Variant of boy or girl paradox

Here is an interesting question my friend brought up:

You are invited to a family party. A Boy opens the door for you. There are two children there. What is the probability that a boy opens door for you next time?

This is my solution:

$$P(\text{boy second time | boy first time}) = \frac{P(\text{boy both time)}}{P(\text{boy first time)}}$$ $$=\frac{1/3+1/3*1/2*1/2+1/3*1/2*1/2}{1/3+1/3*1/2+1/3*1/2}=3/4$$

The $1/3$ comes from the fact that given we have one boy already so the combination can only be {boy, girl}, {girl, boy}, {boy, boy}. The 1/2 comes from for each time there is a 1/2 probability that a boy will open the door if there is one boy one girl.

Is my calculation correct? I feel not confident about the 1/3 argument.

Edited: @Rolf proposed another point that actually the probability of two boys and one boy/one girl should be 1/2, not 1/3 each. Edited again: Actually there are two approaches, as shown below. The key is to whether to assume the prior or not in the calculation.

• That feels off ... if, as you say, the fact that a boy opened the first time means that there is a 1/3 chance that there are 2 boys living at the house, and a 2/3 chance that there is 1 boy and 1 girl at the house, then the chance of a boy opening the door at any time should be 1/3 * 1 + 2/3 * 1/2 = 2/3 – Bram28 Nov 14 '17 at 3:00
• @Bram28: We have further information than the fact that there is at least one boy. We know that a boy was chosen randomly to answer the door, This boosts the chance that there are two boys. – Ross Millikan Nov 14 '17 at 3:10
• @RossMillikan Ugh .. am I caught by the two boys paradox again?! Back to the drawing board ... – Bram28 Nov 14 '17 at 3:12
• Why do we know that a child is chosen at random to open the door? Maybe the oldest or youngest child always opens the door. – pipe Nov 14 '17 at 9:02
• The quoted section is a particularly bad version of this problem. We're left to assume that the other child s equally likely to be a boy or a girl and that, when somebody knocks on the door, a child is chosen uniformly at random to open it. Neither of these assumptions seems to be good. Perhaps the other child is the boy's friend (so much more likely to be male than female). Perhaps the oldest child is always sent to open the door. Perhaps normally an adult opens the door, which would seem safer. – David Richerby Nov 14 '17 at 15:11

There's definitely one boy, so half of the possibilities have a probability of one:

$$\frac{1}{2}\times 1=\frac{1}{2}$$

Assuming an equal chance of either boy or girl, the other half will have an equal chance of being a boy:

$$\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$$

$$\frac{1}{2}+ \frac{1}{4}=\mathbf{\frac{3}{4}}$$

Done.

• This is the best answer. There's 1/2 chance the same kid will open the door, who will definitely be a boy. There's 1/2 chance a different kid will open the door, and we don't know anything about their gender, so there's 1/2 chance they will be a boy. – jwg Nov 14 '17 at 10:40
• Yes, there is 1/2 chance that the same kid will open the door. But there is 2/3 chance that the other kid is a girl. – fishinear Nov 14 '17 at 11:53
• Simplest answer – Timtech Nov 14 '17 at 16:13
• @fishinear why? – NeutronStar Nov 14 '17 at 23:10
• @fishinear In the traditional problem we don't know which child is a boy, so there are 3 possibilities, but in this case we know that the child who opens the door is a boy, so we only have 2 possibilities. So the probability is 1/2. – RJM Nov 14 '17 at 23:17

Let's suppose we know that there will be two children at the party, and we make the rather doubtful assumptions that each is equally likely to be boy or girl, and each time someone opens the door it is equally likely to be each of the two children, independent of who opens the door on other occasions.

With probability $1/4$ both children are boys, and the door is opened by a boy both times.

With probability $1/4$ both are girls, and the door is opened by a girl both times.

With probability $1/2$ one is a boy and one is a girl, and the door openers are equally likely to be (boy,boy), (boy,girl), (girl,boy), (girl,girl).

Then

$$\frac{P(\text{boy both times})}{P(\text{boy first time})} = \frac{(1/4) \cdot 1 + (1/2) \cdot (1/4) + (1/4) \cdot 0}{(1/4) \cdot 1 + (1/2) \cdot (1/2) + (1/4) \cdot 0} = \frac{3}{4}$$

• Is it a coincidence that I got 3/4? – James Nov 14 '17 at 3:13
• @stressed-out I think Rob's approach is somehow similar to mine since he has a zero times 1/4 in the end, where I just use 1/3. I'm not sure though.. – James Nov 14 '17 at 3:15
• Isn't the probability that both are girls zero? – David Schwartz Nov 14 '17 at 9:41
• It does not matter how big a chance there was for a boy the first time. It is a given fact that the first one was a boy. The only information you get from that, is that there is a boy in the house. – fishinear Nov 14 '17 at 12:13
• @fishinear I'm uncertain if you are disagreeing with someone or not. But you get more than just the information that there is a boy in the house, if you also assume a random child opens the door and that children have an equal chance of being boys or girls. If you don't assume a random child opens the door and that children have an equal chance of being boys or girls, or something similar, this question is unanswerable. – Yakk Nov 14 '17 at 16:10

