[EM] Flipping a coin

[MIKE OSSIPOFF]

  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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