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**Thurhame****Member**- Registered: 2013-03-03
- Posts: 6

MathsIsFun wrote:

"I tell you I have two children and that (at least) one of them is a boy, ... your right answer to my question is not 1/2 but 1/3." Agree!

"I tell you I have two children, and (at least) one of them is a boy

born on a Tuesday. What probability should you assign to the event that I have two boys?" he gives the answer13/27... Really?

Assume that boy/girl are equally likely and each day of the week is equally likely for each child.

Case 1: Both children were born on the same day of the week (1/7).

This day could be Tuesday (1/7) or NotTuesday (6/7).

Genders could be BB (1/4), BG (2/4), GG (1/4).

Case 2: The children were born on different days of the week (6/7).

Case 2a: Exactly one of the children was born on a Tuesday ((6 choose 1)/(7 choose 2) = 2/7).

Gender of the Tuesday child could be B (1/2) or G (1/2).

Gender of the other child could be B (1/2) or G (1/2).

Case 2b: Neither child was born on a Tuesday ((6 choose 2)/(7 choose 2) = 5/7).

Genders could be BB (1/4), BG (2/4), GG (1/4).

This is the sample set. Applying the restriction "one of the children is a boy born on a Tuesday" leaves us with

3/196 from Case 1, including 1/196 that both are boys.

24/196 from Case 2, including 12/196 that both are boys.

The probability of both being boys given the restriction is thus (1/196 + 12/196)/(3/196 + 24/196) = 13/27.

To put it another way, saying the boy mentioned was born on a Tuesday has a 12/13 chance of identifying the boy mentioned (1/2 chance of the other being a boy), and a 1/13 chance of leaving the boy unidentified (1/3 chance of both being boys).

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**JeffJo****Member**- Registered: 2013-03-19
- Posts: 1

"I tell you I have two children and that (at least) one of them is a boy, ... your right answer to my question is not 1/2 but 1/3." Agree!

"I tell you I have two children, and (at least) one of them is a boy born on a Tuesday. What probability should you assign to the event that I have two boys?" he gives the answer 13/27 ... Really?

This problem, as stated, is ambiguous. In most fields of Mathematics, that makes it unanswerable. But Probability can handle it - in fact, you could say that the probability is a quantitative measure of the ambiguity in the scenario. You just need to make reasonable assumptions about the ambiguity.

Unfortunately, most people who answer problems like this don't recognize all of the ambiguities. Without even realizing it, they make assumptions that avoid the ambiguity, but affect the answer. But when you don't know that you are making an assumption, you can't defend whether or not it is a reasonable one.

In this case, the ambiguity is that we need to know why this person told you "I have two children and that (at least) one of them is a boy (born on a Tuesday)."

Case 1: The person was required to tell you that fact, if it was true. If it wasn't, other people would have been chosen until one who could truthfully make the required statement was found.

In this case, the answers as presented are correct: 1/3 (without "Tuesday") and 13/27 (with).

Case 2: Just one person was selected, and with free will offered to tell you (1) the number of his or her children, (2) the gender of (at least) one, and maybe (3) the day of the week a child of that gender was born.

In this case, **it is possible** for the father of a boy born on a Tuesday to tell you he has a boy born on a Thursday, or for the mother of a boy born on a Tuesday to tell you she has a girl born on a Friday. The previous answers are wrong, because they count every who parent could tell you about the boy born on a Tuesday, not the ones who would.

If, instead, you count only half of the parents who have a boy born on a Tuesday and another child that fits some other description, you get 1/2 for every possible permutation of the facts.

One of these cases is reasonable to assume, when you aren't told the parent's motivation. The other is not. And your intuition is a good guide to tell which.

BTW, this problem originated at a conference honoring Martin Gardner, the long-time author of the Mathematical Games column in Scientific American. He originally answered 1/3 for the simpler problem, which is all most people recall. They somehow forget that he withdrew that answer, and said that either 1/3 or 1/2 could be correct based on the very ambiguity I described.

*Last edited by JeffJo (2013-03-19 09:33:15)*

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