Question

The original question is:

Describe an algorithm to output a die roll(a random number from 1 to 6), given a function that outputs a coin toss(a random number from 1 to 2). Each possible outcome should be equally likely.

The most popular answer to this question is:

Flip the coin three times, and use the three coin flips as the bits of a three-bit number. If the number is in the range 1 to 6, output the number. Otherwise, repeat.

My question is:

Most of the discussions on Stack Overflow come in the above flavour. I’ve also searched the Internet, finding that there exist many answers of other flavours, which they don’t explicitly dig into. Could someone please share one or two different ideas on this problem?

Answer 1

If you just want other options, not necessarily good ones then how about this:

1. Flip a coin for each possible output value.
2. If there is one or more heads then discard all the possible values that got tails.
3. If you have only one value left then stop. Else goto 1.

I would suspect it is going to have a higher expected number of coin tosses than the method you’ve described and have no advantages at all really.

In general I assume this is why there is not much on the other possible ways of using random numbers. They are just not as good.