Puzzle time - Visit from Xylofon
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A fleet of saucers from planet Xylofon has been sent to bring back the inhabitants of a certain randomly selected house, for exhibition in the Xylofon Xoo. The house happens to contain 5 men and 8 women, to be beamed up randomly one at a time.
Owing to the Xylofonians' strict sex separation policy, a single saucer cannot bring back earthlings of both sexes. Thus, it beams people up randomly until it gets a member of a second sex, at which point that one is beamed back down and the saucer takes off with whoever remains on board. Another saucer then starts beaming people up randomly, following the same rule, and so forth.
What is the probability that the last person beamed up is a woman?
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I think if you imagine a string of 13 people 8 of whom are identitied as women by the sex detection algorithm in the xylofon ship and 5 as men, then you can build a sequence of M or F elements by choosing from 13 then 12 then 11 randomly. You are interested in the probability that number 13 is F, but this sequence should be reversible. Nothing about the sequence would be informative that it was built from one end or the other. So the question becomes, what is that chance that the first person picked is F. Which is an easy question to answer. Nothing in fact is special about any position in the order. Before the selections begin, each position has an identical chance of being F. First, last, and everything in between.
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The complication of the ships taking only one of the sexes seems to be a red herring, which does not change the solution to the simpler question I boiled it down to in my answer.
@Horace said in Puzzle time - Visit from Xylofon:
The complication of the ships taking only one of the sexes seems to be a red herring, which does not change the solution to the simpler question I boiled it down to in my answer.
On second thought, since the sequence accepts runs and rerolls random selections where the previous selection was different, that complication is not a red herring. Taking the simple case of 2 and 1 F and M, a third of the time you’ll get F last (because M was first) and of the two thirds remaining, half the time you’ll get M last - one third. Of the one third remaining, it’s half and half between F and M. So total is half and half, which suppose generalizes to any starting condition.
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@Axtremus said in Puzzle time - Visit from Xylofon:
New answer:
click to showThere's one way to be right, but an infinite number of ways to be wrong.
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Very wordy. I like my answer better
SOLUTION: Let's try some smaller numbers and see what happens. Obviously if the house is all men or all women, the sex of the last person beamed up will be determined. If there are equal numbers of men and women, then by symmetry, the probability that the last person beamed up is a woman would be 1/2. So the simplest interesting case is, say, one man and two women.
In that case, if the man is beamed up first (probability: 1/3), the last person beamed up will be a woman. Suppose a woman is beamed up first; if she is followed by a man (who is then beamed back down), we are down to the symmetric case where the probability of ending with a woman is 1/2. Finally, if a second woman follows the first (probability 2/3 x 1/2 = 1/3), the man will be last to be beamed up. Putting the cases together, we get probability 1/2 that the last person beamed up is a woman. Is it possible that 1/2 is the answer no matter how many men and women are present, as long as there's at least one of each?
Looking more closely at the above analysis, it seems that the sex of the last person beamed up is determined by the next-to-last saucer — the one that reduces the house to one sex. To see why this is so, it is useful to imagine that the Xylofonian acquisition process operates the following way: Each time a flying saucer arrives, the current inhabitants of the house arrange themselves in a uniformly random permutation, from which they are beamed up left to right.
For example, if the inhabitants at one saucer's arrival consist of males Amit and Boris and females Carol, Dina and Esme, and they arrange themselves "Dina, Esme, Boris, Carol, Amit," then the saucer will beam up Dina, Esme, and Boris, then will beam Boris back down again, and take off with just the females Dina and Esme. The remaining folks, Boris, Amit, and Carol, will now re-permute themselves in anticipation of the next saucer's arrival.
We see that a saucer will be the next to last just when the permutation it encounters consists of all men followed by all women, or all women followed by all men. But no matter how many of each sex are in the house at this point, these two events are equally likely! Why? Because if we simply reverse the order of a such a permutation, we go from all-men-then-all-women to all-women-then-all-men, and vice versa.
There's just one more observation to make: If both men and women are present initially, then one saucer will never do, thus there always will be a next-to-last saucer. When that comes — even though we do not know in advance which saucer it will be — it is equally likely to depart with the rest of the men, or the rest of the women.