A boy opened the door the first time around, but we are told there are two children (and we are assuming the boy is one of the two). Assuming there is an equal chance the other child is a boy or girl, that means there is also an equal chance for the household to be a two-boy household or a boy-girl household. So, the sample space is BB or BG with equal likelihood. Hence, a boy opening the door at any time is

$$\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}$$

So ... @James ... you got the right answer ... but the wrong method. In fact, you made two mistakes. The first mistake was the same I originally made (see link to my original answer below) that with a boy opening the door there would be a $\frac{1}{3}$ chance of there being two boys and a $\frac{2}{3}$ chance of there being one boy. No, that is really just $\frac{1}{2}$ and $\frac{1}{2}$. The second mistake is that you used that new sample space, that is based on the fact that a boy opened the door the first time, to calculate the probability of a boy opening the door the first time! No, if you calculate it this way, you should make no assumptions at all and thus you have a sample space of BB, BG, GB, and GG. .. this is what Robert did in his answer.

• What is your sample space? Why are you dividing by $3$ and not $4$? The way you have constructed your sample space, it seems to me that your events aren't equally likely. There are $4$ equally likely cases, not $3$. – stressed out Nov 14 '17 at 3:05
• @stressed-out We were given that a boy opened the door .... and I am assuming that is supposed to be one of the two kids at the house. So, the sample space becomes BB, BG, and GB ... that's why I am dividing by $3$, not $4$ – Bram28 Nov 14 '17 at 3:06
• Thanks. I get it now. I totally missed that information. :) (+1) – stressed out Nov 14 '17 at 3:10
• There are actually even odds between 2 boys and 1 boy 1 girl. If we were given as fact 'at least one boy is at the party' then it would be 1/3 to 2/3, but we are given 'a boy opens the door the first time', meaning that the child that didn't open the door the first time has even odds either a boy or a girl. This is similar to the formulation "you know that the older child is a boy, what are the odds the younger child is a boy" Also, note that in boy-girl houses 1/2 the time the boy won't open the door the first time. – Rolf Hoyer Nov 14 '17 at 3:14
• @James Well, at this point it should be pretty clear that that is the right answer, and of course you can get to the right answer through different approaches ... so it is not that surprising ... but as I just explained in mu updated answer, your approach really shouldn't work ... so I think that you got 3/4 was coincidence, sorry to say. Well, enough probability for today! Thanks for the question!! (underscoring the fact that reasoning with probabilities is hard!) :) – Bram28 Nov 14 '17 at 4:03

This is a classic underspecified problem.

You are invited to a family party. A Boy opens the door for you. There are two children there. What is the probability that a boy opens door for you next time?

We can make many sets of assumptions about facts not mentioned, and get different responses.

As an example, what if we assume we live on an island where there are no girls. Then the probability next time it is opened by a boy is clearly 100%.

What if we assume children always alternate turns opening doors, and each child has a 50% chance of being a boy or a girl. Then the probabilty is 50%.

If we assume a random child opens the door, that children are randomly boys or girls, then the probability is 75%.

If we assume that, if there are two boys, neither opens the door and instead an adult does, and if there is a boy and a girl they take turns, then the probability is 0%.

So unspecified criteria about how the door is chosen to be opened and the distribution of children and how they interact permits the probability to be anywhere from 0% to 100%.

This is the same issue that the classic Monte Hall problem has: there are reasonable assumptions that can set the answer anywhere. People pick some set they consider reasonable and draw a conclusion with them.

Now, on a test, you'd just answer 75% and state your assumptions directly, then draw conclusions from those assumptions. But that is a test taking strategy, not a mathematical truth.

Assume each child is randomly of either gender.

Assume a random child always opens the door.

BB   BG   GB   GG
BB   100%  0%   0%   0%
BG    25%  25%  25%  25%
GB    25%  25%  25%  25%
GG    0%   0%   0%  100%

on the left is the gender of the two children. Each of the rows has equal probability. We could simply divide every probability by 4, but a later step makes that not required as we are going to normalize it anyhow.

On the top is who opens the door 1st and second time. In the middle is the chance that this happens.

We now eliminate every case where a boy doesn't open it the first time:

BB   BG
BB   100%  0%
BG    25%  25%
GB    25%  25%

Now add up percentages:

BB   BG
150%  50%

and normalize:

BB   BG
75%  25%

75% chance that a boy opens the door the 2nd time.

