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Topic: Amusing very tough problem on probability.
Replies: 9   Last Post: Apr 8, 2016 8:04 PM

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Tommy Jensen

Posts: 249
From: Daegu, Korea
Registered: 12/6/09
Re: Amusing very tough problem on probability.
Posted: Apr 5, 2016 4:46 AM
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On Mon, 04 Apr 2016 23:47:47 -0500, quasi wrote:

> quasi wrote:

>>So, I'm back to my original claim. The 50% cutoff strategy the _only_
>>rational optimal strategy. Anything else is just superstition.

> I came to my senses.
> To check it, I computed 50% cutoff vs 51% cutoff.
> As you intuited, 51% wins against 50% (by a slight margin).
> Based on how that worked out, I solved the original problem.
> The optimal cutoff, expressed as a fraction between 0 and 1, is exactly
> (sqrt(5) - 1)/2, or approximately 61.8%. If the initial percent is below
> the cutoff, change; if above, stay; at the exact cutoff, it doesn't
> matter.

The superstition bit is funny though.

The cutoff number sounds right. Is there an easy proof?

How about the version in which each player is given
several rounds of opportunity to change their current holding?
It seems natural to expect that the cutoff should have to
decrease with every new round, and for each fixed round, the
cutoff for that round should increase as more rounds are
allowed in the game; since you would expect that to win the
game when there are many rounds, you need a larger percent
number than when the game has fewer rounds.

In the 3-player version, you also have to have a larger
cutoff, since you expect that the winner needs to finish with
a better holding. Do you see the answer for the n-player game?

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