# Luck or skill in 3-person game

Two friends and I have been playing a game that is a combination of skill and luck. (as most games are). We assume that if the game was all luck and/or we all had the same skill level eventually the win% of all of us would approach 33%.

So what sample size would we need to show that one of us was "better" at the game? That is if we played 50 games, what % of games (or how far from 33%) would someone need to demonstrate skill vs. luck (perhaps at a 95% or 99% likelihood). What if we played 100 games or 200 games?

I would assume we would need a smaller percentage difference from 33% as the sample size went up.

So, you need to add more assumptions. A reasonable model is that you are sampling independently from a distribution on a $3$-element set, and you would like to distinguish the distribution from $( \frac{1}{3},\frac{1}{3},\frac{1}{3})$. The usual way to do this is with a multinomial test. You can also use simpler but less powerful binomial tests to test each frequency against $\frac{1}{3}$.
In practice I would use a normal approximation to the binomial distribution with continuity correction, and a threshold tighter than $0.05$ to reflect that any of the $3$ players might be strongest. Suppose the probability of winning is $\frac13$. The standard deviation after $n$ games is $\sqrt{n (\frac13)(1-\frac13)}$. A result of $40$ in $100$ games would be $6.67$ games above the mean, but $40$ should be viewed as the interval $[39.5,40.5]$, so it should be viewed as $6.17$ above the mean under the continuity correction. This is $6.17\bigg/(10\sqrt{\frac29}) = 1.3$ standard deviations above the mean, which wouldn't be strong evidence that you are stronger than average. At $43 (1.94\sigma), 44 (2.16\sigma),$ or $45 (2.37\sigma)$ you might conclude that you have strong evidence that you are stronger than average.