P.Mean: Another counterintuitive probability problem (created 20100704). 
A recent article in Science News, rekindled the two children problem and offered an odd twist. Let me tackle the basic problem here and briefly outline the twist.
Here's the simple version. Suppose you have two children, one of whom is a boy. What is the probability that both children are boys? The obvious, but incorrect choice is 1/2. The correct answer is 1/3. How does this work? Consider that there are four possible families of two, GG, BG, GB, BB. Each of these is equally likely. Eliminate the first event GG, and the remaining three events, BG, GB, BB are still equally likely.
Wording is important here. You could have said "Suppose you have at least one boy, what is the probability that you have two boys" and all of a sudden the obvious answer doesn't seem so obvious anymore. Also note that if you change the wording to "Suppose the first one is a boy, what is the probability that both are boys" is NOT the same as the original question and with this wording the probability is indeed 1/2.
The correct answer can also be verified by a simple experiment that was conducted by Marilyn Vos Savant. She asked all her male readers who had exactly one sibling what the sex of their sibling was. And about 1/3 of her readers said their sibling was also a male.
One comment about this story posted on the Facebook page of someone linking to this article was typical of the reaction that this sort of puzzle creates.
Are they all on crack, or are they kidding? Birth order? It's never mentioned, and wouldn't be an issue anyway. If you have two children, and one is a boy, the odds are 1/2 that you have two boys. That 1/3 and other insanity wouldn't come into play unless they were part of the original question. These are the kind of "clever" people who get hired because they seem smart, then run a business into the ground.
You can calculate the probability without referring to birth order if you like. The three events, two boys, two girls, and a mix are NOT equally likely, as a mix has probability 1/2, twice as large as either of the other two events. So why would you expect that after the two girl possibility was removed that the two remaining choices would suddenly shift from being unequal to being equal? A mix would still be twice as likely as two boys and since the total probability must equal 1, the answer must be 2/3 mix, 1/3 two boys.
A mathematical calculation provides the same answer. Recall that
P[A  B] = P[A and B] / P[B].
So for this particular problem, you have
P[Two boys  At least one boy] =
P[Two boys AND At least one boy] / P[At least one boy] =
P[Two boys] / P[At least one boy] =
(1/4) / (3/4) =
1/3
Let's look at the twist, now. I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys? They are seven days in the week and four possible families of size 2. You can list the 7*7*4 possible families, using the convention that uppercase is a boy and lower case is a girl.
MO MO
 MO mo

mo MO

mo mo

MO TU

MO tu

mo TU

mo tu

MO WE

MO we

mo WE

mo we

MO TH

MO th

mo TH

mo th

MO FR

MO fr

mo FR

mo fr

MO SA

MO sa

mo SA

mo sa

MO SU

MO su

mo SU

mo su

TU MO

TU mo

tu MO

tu mo

TU TU

TU tu

tu TU

tu tu

TU WE

TU we

tu WE

tu we

TU TH

TU th

tu TH

tu th

TU FR

TU fr

tu FR

tu fr

TU SA

TU sa

tu SA

tu sa

TU SU

TU su

tu SU

tu su

WE MO

WE mo

we MO

we mo

WE TU

WE tu

we TU

we tu

WE WE

WE we

we WE

we we

WE TH

WE th

we TH

we th

WE FR

WE fr

we FR

we fr

WE SA

WE sa

we SA

we sa

WE SU

WE su

we SU

we su

TH MO

TH mo

th MO

th mo

TH TU

TH tu

th TU

th tu

TH WE

TH we

th WE

th we

TH TH

TH th

th TH

th th

TH FR

TH fr

th FR

th fr

TH SA

TH sa

th SA

th sa

TH SU

TH su

th SU

th su

FR MO

FR mo

fr MO

fr mo

FR TU

FR tu

fr TU

fr tu

FR WE

FR we

fr WE

fr we

FR TH

FR th

fr TH

fr th

FR FR

FR fr

fr FR

fr fr

FR SA

FR sa

fr SA

fr sa

FR SU

FR su

fr SU

fr su

SA MO

SA mo

sa MO

sa mo

SA TU

SA tu

sa TU

sa tu

SA WE

SA we

sa WE

sa we

SA TH

SA th

sa TH

sa th

SA FR

SA fr

sa FR

sa fr

SA SA

SA sa

sa SA

sa sa

SA SU

SA su

sa SU

sa su

SU MO

SU mo

su MO

su mo

SU TU

SU tu

su TU

su tu

SU WE

SU we

su WE

su we

SU TH

SU th

su TH

su th

SU FR

SU fr

su FR

su fr

SU SA

SU sa

su SA

su sa

SU SU

SU su

su SU

su su
Now you can copy this list into a text editor or word processor and count the number of occurrences of TU (case being important here). You would find 28 instances, which makes sense. But two of those instances would be on the same line (TU TU). So there are 27 families with a son born on a Tuesday.

MO TU

mo TU

TU MO

TU mo

TU TU

TU tu

tu TU

TU WE

TU we

TU TH

TU th

TU FR

TU fr

TU SA

TU sa

TU SU

TU su

WE TU

we TU

TH TU

th TU

FR TU

fr TU

SA TU

sa TU

SU TU

su TU
and thirteen of these have two sons.

MO TU

TU MO

TU TU

TU WE

TU TH

TU FR

TU SA

TU SU

WE TU

TH TU

FR TU

SA TU

SU TU
Let's try to work it out using the formula for conditional probability.
P[Two boys  At least one boy born on a Tuesday]
= P[Two boys AND At least one boy born on a Tuesday] / P[At least one boy
born on a Tuesday]
Now the days a child are born on and the sex of the child are independent. So we can rewrite the top probability as
= P[Two boys AND At least one Tuesday birth] / P[At least one boy
born on a Tuesday]
= P[Two boys]*P[At least one Tuesday birth] / P[At least one boy born on a
Tuesday]
We also need to compute the probability of "at least one" as the complement of the probability of "none."
= P[Two boys]*(1  P[No Tuesday births]) / P[At least one boy born on a
Tuesday]
The bottom probability is a bit trickier. It can be thought of as a union of two events.
= P[Two boys]*(1  P[No Tuesday births]) / P[First child is a Tuesday
boy OR Second child is a Tuesday boy]
= P[Two boys]*(1  P[No Tuesday births]) / (P[First child is a
Tuesday boy] +P[Second child is a Tuesday boy] P[Both are Tuesday boys])
= (1/4)*(1  (6/7)^2) / (1/14 + 1/14  1/196)
= (1/4)*(13/49) / (14/196 + 14/1961  1/196)
= (13/196) / (27/196)
= 13/27
It may be worth noting that a similar controversial probability problem, the Monty Hall problem, is currently causing trouble at Wikipedia. http://en.wikipedia.org/wiki/Monty_Hall_problem.
Now I personally dislike problems like these because in my mind they are trivializing the field of probability and statistics. In my job, I don't ever talk about two child families or a goat behind a hidden door. I also dislike these types of problems because they rely on subtleties of the English language, sometimes subtleties that veer towards the ambiguous. But enough people find problems like this interesting that I should write a bit about them.
This work is licensed under a Creative Commons Attribution 3.0 United States License. This page was written by Steve Simon and was last modified on 20100706. Need more information? I have a page with general help resources. You can also browse for pages similar to this one at Category: Probability concepts.