How can I calculate the odds of winning successive games of paper/scissors/rock?

Question

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.

Answer 1

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 $\frac 1{512}$.

Answer 2

Starting from an arbitrary point, and assuming both of you are playing perfectly randomly, the probability is $$\frac{1}{3^9} = \frac{1}{19683} \approx 0.00005$$ Which is obviously quite low.

Of course, you're probably replaying until somebody wins. In this case, the probability of winning reaches $$\frac{1}{2^9} = \frac{1}{512} \approx 0.002$$ which is considerably more likely but still pretty out there.

But this isn't necessarily what's going on. More likely, you've been playing for a long time, and you're in the midst of a streak. Streaks in a longer run of attempts are more common than starting to count and then discovering that you've gone 9-0. Over seasons of $100$ trials, winning 9 straight happens in $8\%$ish of seasons. You can use the formula from this answer to calculate this.

Of course it's also possible that you're better at Rock Paper Scissors than he is; if you've partially figured out his strategy, you can tune your own to maximize your probability of victory. That goes deep into game theory territory.

Answer 3

It's not as unlikely as it seems: As fleablood already pointed out, if you replayed any tie, then the chance of you going up $9-0$ is $\frac{1}{512}$ (assuming you both make random moves) ... while the chance of anyone going up $9-0$ is $\frac{1}{256}$.

So, given that you and your friend have probably played this game quite often (say, hundreds of games), something like this was almost bound to happen.