# Monty Hall Problem and Two Boys Riddle [Solutions]

Yesterday, 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]”

1. PalmerEldritch
February 17, 2014 at 5:42 am

The correct answer to the Two Boys Riddle as posted here ( “I have two children, one of which is a boy born on a Tuesday.”) is 1/2.

If instead the question was “Of all 2 children families that have at least 1 boy born on a Tuesday, what proportion have 2 boys?”, then the answer would indeed be 13/27.

There’s a difference between the 2 questions. 🙂

2. February 17, 2014 at 10:30 am

Palmer, correct answer is 13/27. The riddle as posed here is actually the same problem as the wording you offered. Consider, as posed here, the riddle offers no information as to the gender of the child that is not a boy born on Tuesday, only that there is one other child. The variation you suggest also offers no information as to the gender of the other child. In both cases, what is offered is a) the family has two children, and b) (at least) one child is a boy that was born on Tuesday.

• PalmerEldritch
February 18, 2014 at 3:46 am

The riddle as posed assumes that I will tell you ‘ a boy born on Tuesday’ if I have one. Whilst it is a necessary condition for me to have a boy born on Tuesday in order to tell you that fact, it isn’t a sufficient condition as I could tell you another fact, such as ‘girl born on Sunday’ if my 2 children are a ‘boy born on Tuesday’ and a ‘girl born on Sunday’

The wording I gave requires no such assumption to be made – that’s why the 2 questions are different and why you get different answers

• February 18, 2014 at 9:57 pm

Palmer: As posted, we are given two facts:

1) there are two children; and
2) one of those children is a boy born on Tuesday.

No other facts are conveyed, and no other facts should be assumed. Nothing is said about birth order. The child referenced in 2) may be the first born, second born, or a twin. There is nothing said about the child not referenced by 2). The other child may or may not be a boy, may or may not be born on Tuesday, may or may not be born on Wednesday, etc. All we can say about that other child is that it exists.

Thus, as posed, the riddle is really the same as how you worded it, and the probability is 13/27.

• PalmerEldritch
February 19, 2014 at 8:26 am

In the solution you give to this problem you ask the question “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.”? ” , and arrive at a figure of 27 (1+12+14), of which 13 fathers have 2 boys.

The question you should ask instead is “in how many of those families WOULD the father honestly state, “I have two children, one of which is a boy born on a Tuesday.”? “: and the answer to that question is 14: made up of 1 father who has 2 boys both born on Tuesday, 6 (or 50% of) fathers who have a boy born on Tuesday and boy born on NotTuesday (50% would tell you about the boy born on Not Tuesday), and 7 (or 50%) fathers who have a boy born on Tuesday and a girl. (50% would tell you about the girl)

Of those 14 fathers who WOULD state “I have two children, one of which is a boy born on a Tuesday.” , 7 of them have 2 boys. 7/14 = 1/2.

Your solution makes it a requirement that every father of 2 children who has a boy born on Tuesday has to tell you that fact. Since that isn’t stated in the problem definition you can’t assume it.

• February 22, 2014 at 10:26 am

>>Your solution makes it a requirement that every father of 2 children who has a boy born on Tuesday has to tell you that fact.<

• PalmerEldritch
February 24, 2014 at 7:23 am

I don’t need to see your JAVA code. I know what it does (it calculates the proportion of 2 children families that have 2 boys given that they have a boy born on Tuesday) from the answer you got .

Of the 27 parents of two children families who COULD truthfully say “one is a boy born on Tuesday” why do you assume that all parents WOULD say that, when 26 of those parents COULD equally truthfully state a different fact about the gender and birth day of one of their children?.

To analyse the problem it’s not sufficient to simply count cases, you need to calculate the probability of each case producing the observed outcome. Bayes Theory gives an answer of (1+12Q)/(1+26Q) where Q = [Probability parent TELLS you he has a “born on Tuesday”] . If you make it a requirement of the problem then Q = 1, if it’s a random statement of fact about the gender and birth day of one of the children then Q = 1/2.

Since the problem doesn’t state it’s a requirement you can’t simply assume it.

• February 24, 2014 at 10:25 am

I understand what you’re saying. But if we consider the statement as a given (because it was given) and look at conditional probabilities given what we were given, we get the 13/27 answer. (Because the Q = [Probability parent TELLS you he has a “born on Tuesday”] _given_ that [parent TELLS you he has a “born on Tuesday”] is 1.)

Of course, that’s not saying your interpretation is wrong, or that using your interpretation, the answer is not 1/2. But if we really want to split hairs, even that interpretation is leaving out applicable probabilities–the probability the speaker is a liar, the probability the speaker is not a liar but is confused, the probability the speaker is not liar or confused but is being coerced to deceive under threat from a sociopath, etc.

For me, Occam’s Razor cuts away the Q != 1 interpretations.

• JeffJo
April 4, 2014 at 9:05 am

A version of Bertrand’s Box paradox: Four boxes each contain two medallions; one has two bronze medallions, one has two gold medallions, and two have one of each kind.

1) I choose a box at random. What is the probability the box contains two medallions of the same kind?

2) I look in the box, so that I know what kind of medallions it contains. I tell you that there is at least one bronze medallion in it. What is the probability the box contains two medallions of the same kind (which can only be bronze)?

3) Instead of telling you about a medallion, I take one out without showing it to you. I tell you (although it should be obvious) that there was at least one medallion of the removed medallion’s kind in the box. What is the probability the box contained two medallions of the same kind?

The answer to Q1 is obviously 1/2. Since have no more information in Q3 than in Q1, its answer is also 1/2. But the removed coin has one, and only one, “kind.” If it is bronze, Q2 is identical to Q3 (remember what I told you in Q3). If it is gold, Q2 is a symmetric question that must have the same answer. So you have a paradox if you claim the answer to Q2 is not 1/2.

Joseph Bertrand used this apparent paradox in 1889 to warn future probabilists that you can’t base conditional probabilities on the information alone, you have to consider the probability that you would possess that information. The important factor here is P(I tell you there is a bronze medallion|the box has both kinds)=1/2, not P(I tell you there is a bronze medallion|I tell you there is a bronze medallion)=1. The answer to the Two Boy Problem, with or without including “born on Tuesday,” is 1/2.

As further proof, alter your Monty Hall Problem by stating the door numbers chosen by the contestant, and opened by Monty Hall. Use #1 and #3, respectively, to match the problem as discussed on Wikipedia. If we “consider these door numbers as given” as Andy insists we should, then we use P(Monty opens #3|Monty opens #3)=1 and the answer is 1/2, not 1/3. If we use P(Monty opens #3|Car is behind #1 & contestant chose #1)=1/2, the answer is 1/3.

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