How can you decide the winning percentage of a player in a game based on PMFs for each player?

Question

In a two player game, each player is trying to score the higher number of points. We can assume that the probability that a player scores $n$ points is independent of how many points his opponent scores. For example, if this were the setup for player one and player two:

$P_1(4) = 0.8$

$P_1(5) = 0.1$

$P_1(6) = 0.1$

$P_2(6) = 0.1$

$P_2(7) = 0.9$

In this example, the two players will tie $10\%$ of the time, and player two will win 90% of the time. For more complicated PMFs, how can we solve this problem in general?

Answer 1

As @InterstellarProbe commented above, in the example you've given a tie only occurs $1\%$ of the time and player $2$ wins the other $99\%$ of the time, since the only score they overlap on is $6$ and so a probability of getting a tie is $P_1(6)\cdot P_2(6) = 1\%$. For all other states, player $2$ wins.

In order to solve this problem in general, let's first consider that to find the probability of getting a tie, you must simply find all of the scores that the players can possibly get in common, multiply together all of the pairs, and sum them. For example, if $P_1(5) = 0.3$, $P_1(6) = 0.2$, $P_2(5) = 0.1$, and $P_2(6) = 0.4$ (not exhausting their entire pmfs) where $5$ and $6$ are the only potential scores they have in common, then the probability of yielding a tie is $P_1(5)\cdot P_2(5) + P_1(6)\cdot P_2(6)$.

Likewise, you can find all the pairs of numbers where player $2$ is greater than player $1$, or vice-versa, and take the sum of the products of the pairs to find those probabilities, too.

Answer 2

You have the idea. Assuming independence $P_{12}(a,b)=P_1(a)P_2(b)$. If the distributions are discrete you can just sum over all the cases where $b \gt a$ to get the chance $2$ wins. If they are continuous you integrate. In your example the joint pmf has six pairs of values where the probability is greater than $0$, so you can just add them up.