# Probability that you win in rock scissors paper with 3 players?

Let's say person A, person B and person C are playing rock paper scissors.

Clearly $$n(S)=3\times 3\times 3=27$$. We want to find the probability that person A wins.

$$n(\text{only A wins})=3$$ (A wins with rock/scissors/paper).

$$n(\text{A and one other person wins})=2\times 3=6$$ ($$2$$ ways to choose the other winner, $$3$$ ways to win (rock/scissors/paper))

$$\therefore P(\text{A wins})=\frac{3+2\times3}{27}=\frac{1}{3}$$, but this is flawed but this because it doesn't account for the probability where you draw. (i.e. if the probability of each person winning is $$\frac{1}{3}$$, then the total probability is already $$1$$, when we haven't accounted for the probability of drawing yet.)

What is the correct probability that you will win in a game of rock scissors paper with 3 players? Thanks.

• You count the case where A and B win, and also the case where B and A win, though they're the same case. Instead of $12$ cases where two people win, there are only $9$ ($3$ ways to choose the two winners, and $3$ ways to win.) – saulspatz Jan 25 at 6:24
• Doubt - Lets say the three players put rock, paper and scissors respectively. Would you consider that as a win for all 3 participants, or just a draw? – Smriti Sivakumar Jan 25 at 7:20
• @saulspatz Yes, you are correct.. But what is the probability that one person wins? – Freddie Jan 25 at 8:48
• @SmritiSivakumar That would be a draw. Draw is when all three people put rock/scissor/paper, or there is one of each. – Freddie Jan 25 at 8:49
• What if two players choose "sciccors" and one "paper" ? Do then two players win ? – Peter Jan 25 at 11:29

There is nothing wrong with your calculation; each player wins with probability $$1/3$$. The point is that these are not disjoint events, so $$P(A\text{ wins})+P(B\text{ wins})+P(C\text{ wins})\neq P(A\text{ or }B\text{ or }C\text{ wins}).$$ In fact, the sum of the probabilities is the expected number of players who win. (When there are no multiple winners, so the events are disjoint, this number is always $$0$$ or $$1$$ and so its expectation is the probability someone will win, but this is not true in general.)

Here $$P(\text{no-one wins})=3\times\frac{1}{27}+\frac{3!}{27}=\frac13$$, $$P(2\text { people win})=3\times 3\times\frac1{27}=\frac13$$, and $$P(1\text{ person wins})=\frac13$$, so the expectation is exactly $$1$$, consistent with your answer.

The sample space contains $$3\times 3\times 3=27$$ outcomes, as you say.

There are different situations. If all players pick the same symbol, there is not a player that stands out (3 outcomes). If the three players pick three different symbols, there is also not a player that stands out (6 outcomes).

You call these $$3+6=9$$ outcomes draws.

The remaining situations corresponds to two of the players having the same symbol, and the last player having a different symbol. If the symbol held by the two players beats the symbol of the remaining player, you use the terminology that the two players win together (9 outcomes). However, if the symbol shared by the two players is weaker than the symbol chosen by the last player, then the last player wins alone (9 outcomes).

You count the number of outcomes where player A wins alone as 3, corresponding to probability that A wins alone on $$\frac{3}{27}=\frac19$$.

You also correctly count the number of outcomes where player A wins together with another player as 6, corresponding to probability $$\frac{6}{27}=\frac{2}{9}$$. But in 3 of these 6 outcomes, player B also wins together with another player, and in the other 3 it is player C which also wins. So this overlaps.

This why you cannot say that the (correct!) probability $$\frac13$$ that A wins either alone or together with one other player does not leave room for the situation where we have a draw.

If we denote a draw with three different symbols $$ABC$$, the situation where A and B win together as $$AB$$, and the situation where A wins alone with $$A$$, and finally a draw with a single symbol repeated threes times $$O$$, then we have: $$P(ABC) = \frac{6}{27} \\ P(AB)=P(BC)=P(CA)=\frac{3}{27} \\ P(A)=P(B)=P(C)=\frac{3}{27} \\ P(O)=\frac{3}{27}$$ and these eight events are disjoint and make up the entire sample space, and $$\frac{6}{27} + \left(\frac{3}{27}+\frac{3}{27}+\frac{3}{27}\right) + \left(\frac{3}{27}+\frac{3}{27}+\frac{3}{27}\right) + \frac{3}{27} = 1.$$