# Long term probability of a game with multiple outcomes?

I'm not a stats expert so please excuse me if this is a really basic question and I've simply missed the reference...

So let's say this person is playing a game with a small chance of winning, say $P_w$, every time they play. If they lose, they must play again. This game is rather frustrating, so after $m$ plays, there is also a distinct probability $P_q$ that the player will quit.

What I'm trying to do is determine the probabilities of the player winning or quitting after a large number of rounds $N >> m$.

Probably very basic, but I'm not seeing it somehow...

• You need to specify better the quitting process! Is quitting possible only at round $m$, or at any round after round $m$? With the same quitting probability? Aug 18, 2015 at 12:21
• Sorry, unclear on my part. For the purposes of this model, it's only possible to quit after participating in m rounds, but alternatively I'd also like to be able to model the (simpler) case that the participant can quit at any time, but this would require a modification of p_q.
– Matt
Aug 18, 2015 at 12:31

We assume the per round win probability is $p_w > 0$. Winning at each round is independent of all other rounds. Then let $$X_1, X_2, \dots, X_m, \dots, X_n, \dots$$ be the result of each round, $X_i$ is a bernoulli random variable with distribution $P(X_i=1)=p_w, P(X_i=0)=1-p_w$. Quitting exactly at round $m$ is a bernoulli variable with quitting probability ($Q=1$) equal to $p_q$. Then the probability of quitting before winning is $$P(X_1+X_2+\cdots+X_m=0, Q=1) = (1-p_w)^m \cdot p_q$$ while the probability of winning before quiting is $$1-p_q + p_q \cdot (1-(1-p_w)^m)$$ In the last formula we are assuming that, if not quiting at round $m$, we will play forever so certainly will win, sooner or later.