# Odds of drawing multiple rounds of rock paper scissors.

OK, so the back story for this is me and my friend often decide things on a quick game of rock paper scissors. I think on this occasion it was for who would get up and answer the door when the pizza came. The point is, on this occasion, we drew 17 times in a row before we broke the dead lock.

How do I find the odds of drawing n games of rock, paper, scissors?

We were playing the normal rules of rock, paper, scissors and, for arguments sake, there were no tactics or mind games going on - just random choices by both my friend and I (though I think some kind of subconscious thing must have been at play because my gut tells me the odds of this are astronomical).

• In any one round P(win) = P(loss) = P(draw) = 1/3. If you believe $n$ successive rounds are independent, then the probability of drawing $n$ times is $1/3^n$. For $n = 17$ that is highly unlikely! – Simon S Apr 27 '15 at 14:10
• About $1$ in $129$ million. If everyone in the world paired up to play a game of r-p-s ($\sim 3.62$ billion matches), only $\sim 28$ matches or so would run at least this long. – Travis Willse Apr 27 '15 at 14:14
• For real life players the assumption of independence may not be justified, however. Once both picked paper thrice, say, they may both try to outwit each other with the same ideas ("Now he'll think I take paper again and takes scissors; hence I take rock") and draws continue with higher than usual probability. – Hagen von Eitzen Apr 27 '15 at 14:20
• I'd agree, there must have been something like that going on - though I imagine, even taking this into account, the odds were high for 17 in a row. My other friend, who was also present, flat out wouldn't believe that my friend and I hadn't staged it. – Andrew Jones Apr 27 '15 at 14:32

The chances of a draw when both parties chosing uniformly are exactly $\frac13$ (it's a fair game because of this). Now for that to happen $17$ times in a row, the chances are $$\frac1{3^{17}} \approx 7.74\cdot 10^{-9}$$ but keep in mind that these are a-priori. That means for example after two draws, the chances of another draw stay at $\frac13$ and the chances for the $15$ remaining draws consecutively are $\frac1{3^{15}}$.