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Timeline for Nepero game (by Yacov Perelman)

Current License: CC BY-SA 3.0

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Jan 8, 2015 at 23:03 answer added Robert Israel timeline score: 4
Jan 8, 2015 at 21:33 comment added user65782 @ Steven Landsburg: this would seems to be the most plausible option, but couldn't find any counterexample. Also Jakov Perelman in his book states that for $n$ integer the theorem holds, but doesn't give proofs (it's not an academic book) or references, so it's not clear if he did really know for sure it was true or if his was just a guess.
Jan 8, 2015 at 21:10 comment added Steven Landsburg continued --- so you might expect to find rare counterexamples when $n/e$ is just slightly less than a half-integer.
Jan 8, 2015 at 21:09 comment added Steven Landsburg Suppose just for a moment that instead of taking $n$ to be an integer, you take $n=ke/2$, where $k$ is an odd integer. Then $n/e$ is a half-integer, so the two choices for $m$ are equally close, but the higher of the two choices always gives a (slightly) better payoff --- in fact the ratio of the two payoffs goes rapidly to $1$ from above. This suggests that when your $n/e$ is very close to a half-integer, there should be a slight preference for the larger of the two $m$'s, which is almost always outweighed by the strong preference for the closer one.....continued
Jan 8, 2015 at 20:05 history edited asdfghj
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Jan 8, 2015 at 19:06 comment added asdfghj Yes, indeed, even by not mentioning that, it follows from the geometric-arithmetic means inequality that the parts must be equals. But still have no idea of how to prove the theorem I'm asking for.
Jan 8, 2015 at 17:39 comment added asdfghj Yes, they are $m$ equal parts (but I think it could be proved this other fact: to maximize the product they have to be all equals). The single parts are not necessarily integers, for example for $n=10$ and $m=3$ you get $(10/3)^3$.
Jan 8, 2015 at 17:34 comment added Alex Degtyarev What exactly is the game? Are these supposed to be $m$ equal parts? Are they supposed to be integers? Then, you also have a divisibility condition.
Jan 8, 2015 at 17:32 review First posts
Jan 8, 2015 at 17:34
Jan 8, 2015 at 17:28 history asked asdfghj CC BY-SA 3.0