Given probability of A not B and probability of B not A, what is probability of A occurring before B?

Question

Context: In a game I sometimes play, players roll percentile dice against one another to win points. Each player has a unique threshold they have to roll below to score a point, but if both players roll below their threshold, neither scores a point and they roll again. The first player to three points wins.

Question: I'm trying to figure out the probability of player A scoring three points before player B.

$p$ = the probability of player A rolling below their scoring threshold

$q$ = the probability of player B rolling below their scoring threshold

$x$ = the probability of player A scoring a point

$y$ = the probability of player B scoring a point

Since $p$ and $q$ are independent events, I know that you can calculate the probability of a given player scoring points by $x = p*(1-q)$ and $y = q*(1-p)$

However, I'm not sure what formula to use to determine the probability that $x$ occurs 3 times before $y$ occurs 3 times if I assume an open-ended number of trials.

Example: To give some solid numbers, assume $p = 0.56$ and $q = 0.54$; this gives us $$x = 0.56*(1-0.54) = 0.2576$$ and $$y = 0.54*(1-0.56) = 0.2376$$ which means that for each individual round of rolling, there is a 25.76% chance that A will score a point, a 23.76% chance that B will score a point, and a 50.48% chance that neither will score a point (either because neither of them will roll below their threshold or because both will roll below their threshold, prompting another roll).

Given all of this, how do I go about calculating the probability that A will score 3 points before B scores 3 points?

Answer 1

First a simplification: we can throw away the two neutral events (both succeed or both fail) and consider only the two scoring events with probability $p(1-q)$ and $q(1-p)$ as the "real" outcomes. In this perspective, those two events occur with probabilities $$ a = \frac{p(1-q)}{p(1-q)+q(1-p)} \quad\text{and}\quad b = \frac{q(1-q)}{p(1-q)+q(1-p)}. $$ (Think of this as changing a game "turn" into "keep rolling until someone scores a point"; then these are the probabilities of $A$ and $B$ scoring in each "turn". One can formally derive these quantities with infinite series, but hopefully the idea is reasonably clear.)

Now "the first to score $3$ points" is the same as "scoring more points when $5$ turns are played" (verify this). And this is a job that generating functions are great for. If we let $c_k$ be the probability that player $B$ scores exactly $k$ points in $5$ turns, then $$ \sum_{k=0}^5 c_k x^k = (a+bx)^5 = \sum_{k=0}^5 \binom5k a^{5-k}b^k x^k, $$ and so the probability that player $B$ wins is $$ \sum_{k=3}^5 c_k = \sum_{k=3}^5 \binom5k a^{5-k}b^k. $$ In the given example, with $p=0.56$ and $q=0.54$, we end up with $a=322/619$ and $b=297/619$, and so the probability of $B$ winning is \begin{align*} \sum_{k=3}^5 \binom5k \biggl( \frac{322}{619} \biggr)^{5-k} \biggl( \frac{297}{619} \biggr)^k &= \frac1{619^5} \big( 10\cdot322^2\cdot297^3 + 5\cdot322\cdot297^4 + 297^5 \bigr) \\ & = \frac{42001248396987}{90876845839099} \approx 0.462178 \end{align*} (and so the probability of $A$ winning is about $0.537822$).