Monty Hall Problem and Two Boys Riddle [Solutions]

FAS Talk Puzzler: Monty Hall Problem and Two Boys RiddleYesterday, I presented two of my favorite probability puzzlers: the Monty Hall Problem and the Two Boys Riddle.  If you didn’t see those, you might want to read yesterday’s post before reading the solutions below.

None of my millions of readers (okay, tens) got the correct answer to the Two Boys Riddle, but I offer a hat tip to Jim Peurach who was the first with the correct solution for the Monty Hall Problem.

So, without further ado, the solutions…

Solution: Monty Hall Problem

The solution to the Monty Hall Problem is that the contestant should always switch doors after being shown a dud prize.  By adopting this strategy, the contestant will win the grand prize 2/3 of the time.

There are several ways to explain this solution, but the one I find most compelling is this:  If you switch doors after being shown a dud prize, you reverse whatever outcome you would have had.  That is, if you had originally chosen the grand prize, switching will leave you with a dud.  But if you had originally chosen a dud, switching will leave you with the grand prize.  So, by always switching, you win whenever you would have originally lost, which is 2/3 of the time.

It is important that you always switch for this strategy to work.  Many people fall into the trap of believing that once one “dud” door is eliminated, you’re left with two doors and thus a 50-50 chance.  This would be true if you ignored your initial choice and chose randomly between the two remaining doors.  But doing so ignores important information:  That the door you originally chose had a 1/3 chance of winning.  Eliminating one of the doors you did not originally pick does nothing to change this.  That is, if you always stick with the original door, you’ll have a 1/3 chance of winning.  Thus, the other remaining door must have a 2/3 chance of winning.

Solution: Two Boys Riddle

The solution to the Two Boys Riddle is that there is a 13/27 chance that the man has two boys.

It’s easy to be led astray by this simple riddle.  When I first encountered this problem, I was convinced the answer was 1/2, reasoning that since the day on which one child is born is independent of the gender of another child, the information that the boy was born on Tuesday was irrelevant and the riddle reduced to “What are the odds the other child is also a boy?”

But that reasoning is flawed, as we shall see.

Another common incorrect answer is 1/3, based on the argument that there are four equally likely permutations for any family with two children: BB, BG, GB, and GG.  Since we know at least one child is a boy, we can eliminate the GG case, leaving three possible permutations, one of which has two boys.

But that reasoning is also flawed.

Let’s see how we arrive at the correct value of 13/27.  First consider a large population of families with exactly two children.   Assuming the gender of the first child is independent of that of the second child (which may or may not be exactly true), then each family falls into one of four equally likely groups:  BB, BG, GB, and GG

Since these are equally likely permutations, we can expect the same number of families in each group.  Lets call that number N.

Now, lets calculate the number of families in each group that have a boy born on a Tuesday.

Family Type Number of Families Expected Number of Families with a Boy Born on Tuesday
BB N N * 13/49
BG N N * 1/7
GB N N * 1/7
GG N 0

The BB group is the most complicated.  Within this group, each of the two boys could have been born on any day of the week, making for 49 possible permutations.  Of these permutations, 13 correspond to a boy being born on a Tuesday (6 where the first boy was born on a Tuesday and the second was not, 6 where the second was born on a Tuesday and the first was not, and 1 where both boys were born on Tuesday).

The BG and GB groups are simpler.  In each of these groups, the one boy could have been born on any day of the week, so we can expect that 1/7 of them were born on a Tuesday.

The GG group is trivially simple.  There are no boys at all, so there are no boys born on a Tuesday.

So, across our entire population of families with two children, now many could possibly be the family of the man in our riddle?  That is, in how many of those families could the father honestly state, “I have two children, one of which is a boy born on a Tuesday.”?  Adding the values from the table above, we see the total number of families having a boy born on a Tuesday is:

N * [13/49 + 1/7 + 1/7]

which equals

N * 27/49

So, this is the starting population for the probability for which the riddle is asking.  Out of this population, how many families have two boys?  That’s easy; that is the number from the BB row in the table above: N * 13/49.

So, the answer to the riddle—What is the probability of having two boys given a family of two children containing a boy born on Tuesday?—is:

(N * 13/49) / (N * 27/49)

which equals

13/27

Do you have another favorite probability riddle?  Please share it in the comments.

9 comments for “Monty Hall Problem and Two Boys Riddle [Solutions]

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.