Today’s puzzle
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wrote on 23 May 2020, 13:23 last edited by
When two pixo-bacteria mate, a new bacterium results; if the parents are of different sexes the child is female, otherwise it is male. When food is scarce, matings are random and the parents die when the child is born.
It follows that, if food remains scarce, a colony of pixo-bacteria will eventually reduce to a single bacterium. If the colony originally had 10 males and 15 females, what is the probability that the ultimate pixo-bacterium will be female?
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wrote on 23 May 2020, 14:12 last edited by
OK, that was easier than I thought it would be.
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wrote on 23 May 2020, 14:22 last edited by
You can only subtract females two at a time so you'll inevitably end up with one, at which time either they are the last one left or there are one or more males left with them. With one or more males, the mating iterations reduce male count by 1 every time, leaving the female as the last one.
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wrote on 23 May 2020, 14:26 last edited by
Yep. It's a little disappointing that the solution is so "discrete".
It would be more interesting to reason about the distribution if the parents wouldn't die.