In a recent reply I spoke of there being approximate ways to choose between 3 alternatives by flipping a coin. Of course it needn't be approximate. Obviously one could choose 3 of the 4 possible outcomes of 2 coin-flips, one outcome representing each alternative. If the other outcome occurs, flip again. So one repeats the 2-coin-flip procedure till it gives one of the 3 outcomes that one has chosen.
I should have known that, because I've used coin-flipping to choose from several alternatives. Say there are N alternatives. If N is a power of 2, if N is 2^p, then flip the coin p times, with each of the possiblel 2^p outcomes representing one of the alternatives.
If, as is generally the case, N isn't a power of 2, then pick the smallest power, p, of 2 that is larger than N. Flip the coin p times, to make a p-digit binary number from 0 to 2^p - 1. Let the binary numbers from 1 to N represesnt the N alternatives, and repeat the procedure till you get one of those numbers. In other words, if you get 0 or a number greater than N, disregard it and repeat the procedure till you get a number from 1 to N.
For example, with 3 alternatives, flip a coin twice to get a binary number from 0 to 3, having numbered the alternatives from 1 to 3. Repeat till you get a number from 1 to 3, disregarding and repeating if you throw a 0.
Anyway, since I used to use that, of course I should have mentioned it. It's obvious, but I mention it anyway.
If there are a lot of alternatives it might be easier to draw a number or a few decimal digits from a paper bag.
Mike Ossipoff
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