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Is it possible to 'split' coin flipping 3 ways?

I just was about to flip a coin when a third choice came along and I wondered if I could flip a 3 sided coin by flipping twice, assigning the four outcomes to 3 of the choices and designating the fourth as a do-over?

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marked as duplicate by Byron Schmuland, Henning Makholm, Austin Mohr, Pete L. Clark, Zev Chonoles Feb 10 '12 at 2:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Yes. The three acceptable outcomes are all equally likely. – Austin Mohr Feb 10 '12 at 1:05
3-Sided coin ... Not sure what that is? – NoChance Feb 10 '12 at 1:09
@Emmad: Probably British. – Henning Makholm Feb 10 '12 at 1:19

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Yes! It's a little more complicated in general, but you can choose between any events with probabilities $p_1, p_2, p_3, ..., p_n$ in the following way:

Partition the interval $[0,1]$ into disjoint subintervals of size $p_1, p_2, ..., p_n$. For your example, the subintervals would be $[0, 1/3], [1/3, 2/3]$, and $[2/3, 1]$.

Assign heads = $1$ and tails = $0$, and start flipping the coin to generate successive digits, after the decimal point (binary point?), of a binary number . Stop whenever you're sure which interval the number will land in.

For instance, in your case, if you get heads, tails, and then tails, your number will be $.100$. You can stop flipping the coin now, because every binary number that starts with $.100$ must be in the interval $[1/3, 2/3]$, and you can confidently choose the second of your three events without having to flip the coin infinitely many more times.

Interestingly enough, this means that with just a coin, you can pick an event with probability $1/\pi$, or any number you want!

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A three-sided coin would need to be like a pencil with a triangular cross-section, where the assumption is that the pencil will not balance on its tip but always come to a rest with one side down (this ius just like hte assumption that an ordinary coin will not come to rest standing straight up on edge. The face on the bottom is defined to be the outcome. Similarly, for a tetrahedral die, the outcome is the number marked on the bottom face. (For what it is worth, tetrahedral dice are not popular because they don't roll too well, and the rolling of the dice is important in randomizing the outcome: dice are not supposed to "stick" a landing the way a gymnast does.)

Turning to your question, toss a fair coin three times, and assign HHT, TTH to one choice, HTH, THT to the second choice, and HTT, THH to the third choice. If the outcome is HHH or TTT, flip three times again. The three choices will have equal probability $\frac{1}{3}$ of occurring, and on average, you will need to do $\frac{4}{3}$ triple coin tosses.

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