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After some time has passed (and the posting is some pages back) we might possibly discuss the brain teaser?

This type of problem is really interesting because it defies "intuition" so much.

It is related to the Roulette illusion: "OMG it was red for the last 5 times, it now just HAS to be black".
So the problem here could also be translated to this: If everyone played until she won, would the Casino make a loss? (not considering the Zero for now)
And we know the answer ;).
Always doubling the bet +1 of course works - theoretically, in reality you get caught by the next illusion: underestimating exponential growth.
We know the story with the chess board and the rice grains...
So the thing that actually messes with our minds is probably that the probability of the sequence of 6 reds is very low, still the draws are independent of each other.

The teaser here is also a bit of an illusion.
One might think that as there exist families with more than one boy, but no families with more than one girl, the boy population must be higher.
On the other hand if you consider a population of N families where N/2 families get a daughter and N/2 families get a boy - then the ratio is 50:50 and the N/2-boy-subpopulation continues to get children.
Now you have the same problem as before just with population N/2, after which the ratio is still 50:50.
Suddenly the problem looks completely different.

Thanks for posting this ;).

As for a simulation, here is one I clobbered together in Prolog (runs in SWI-Prolog):

family --> { random_float >= 0.5, ! }, [boy], family.
family --> [girl].


make_family(Boys, Girls) :-
    phrase(family, Family),
    length(Family, Kids),
    Boys is Kids - 1,
    Girls = 1.


simulation(N) :- simulation(N, 0, 0).


simulation(0, Boys, Girls) :- format('~w boys, ~w girls~n', [Boys, Girls]).
simulation(N, B, G) :-
    N > 0,
    make_family(Boys, Girls),
    N1 is N - 1,
    B1 is B + Boys,
    G1 is G + Girls,
    simulation(N1, B1, G1).

Example runs:

?- between(1, 10, _), simulation(1000), false.
1020 boys, 1000 girls
1003 boys, 1000 girls
981 boys, 1000 girls
953 boys, 1000 girls
1013 boys, 1000 girls
986 boys, 1000 girls
1004 boys, 1000 girls
958 boys, 1000 girls
1024 boys, 1000 girls
1144 boys, 1000 girls

Prolog, how nerdy

Considering simplified imperative python:

for generation in range(num_generations):  # assuming uniform distribution
  while random.randint(1,2) != 1:
    boys += 1
  girls += 1

it can also be seen that the rand-loop will run once on average
-> ( (0.5 * num_generations) * 0 + (0.25 * num_generations) * 1 + (0.125 * num_generations) * 2 + ... ) / num_generations ~= 1

Still, the first thought might again be: "but sometimes the loop runs very, very often but it never gets less than 0:shout:"

I like that, it's like an optical illusion without optics

Man kann das auch mathematisch beweisen. Man braucht dazu gar kein Programm zu schreiben.

Der Erwartungswert für das Verhältnis der Knaben zu den Mädchen entspricht folgender Reihe
1/2 * 0 + 1/2^2 * 1 + 1/2^3 * 2 + 1/2^4 * 3 + 1/2^5 * 4 + ...
= Sum((x - 1) / 2^x)
unter der Annahme, dass die Wahrscheinlichkeit für eine Geburt eines Knaben gleich der eines Mädchen ist (also 1/2).

Wie man sieht (http://www.wolframalpha.com/input/?i=Sum%28%28x+-+1%29+%2F+2^x%29+plot+from+1+to+100), konvergiert diese Reihe gegen 1.

Es ist also, auch wenn es zunächst nicht intuitiv erscheint, bei der angegebenen Strategie zu erwarten, dass das Verhältnis von Mädchen zu Knaben auf lange Sicht annähernd gleich 1 sein wird.