A friend and I have a game we play, and when we draw or call at the same time we use one round of paper/scissors/rock to determine the winner. I am currently 9-0 up on him and was wondering what the mathematical odds are of that? Is it as simple as $\frac{1}{3\times9}$? Or is there more to it?
:edit: they were all individual times, we have only played 9 times, separated by weeks, each time there were no draws, I beat him first time every time, I'm not concious of any "tactics" that help win games of p/s/r, to my knowledge we are both playing completely randomly.
Best Answer
Each game has three possible outcomes: win, lose or draw. If you play two games the first game could be any of the three and for each of those three outcomes the second game can have three outcomes. So if you play two games there are $3\times 3 = 9$ \possible out comes. They are
(win,win), (win,lose),(win, draw)
(lose,win), (lose, lose), (lose, draw)
(draw,win), (draw, lose), (lose, lose)
You play three games and each of those $9$ will have $3$ outcomes or a total of $9\times 3 = 27$. Outcomes.
If you play $k$ games there are $3^k$ possible outcomes.
So if you play $9$ games there are $3^9 = 19683$ possible outcomes. And (win,win,win,win,win,win,win,win,win) is one of $19,683$ possible outcomes.
If we assume all outcomes are equally likely the probability of winning nine out of nine games is $\frac 1{19,683}$.
Frankly, I think you are just a better player.
If you don't count draws. Then each game has $2$ outcomes and for $9$ games there are $2^9 = 512$ possible outcomes and the probability of nine out of nine wins is $\frac 1{512}$.