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JohnM2433
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Milwaukee, Wisconsin
Jul 2002
time: 21:23
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If there are two amoebae to start with, it will take 59 minutes to fill the jar, as the situation at the start is the same as it was after one minute in the first instance (59 minutes remaining).
It was fairly obvious to me how that "trick" works. I checked just to make sure. With some sleight of hand, you could probably do the same thing with real cards.
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JohnM2433
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Milwaukee, Wisconsin
Jul 2002
time: 21:23
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OK, this same basic problem, only in another form, was covered in another thread. But I can't think of anything better, so...
You are on a game show, and you are shown three doors. The grand prize is behind one of the doors. You are asked to select a door. Not knowing which door the prize is behind, you pick one at random. Then the host points to one of the remaining doors and tells you that the prize is not behind that door. Now he asks you to pick one of the doors again, and this time you really will get the prize if it is behind the door, and will lose it if it is not.
Assuming that the host is telling the truth, and that the same thing would have happened no matter which door you picked, which of the remaining doors is more likely to have the grand prize behind it: the one you selected or the other door that the host didn't point out?
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JohnM2433
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Milwaukee, Wisconsin
Jul 2002
time: 21:23
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CyberGnu gave the correct answer to my problem. Sagacious Dolphin's observation that the same principle holds for any number of doors (more than two) is also correct; the odds of the original door containing the prize will not change, and indeed cannot, since you must get the same basic information about that door (i.e., none) each time. Or to put it another way, the probability of the door you picked being the correct one when you pick it is the average of the probabilities of it being correct for each other door the host might point out, since he is equally likely to select each one. But since the cases are all symetrical, the probability for the chosen door must be the same in each case, and thus the same as it was originally. Only the odds for the doors you did not select change.
There was another thread in which someone presented the following senario: Of three prisoners, one will die and the other two will go free, but none of them knows which is which. One prisoner asks a guard to give a letter to one of the other prisoners who will go free. The guard does so, and tells him which one. How is it that the prisoner's odds of dying have now increased from 1/3 to 1/2, since they would have been the same if the other of the other two prisoners had gotten the letter?
Well, his odds of dying actually stay the same, and the odds of the other prisoner who didn't get the letter dying go up to 2/3. I posted as much. The comparison to the game show problem is obvious. But now I can't seem to locate that thread. I can't remember its name or what forum it was in, so I would appreciate if anyone could tell me which it was, as I wanted to check in on it.
I'm still trying to get the answer to CyberGnu's latest riddle...
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01-11-2002 03:09
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