```Date: Feb 1, 1999 8:41 AM
Author: John Conway
Subject: Re: coins problem

On Mon, 1 Feb 1999, Helena Verrill wrote:> By the way, you said that an information count would make you> think you could do (3^n-1)/2 --- I suppose you mean that there are 3> possibilities for weighing, so n weighing gives 3^n outcomes,> and k coins means 2k possibilitis, (anyone light or heavy),   I'd prefer to note the all-balanced case can't happen if there'sa fake coin, but does happen if there's no fake, which increasesthe number of possibilities to  2k+1.> so, max k has 2k=3^n, and since that k is not an integer, > you get (3^n-1)/2.... so how do you prove you can't do this many?    I found a very easy argument to prove this when I considered this problem as a student, but can't remember it at this moment.> And why is it, that if you just have the extra information> that you know the odd coin is light, then the 'information> count' does work, and you can do 3^n in n weightings?> what's the difference?   Since 3^n is what you'd expect, it requires no explanation,unlike (3^n - 3)/2, for which I'll try to reconstruct theargument.    John Conway
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