If you prefer more math, we can use Bayes's theorem. Here are the initial probabilities (same chart as above, except divided by 4):

BB   BG   GB   GG
BB    1/4   0    0    0
BG    1/16 1/16 1/16 1/16
GB    1/16 1/16 1/16 1/16
GG    0    0    0    1/4

now we want the probability of boy boy given boy first, or:

P(BB|BX) = P(BX|BB) * P(BB) / P(BX)

P(BB|BX) = 100% * (3/8) / (1/2)
= 75%

Alternatively, eliminate the cases where a boy doesn't open the door first:

BB   BG
BB    1/4   0
BG    1/16 1/16
GB    1/16 1/16
GG    0    0
---------------
3/8  1/8  = 1/2

Giving is P(XB|BX) = (3/8)/(1/2) = 3/4

• This is mathematics. – jwg Nov 14 '17 at 20:45
• @jwg This is comment field. – Yakk Nov 14 '17 at 21:11
• You don't need to laboriously exclude tendentious possibilities of 'facts not mentioned'. Mathematics is the science which consists of solving problems under simplifying assumptions. – jwg Nov 15 '17 at 11:27
• @jwg Mathematics is the art of spreading truth via abstraction. Solving problems by assuming facts not supported is called making mistakes. You can say "if X Y and Z, then conclusion" and generate something of use; simply blindly assuming X Y and Z then asserting conclusion is an error. By looking at the range of answers as a function of which assumptions you can determine what about the question itself is causing the apparent paradox. – Yakk Nov 15 '17 at 12:15
• @Yakk: Granted. But there is a social convention aspect to these kinds of assumptions, and some are generally accepted, and others are not. It would generally be accepted that children are born boy or girl with equal likelihood, so the scenario where they're on an island with no girls would be considered unnecessary caution. But the idea that the child who answers the door is randomly, uniformly, and independently selected each time the bell rings is generally (in my experience) not considered a safe assumption. I don't know that there is a rational basis for that, of course. – Brian Tung Nov 15 '17 at 21:09

I would think about it differently. Next time, the chance that second kid (no matter it's sex) opens door is $P(second) = \frac{1}{2}$.

Chance of second kid being girl is $P(girl) = \frac{1}{2}$.

Therefore, you have chance $P(second) * P(girl) = \frac{1}{2} * \frac{1}{2} = \frac{1}{4}$ girl opens the door. Therefore $P = 1 - P(second) * P(girl) = \frac{3}{4}$

While everyone arrives at the same answer, I have a problem with some of the reasoning:

"There's definitely one boy, so half of the possibilities have a probability of one:" I think that there are three possibilities, with probabilities of 1, 1/2 and 1/2.

"there is also an equal chance for the household to be a two-boy household or a boy-girl household. So, the sample space is BB or BG with equal likelihood." I think it is twice as likely to be mixed-sex as not. Here is my solution:

The possibilities for the children are BB, BG, and GB, since they are distinguishable by age. By Bayes theorem, given that the door was opened by a boy the probabilities for these combinations are 1, 1/2 and 1/2, each divided by their sum, or 1/2, 1/4 and 1/4. Thus, there is a probability of the door being opened by a boy the second time of 1 in the first case and 1/2 in each of the other two cases. Multiplying these probabilities by the probabilities of the cases and adding gives 1/2 + 1/8 + 1/8 = 3/4.

We assume the children take turns opening the door or that they respond at random.

• The argument is: given that a boy did answer this time, for one to answer next time it needs be either the same child (definitely a boy) or the other child and that turn out to be a boy. So the probability is $\tfrac 12\cdotp\tfrac 11+\tfrac 12\cdotp\tfrac 12$. That is an elegant solution; still however, the Bayes' approach will also work. – Graham Kemp Aug 28 at 1:01

There are either 1 or 2 boys in the family; call that count $$X$$.

You want the probability that a boy opens the door next time, given one did this time. Call those events $$B_N, B_T$$ respectively. Bayes' Rule gives us that:

\begin{align}\mathsf P(B_N\mid B_T)&=\dfrac{\mathsf P(X{=}1, B_T, B_N)+\mathsf P(X{=}2, B_T, B_N)}{\mathsf P(X{=}1, B_T)+\mathsf P(X{=}2, B_T)}\\[2ex]&=\dfrac{\tfrac 24\cdot\tfrac 12\cdot\tfrac 12+\tfrac 14\cdot\tfrac 11\cdot\tfrac 11}{\tfrac 24\cdot\tfrac 12+\tfrac 14\cdot\tfrac 11}\\[2ex]&=\dfrac 34\end{align}

Alternatively, as has been proposed. A boy will open the door next time (given one did this time) if it is the same boy, or its the other child which turns out to be a boy$$\mathsf P(B_N\mid B_T)=\tfrac 12+\tfrac 12\cdot\tfrac 12=\tfrac 34$$

So both approaches give the answer of $$~\dfrac 34~$$.

To test that we are not just accidentally getting the same answer, consider if it was that there were three children in the family.   The first approach gives us $$\tfrac{\tfrac 38\tfrac 19+\tfrac 38\tfrac 49+\tfrac 18\tfrac 99}{\tfrac 38\tfrac 13+\tfrac 38\tfrac 23+\tfrac 18\tfrac 33}=\tfrac 23$$, while the second approach gives us $$\tfrac 13+\tfrac 23\cdot\tfrac12=\tfrac 23$$.