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Let's say that I wanted to generate 4 random numbers using a coin toss. I could toss the (unbiased) coin 4 times to generate one of 16 possible numbers (e.g. TTHH=0011=3) and just ignore any results higher than 9. Would this method be entirely impartial? or is it more probably to get a certain number?

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If you completely start over afresh with four new coin tosses if you get a number $>9$, then this method is impartial (and will eventually succeed almost surely, the expected number of 4-rounds is less than $2$). Each o fthe $10$ used outcomes is equally likely because each of the $16$ outcomes in total is equally likely.


If you did not throw away all previous coin tosses, then you would introduce some bias. Say, you just remove the first coin in order to hopefully get a digit with just one additional tossing. Then in such a second round, you could never produce $0,1,2,3$ any more, because the lowest sequence would be $THTT=4$ from a rejected $HTHT$. (Then again, you could at least reuse the last coin and save one toss per repeat; deeper tricks are possible)

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This works fine. No different than if you needed numbers from 1-5 and used a six sided fair die. You would discard the 6 result.

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