In the game of RPS, a tie happens when every player chooses the same symbol (i.e all rock) or there is at least 1 rock, 1 paper, and 1 scissors chosen. A tie results in the reply of the round. What is the probability of a tie happening when 2, 3, 4...n people play the game? Is there any general formula to be used? Assume that the probability of each symbol is 1/3.

What I have thought so far is with 4 people. By assuming that P1 plays rock, there are 3 players left yet to choose their symbol. My line of thinking is to find the probabilities of having at least 1 paper and at least 1 scissors using the binomial theorem and then multiplying together to find their 'and' probability. However, I suppose that my method uses 2 different sets of data, 1 for the paper and 1 for scissors.

This has boggled my brain.

Thanks very much


1 Answer 1


Hints, for general $n$:

  1. Find the probability that nobody plays rock.
  2. Find the probability that nobody plays scissors.
  3. Find the probability that nobody plays paper. (I know 2 and 3 are trivial once you can do 1.)
  4. Find the probability that all the players play the same symbol.
  5. Justify the assertion that the probability of not-a-tie is the sum of the probabilities from 1, 2, and 3, minus the probability from 4.

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