Two related puzzles, one not so hard, one hard
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Each prisioner agrees that if the two other prisioners have the same color of skullcap, that he (pardon my sexism) will switch his switch to indicate that he has the other color. Assuming a random distribution of colors, this will work three out of four times.
On first glance, it seems that any prisoner would gain no information on seeing the other prisoner's caps. Therefore, it's just a 50/50 shot, so the best strategy for the prisoners is to pick one ahead of time who will flip his switch (and the others will not).
However, when you enumerate the possibilities (BBB, BBW, BWB, etc.), you can see that following frankus's strategy will provide a 3/4 chance of survival.
This apparent paradox goes away once you recognize that, in fact, seeing the 2 prisoners caps does, in fact, provide information to the 3rd prisoner. That information is not the color of the cap, but whether another may have flipped his switch.
For example, if I am a prisoner, and I see two black caps, there are two scenarios:
- I'm wearing a white cap. If I don't flip the switch, we all die. If I do, we live.
- I'm wearing a black cap. If I flip the switch, we die, but if I do not, we die anyway (since other prisoners incorrectly flipped the switch).