Born as the seventh month dies ...

by Rudi C1 min read10th Jul 20208 comments

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Probability & StatisticsBayes' TheoremRationalityWorld Modeling
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Epistemic status: Mathematical reasoning by an amateur, but I feel confident that it’s mostly correct.

Update: There is a Wikipedia page on this with lots of details and other similar problems with different answers. This paper referenced there seems a good summary.

The problem

I was reading The Equation of Knowledge, and it starts with this little cute problem:

Suppose a dad has two kids. At least one of them is a boy born on Tuesday. What's the odds of his sibling being a boy?

Generalizing the problem:

P(2 boys | 1Bn := At least one boy with an independent characteristic (named N here) that has the probability 1/n)

The original problem can now be seen as an instance of P(2 Boys | 1B7).

Simple Bayesian solution

I started solving this with a simple application of Bayes:

P(1Bn | 2 boys) = P(First child being a boy having N | 2 boys) + P(Second child being a boy having N | 2 boys) - P(Both children being boys having N | 2 boys) = 1/n + 1/n - 1/(n^2)

P(1Bn) = P(First child being a boy having N) + P(Second child being a boy having N) - P(Both children being boys having N) = (1/2)(1/n) + (1/2)(1/n) - ((1/2)(1/n))^2 = 1/n + 1/4(n^2)

BayesFactor(2 boys | 1Bn) = P(1Bn | 2 boys)/P(1Bn) = (8n - 4)/(4n - 1)

P(2 boys | 1Bn) = BayesFactor(2 boys | 1Bn) * P(2 boys)=((8n - 4)/(4n - 1)) * (1/4) = (2n - 1)/(4n - 1)

We have:

P(2 boys | 1B1 == At least one boy) = 1/3

P(2 boys | 1B7 == At least one boy born on Sunday) = 13/27

lim{n -> +Inf}[P(2 boys | 1Bn == At least one boy born exactly x seconds after the big bang)] = 2n/4n = 1/2

So ... I am somewhat confused. It's intuitively obvious that having two boys creates more opportunity for specific independent phenomena to happen. But, at first blush, my intuition was firmly suggesting that I throwaway the additional information as useless, and only careful thinking lead me to the (hopefully) correct answer. I also can't quite think of any practical examples for this epistemic error. Your thoughts appreciated.

Generalizing more

Repeating the same analysis, but generalizing the probability of "being a boy" to 1/k,

BayesFactor(2 boys | P(boy)=1/k, 1Bn) = (2n(k^2) - (k^2))/(2nk - 1) 

lim{n -> +Inf}[P(2 Boys | P(boy)=1/k, 1Bn)] = 1/k

Generalizing to random variables

Suppose we have two independent, identically distributed variables X1 and X2, and another two i.i.d variables Z1 and Z2. All of these variables are mutually independent. Repeating the exact same calculations, we'll have:

Px := P(X1=x) = P(X2=x)
Pz := P(Z1=z) = P(Z2=z)

BayesFactor(X1=X2=x | (X1=x, Z1=z) or (X2=x, Z2=z) ) = ... = (2Pz - Pz^2)/(2PxPz - (PxPz)^2)
P(X1=X2=x | (X1=x, Z1=z) or (X2=x, Z2=z) ) = BayesFactor(X1=X2=x | (X1=x, Z1=z) or (X2=x, Z2=z) ) * P(X1=X2=x) = ... = (2Px - PxPz)/(2 - PxPz)

lim{Pz -> 0+}[P(X1=X2=x | (X1=x, Z1=z) or (X2=x, Z2=z) )] = 2Px/2 = Px

If we set Pz = 1 (basically nuking the Z variables), we'll have:

P(X1=X2=x | (X1=x, Z1=z) or (X2=x, Z2=z) ) = P(X1=X2=x | X1=x or X2=x) =  Px/(2 - Px)

So the independent information provided by the Z variables can, maximally, improve the odds by a ratio of 2 - Px >= 1.

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8 comments, sorted by Highlighting new comments since Today at 2:48 PM
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I don't think you have the dependencies quite right, because you can actually use more of the information than you do above to restrict the population from which you draw.

The real underlying population you should draw on seems to to be the population of fathers with exactly two children, of which one might be a boy born on Tuesday.

p(a two boy family given one brother was born on Tuesday) = (p(one brother born on Tuesday in a in two-boy family)) (p(two boys in 2 person families)) / p(out of all two person families, having one be a boy born on Tuesday)

which is if we say Tuesday birth is 1/7 and boy is 1/2,

(2/7) (1/4) / (2/14) = 1/2 so the Tuesday datum drops out.

You're not stating what probability rules (theorems/axioms) you are using (you're probably going by intuition), and you have made mistakes. p(one brother born on Tuesday in a in two-boy family) is not 2/7; It's 1/7 + 1/7 - (1/7)(1/7) because you're counting the two children both being born on Tuesday twice. The same mistake has been made in calculating p(out of all two person families, having one be a boy born on Tuesday); The correct answer is (1/2)(1/7) + (1/2)(1/7) - (1/2)(1/7)(1/2)(1/7).

