - birth order is irrevelant to the problem. The answer is 1/2.

[Toskrin]

I would disagree. Odds are 1/3 no matter if the day of the week is specified.

This is a variation of the Monty Hall Problem:

http://en.wikipedia.org/wiki/Monty_Hall_problem

[drangundsturm]

This is simply an ambiguously worded problem. The grammar used does not absolutely rule out that the other child is also a son born on a Tuesday. To be unambiguous, it should have said, "I have two children, one and only one is a son born on a Tuesday."

This is less a "tricky" probability question than a tricky and ambiguous grammar question. So I cry foul.

[dangus]

No, It’s not. For two kids, there are four possibilities:

Boy then boy
Boy then girl
Girl then boy
Girl then Girl <— Only one eliminated.

The nonsense about Tuesday is a red herring to throw you off, and its very effective.

[dangus]

That ambiguity is irrelevant, and does NOT change the answer: 1/3.

[djf]

Whatever the other child is, it is NOT a son born on Tuesday.

So what are the possibilities?

It could be a son born on one of the other six days of the week.
Or it could be a daughter born on any of the seven days of the week.

So that’s 13 possible outcomes. Six of those outcomes leads to two sons.

Whatever the percentage is, the answer is 6/13

[KarlInOhio]

I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

13/27 if you mean at least one boy born on a Tuesday. 6/13 if you mean one and only one boy born on a Tuesday.

Just figure the odds of each of the three sex/day combinations:

Boy Tuesday (bt) = 1/14
Boy not Tuesday (bnt) = 6/14
Girl (g) = 7/14 = 1/2

Form a table of all nine ordered combinations (bt*bt = 1/196, bt*bnt = 6/196, bnt*bt = 6/196, etc.) and throw out the ones which don't have at least one bt (maybe also throw out the one with two bt depending on whether you want one and only one boy born on Tuesday). Then take the sum of the pairs with two boys (13/196) divided by the sum of all allowed combinations (27/196) and get the answer (13/27).

[pelican001]

Incorrect. There is no equivalent in the current problem to the second required action in the Monty Hall Problem: “The game show host, Monty Hall...now has to open one of the two remaining doors”. There is no equivalent, second condition in the curent problem. The answer is (approximately) 1/2.

[KarlInOhio]

If your son was born last Tuesday, the probability you have two sons is 2/3.

In that case it would be 1/2 because the restriction "last Tuesday" forces an ordering (except in the case of twins, which I will ignore). The two kids must be either girl then boy or boy then boy. It is impossible for the second child to be a girl if you had a boy just last Tuesday.

[John Valentine]

As I read the problem, strewn with red herrings, it boils down to the sex of an unknown child. Either boy or girl, the chance is 1/2.

The only factor not in play in this anlysis is the ratio of boys to girls in the child population, but that issue is one that has generally been ignored in the discussions here, in favor of stipulating conditionalities that are nowhere found in the problem, like the second boy is NOT born on a Tuesday. Who says so?

[laker_dad]

Hey! I’m getting ready to go on vacation tomorrow. Now my head hurts...

[PreciousLiberty]

“The answer is (approximately) 1/2.”

Right. According to the stats I laboriously tracked down (try doing a search on ‘infant gender statistics in the United States’), there are 1.05 male births for every female in the US. Although we know one birth was male in this case, one of the first principles of probability is that one experiment (event) doesn’t affect the next.

Therefore the probability is 1.05/2, .525, or 52.5%. In the United States, of course. (It is likely very slightly smaller than this actually due to rare abnormalities such as hermaphroditism or babies born with no genitalia).

[pelican001]

Thank you.

[Leftism is Mentally Deranged]

There is no probability, since both have already been born.

[Gil4]

One child has been identified as a boy. There is only one variable - the “other child.” If you play the birth order game, it's still 50/50:

Boy, Girl
Boy, Boy

Girl, Boy
Boy, Boy

The “born on a Tuesday” is irrelevant to the question asked.

[ml/nj]

If the problem were: I have two children. The first one was a boy. What is the probability that I have two sons? That would be a different problem, don't you think?

ML/NJ

[John Valentine]

You are wrong. The stipulated male child is either the eldest or the youngest. He can’t be both or neither.

If eldest the choice count is:

Boy-Boy
Boy-Girl, i.e. 1/2

If the youngest, then the choice count is:

Boy-Boy
Girl-Boy, again 1/2

Case closed tight, not to be re-opened.

[sig226]

Welcome to probability. Each outcome has an equal possibility - BG, GB. Briggs is right.