Cool math problem for those that like such things
So, I was out to dinner with sunshine__girl, browascension, and lightling. We were going to play a game and we decided to play paper-scissors-rock to see who would go first. Instead of pairing off, we decided to keep playing until it was unambiguous (i.e. if two people did scissors, and two did rocks, the two scissors would be eliminated, but if there were one of each we'd all go
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lets say 2 people do paper, one person does scissors and one person does rock (out of 4 people). Nobody has to leave the game, because there was no clear winner. So if everyone chooses the same, or all of paper, scissors, and rock are present, the game must continue.
With 2 players, there is a 2/3 chance of the game ending and a 1/3 chance of the game continuing. So you get:
F(2) = 1 + 1/3 * F(2)
which is
1+1/3+1/9+1/27+... = 1 / (1-(1/3)) = 1.5 rounds
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Next, if 0 < i < n, then p_n(i) = 3 * [n choose i] / 3^n,
since, to get from n to i, you need i people to choose (rock / paper / scissors) and n-i people to choose (scissors / rock / paper), which can happen 3 * [n choose i] ways. Since there are 3^n possible "throws" that does it.
And, otherwise, you stick at n, which (just by summing) gives
p_n(n) = [3^n - 3(2^n - 2)] / 3^n.
Alternatively, to leave n, you need _exactly_ 2 of the 3 items to appear, which can happen
3 * (2^n - 2) ways
(since, for instance, there are 2^n - 2 ways people can throw only rock and paper, but at least one of each)
which means that there are
3^n - 3* (2^n -2) ways to stay at n.
That's an ugly expression and makes me doubt that you'll find a good closed-form solution for the original problem.
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