The rule you're not following is:

P(A or B) = P(A) + P(B) - P(A and B)

When these mistakes are corrected, the correct answer comes out:

((1/7 + 1/7 - (1/7)*(1/7))*1/4)/((1/2)*(1/7) + (1/2)*(1/7) - (1/2)*(1/7)*(1/2)*(1/7)) = 13/27 =~ 0.4814

This is correct, if you are excluding the case where both are boys both born on Tuesday. Otherwise you would not subtract p(A and B). But, you did not say only one, you said _at least_ one.

It’s not about excluding that case. It’s about not counting it twice. Search for the inclusion-exclusion principle to see the reasoning behind it. 

I don’t think you’re modeling your problem correctly, unless I misunderstood the question you’re trying to answer. You have those following random variables :


X_1 is bernoulli, first child is a boy


X_2 is bernouilli, second child is a boy


Y_1 is uniform, weekday of birth of the first child


Y_2 is uniform, weekday of birth of the second child


D is a random variable which corresponds to the weekday in the sentence "one of them is a boy, born a (D)". There is many ways to construct one like this, but we only require that if X_1=1 or X_2=1, then D=Y_1 or D=Y_2, and that D=Y_i implies X_i=1.


Then what you're looking for is not P(X_1=1,X_2=1 | (X_1=1,Y_1=monday) or (X_2=1,Y_2=monday)) (which, indeed, is not 1/3), but P(X_1=1,X_2=1 | ((X_1=1,D_1=D) or (X_2=1,D_2=D)) and D=monday). This is still 1/3, as illustrated by this Python snippet (I’m too lazy to properly demonstrate this formally) : https://gist.github.com/sloonz/faf3565c3ddf059960807ac0e2223200


There wass a similar paradox presented on old lesswrong. If someone can manage to find it (a quick google search returned nothing, but i may have misremembered the exact terms of the problem…), the solution would be way better presented there :


Alice, Bob and Charlie are accused of treason. To make an example, one of them, chosen randomly, will be executed tomorrow. Alice ask for a guard, and give him a letter with those instructions : "At least Bob or Charlie will not be executed. Please give him this letter. If I am to be executed and both live, give the letter to any one of them". The guard leaves, returns and tell Alice : "I gave the letter to Bob".

Alice is unable to sleep the following night : "Before doing this, I had a 1/3 chance of being executed. Now that it’s either me or Charlie, I have a 1/2 chance of being executed. I shouldn’t have written that letter".

I don’t think your analysis is right; For one, if you know that the random variable D is Monday, your problem reduces to mine; In the simulation, you have set D incorrectly ( D=Y1in the case both are boys. This makes no sense. You are ignoring the case where the first boy is not born on Monday, but the second one is.), and it is not a simulation of the probability you have written. 

The second example; I’m not sure what your conclusion on it is. It seems like the Monty Hall problem to me, i.e., Alice is still having a chance of 1/3, but Charlie now has 2/3 chance to die. Because:

P(Bob | Alice) = 1/2

P(Bob | Charlie) = 1

if you know that the random variable D is Monday

Yes, that’s kind of my point. There’s two wildly different problems that looks the same on the surface, but they are not. One gives the answer of your post, the other is 1/3. I suspect that your initial confusion is your brain trying to interpret the first problem as an instance of the second. My brain sure did, initially.

On the first one, you go and interview 1000 fathers having two children. You ask them the question "Do you have at least one boy born on a Monday ?". If they answer yes, you then ask then "Do you have two boys ?". You ask the probability that the second answer is yes, conditioning on the event that the first one is yes. The answer is the one of your post.

On the second one, you send one survey to 1000 fathers having two children. It reads something like that. "1. Do you have at least one boy ? 2. Give the weekday of birth of the boy. If you have two, pick any one. 3. Do you have two boys ?". Now the question is, conditioning on the event that the first answer is yes, and on the random variable given by the second answer, what is the probability that the third answer is yes ? The answer is 1/3.

My main point is that none of the answers are counter-intuitive. In the first problem, your conditioning on Monday is like always selecting a specific child, like always picking the youngest one (in the sentence "I have two children, and the youngest one is a boy", which gives then a probability of 1/2 for two boys). With low n, the specificity is low and you're close to the problem without selecting a specific child and get 1/3. With large n, the specificity is high and you’re close to the problem of selecting a specific child (eg the youngest one) and get 1/2. In the second problem, the "born on the monday" piece of information is indeed irrelevant and get factored out.

You’re correct, but I still don’t find the scenario I am describing intuitive by my system 1. If I think about it (especially now that I have analyzed the problem rigorously), yes, I’ll feel that the probability should increase (though of course, with nowhere near the precision of the Bayes rule), but if you just asked me this problem yesterday, and I wasn’t watching for trap questions, I’d give you a wrong answer.

I just found Wikipedia has a whole page on this, named the girl and boy paradox. They cover lots of details there.