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#11
10-07-2003, 04:11 PM
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Re: interesting problem
yep
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#12
10-07-2003, 04:12 PM
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Zero,
of course. This is not a math problem; it's a logic problem. As each player tries to anticipate what the other 19 players will choose, they will constantly adjust their selection downward until reaching the lowest possible number. In this case that is zero.
BR
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#13
10-07-2003, 04:17 PM
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Re: Zero,
not with the stipulation that numbers are chosen without replacement.
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#14
10-07-2003, 04:42 PM
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replacement/no replacement
English is not my first language, I think I understood the word replacement wrong. However I meant that everyone picks their number without knowing what the others pick and so people may pick same numbers.
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#15
10-07-2003, 04:52 PM
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Re: Zero,
but if everyone thinks about what everyone else is thinking before they pick, like all those poker books tell us to, then everyone will pick zero. if they just guess then god knows what will happen. this solution assumes thinking, competent opponents.
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#16
10-07-2003, 05:11 PM
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The answer is 16
Here's what will happen.
You will choose the right answer, 16. 4 dummies will pick random numbers, so they will avg 50. 5 people will decide that the avg is 50 and 2/3 of that is 33, so they will pick 33. 5 people will think that everyone will pick 33 for the reason above, so they will pick 22. The remaining 5 people will think that everyone will adjust downward and thus all end up at 0. So they will pick 0. 1*16 + 4*50 + 5*33 + 5*22 + 5*0 = 491 491/20 = 24.55 (2/3)*24.55 = 16.4 close enough Lots of people are dumb. And lots of people who are very smart in a lot of other ways are pretty bad at math and logic problems, thus will not come up with any rational answer here. The key to solving this problem is having a good grasp on exactly how dumb a given sample of people are.
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#17
10-07-2003, 05:17 PM
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Re: replacement/no replacement
"without replacement" means no two can pick the same number.
If the game were "with replacement", ie everyone can pick the same number, then the game is pointless until you get few enough players (n) that one player can influence 2/3 of the mean by an amount greater than n. For example, at the extreme of two players, one player can shift the mean to anywhere from 1 to 50 if the other player picks 0, and from 25 to 75 if he picks 50. There would be no guaranteed +EV strategy though, since you are always on a guess as to what the other player will pick, and zero is not automatic. Zero becomes automatic (and break even only) when n is greater than the critical point, ie when one player can no longer influence 2/3 of the mean by > 1/n.
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#19
10-07-2003, 06:34 PM
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Re: interesting problem
If there's no replacement, and two people pick the same number, do they just throw away one person's entry? There doesn't appear to be any arbitration scheme or way for the second person to submit a second guess.
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Linear Mode
