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Assume that you have a fair coin and you want to choose from 3 options fairly. The way that I found is you always flip it twice and read it off as binary. If you got 0, then you retry. If you got 1, then you choose option 1. If you got 2, then you choose option 2. If you got 3, then you choose option 3. This approach gives an average of $\frac{8}{3}$ flips. Is there any way that is better on average?

mathlander
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  • This feels oddly close to the geometric distribution... – Sean Roberson Oct 22 '22 at 02:39
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    Your method is valid. Asking about "better on average" leaves open the possibility that you might have in mind choosing from 3 options more than once. If you are only choosing once I don't think your approach can be improved upon. – hardmath Oct 22 '22 at 02:45
  • As an aside, it is well known from results of elementary number theory that you can not perfectly simulate a 1/3 probability with a fair coin using a fixed finite number of tosses. – JMoravitz Oct 22 '22 at 03:35
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    @mathlander In the target of the duplicate link, I've added an answer proving that $\frac83$ is best possible. – Misha Lavrov Oct 22 '22 at 03:44

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