The Tuesday Birthday Problem

An anonymous reader sends in a mathematical puzzle introduced at the recent Gathering 4 Gardner, a convention of mathematicians, magicians, and puzzle enthusiasts held biannually in Atlanta. The Tuesday Birthday Problem is simply stated, but tends to mislead both intuitive and mathematically informed guesses. "I have two children, one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?" The submitter adds, "Believe it or not, the Tuesday thing is relevant. Well, sort of. It's ambiguous."

Comments

Re:Let's rephrase that.

[joss]

> What if I say: "I have two pieces of fruit. This one is a banana". Do you feel this statement implies anything about the other piece of fruit? Can't that also be a banana?

Yes sure.. the other piece can be a banana. But it is different to say "one of them is a banana" than "this one is a banana", because in the first case the number "one" applies to the subset of the fruits which are bananas while in the second it applies to the subset of the single chosen fruit.

For this reason the answer given in TFA is wrong. The correct answer is 12/27.

Sorry, it's 50/50, period.

[drumcat]

Given that it's actually 49/51, here's the deal. The mentioned boy is irrelevant, absolutely. The other child's outcome was not or would not have been influenced by the mentioned boy, short of all the twins, conjoining, BS exceptions. End of story. Unless the story problem says that the mentioned has an effect on the unmentioned, the other is unaffected. "Oh, but I was so close to three cherries on that last slot machine pull - I'm so close to winning!"

Re:Well?

[A. B3ttik]

You don't take into account which one is older/younger.

Older Girl on Monday, Younger Boy on Tuesday
Older Girl on Tuesday, Younger Boy on Tuesday
Older Girl on Wednesday Younger Boy on Tuesday
Older Girl on Thursday, Younger Boy on Tuesday
Older Girl on Friday, Younger Boy on Tuesday
Older Girl on Saturday, Younger Boy on Tuesday
Older Girl on Sunday, Younger Boy on Tuesday

Younger Girl on Monday, Older Boy on Tuesday
Younger Girl on Tuesday, Older Boy on Tuesday
Younger Girl on Wednesday Older Boy on Tuesday
Younger Girl on Thursday, Older Boy on Tuesday
Younger Girl on Friday, Older Boy on Tuesday
Younger Girl on Saturday, Older Boy on Tuesday
Younger Girl on Sunday, Older Boy on Tuesday

Older Boy on Monday, Younger Boy on Tuesday
Older Boy on Tuesday, Younger Boy on Tuesday*
Older Boy on Wednesday Younger Boy on Tuesday
Older Boy on Thursday, Younger Boy on Tuesday
Older Boy on Friday, Younger Boy on Tuesday
Older Boy on Saturday, Younger Boy on Tuesday
Older Boy on Sunday, Younger Boy on Tuesday

Younger Boy on Monday, Older Boy on Tuesday
*[Already counted in the above, and thus not repeated here]
Younger Boy on Wednesday Older Boy on Tuesday
Younger Boy on Thursday, Older Boy on Tuesday
Younger Boy on Friday, Older Boy on Tuesday
Younger Boy on Saturday, Older Boy on Tuesday
Younger Boy on Sunday, Older Boy on Tuesday


Note that "Older Boy on Tuesday, Younger Boy on Tuesday" would, by intuition, be counted twice. But it's not. So you end up with 13/27 probability that the unmentioned child is a Boy, 14/17 probability that it is a girl.

Occums razor!

[djsmiley]

Occums razor.

You want to know the likelyness that something is male - as long as its a "normal" reproduction -> we presume it is as its not mentioned to any more likely one way or the other.

So, 50%.

Hm... reading too much into it peoples??

[s0litaire]

Quote:"I have two children, one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?"

He has 2 kids:
1 Male
1 Unknown

So possible outcomes: (0=Male 1=Female)
0,0
0,1

(Note 1,0 is the same as 0,1)

So probability is 50%

Case closed...

The original Birthday Problem is wrong

[mbone]

From the original article

Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?

Intuition would suggest that the answer should be 1/2, since the sex of one child is independent of the sex of the other. And indeed, had he been told which child was a boy (say, the younger one), this reasoning would be sufficient. But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth:

Boy, girl
Boy, boy
Girl, boy

Since one child is a boy, we know that girl, girl isn’t a possibility. Of the three approximately equally likely possibilities, one has two boys and two have a girl and a boy — so the probability of two boys is 1/3, not 1/2, Devlin concluded.

But this makes no sense whatsoever. For simplicity, I will assume that the human sex ratio is 1:1, and ignore the possibility of identical twins. I will also ignore the possibility of the "Monty Hall effect" - i.e., that there is other information that skews reporting of the first child. (Without this, anything can go. Suppose that I belong to a religion that believes that the birth of two boys in a row is deeply shameful and should never be mentioned. If you know that, saying that I have one boy make the probability of a girl 100%. Any other probability you might want is also possible, and so this has to be ruled out.)

To make the English clearer, suppose that I am on a beach with equal numbers of blue (B) and green (G) pebbles, and I pick two up, carefully noting the color of each.

So, an observer who just knows I have picked up two pebbles (but not the color of either) has the following probability table

BG - 25 %
BB - 25 %
GB - 25 %
GG - 25 %

Ok, I continue to walk along and meet up with the observer, and as I do so one pebble drops at random out of my bag. It is Blue (B). The observer sees this, and modifies his probability table as follows :

New probability = old probabiity x probability of having a Blue pebble fall out (renormalized to sum to unity, if necessary). This yields

BG - 25 %
BB - 50 %
GB - 25 %
GG - 0 %

So, the problem with the original "Birthday Problem" analysis is that, while there are indeed three choices, one with a second boy, and two with a girl, the probability of these choices are not equal ! With the correct probabilities, the chance that the second child is a boy is 50%

I would argue that, in the universe of these problems (where you are not supposed to need additional, unstated, information beyond basic things like people want to minimize jail time and maximize revenue), this is the correct analysis of the original "Birthday Problem," and the one presented in the original article is wrong.

The original "Tuesday Birthday Problem" analysis is likewise wrong, in the same fashion.