# Born as the seventh month dies …

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`.

• 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 17 and boy is 12,

(2/​7) (1/​4) /​ (2/​14) = 12 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 27; 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 13), 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 13, 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 13 chance of be­ing ex­e­cuted. Now that it’s ei­ther me or Char­lie, I have a 12 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 13, but Char­lie now has 23 chance to die. Be­cause:

P(Bob | Alice) = 12

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 13. 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 13.

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 12 for two boys). With low n, the speci­fic­ity is low and you’re close to the prob­lem with­out se­lect­ing a spe­cific child and get 13. 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 12. 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.