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.
Best Answer
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.