# Probability Of Winning For Player 1 and Player 2

Player 1 and Player 2, henceforth called P1, and P2 respectively are in competion with each other. The winner of the competion is the player who wins the most number of games (not the player with the cloestest ratio of wins:loses being 1:1). P1 and P2 play the games on independent devices from each other, start at the same time, and stop when they played all the games they can within the time limit.

• P1 choses to play the games as fast as possible, meaning he can play as many games as P2, plus some number $A$. Therefore, the number of games P1 can play is $N+A$. The probability of him winning a game however is only $ℙ$ $(0 < ℙ < 1)$.

• P2 chooses to play slower, meaning the probability that he wins is just the same as P1, plus some number $B$. Therefore, probability of P2 winning a game is $ℙ+B$ $(0 < ℙ+B < 1)$ & $(ℙ < ℙ+B)$. However, the number of games P2 can play is only $N$.

What is the probabilities for each player winning the competition?

• How far have you gotten with this? Where are you stuck? Apr 14 '18 at 4:34
• It is not simple. Can you compute the expected number of wins for each player? Can you compute the probability $A$ wins exactly $k$ games? Apr 14 '18 at 4:37
• The P(player 2 winning every single game) = (N choose N)(P+B)^(N) * (1-(P+B))^(N-N) @RossMillikan
– QHZ
Apr 14 '18 at 4:45
• From the last comment, it looks like you can compute the probability that either player wins a specified number of games. To compute the probability that $P_1$ wins, you have to add up all the probabilities of all the outcomes where $P_1$ wins more games than $P_2$. Hint: You will need a double sum. Can you compute the probability that $P_1$ wins $k$ games and $P_2$ wins $j$ games? Apr 14 '18 at 4:51
• If there are k games, the # of situations where P2 wins is (k^2)-1. I understand that I would be taking the union of all the probabilities where P2 wins, but I don't know where to go from here. @saulspatz
– QHZ
Apr 14 '18 at 5:05

We have that $P_1$ plays $N+A$ games and the probability of his winning a game is $p$. Also, $P_2$ plays $N$ games and his probability of winning in $p+B>p$. To make the the formulas more compact, I will write $M=N+A, q=p+B$

Let's compute the probability that $P_1$ wins. This means that if $P_2$ wins $j$ games, $P_1$ must win anywhere form $j+1$ to $N+A$. Now the probability that $P_2$ wins exactly $j$ games is $$\binom{N}{j}(q)^j(1-q)^{N-j}$$ $P_1$ wins the match if he wins $j+1$ gamesor, $j+2$, etc. up to $M$ games. That is, when $P_2$ wins exactly $j$ games, the probability that $P_1$ wins is $$\sum_{k=j+1}^M{\binom{M}{k}p^k(1-p)^{M-k}}$$

We have to multiply this by the probability that $P_2$ wins $j$ games, of course, and we have to add the probabilities up for all possible values of $j$. We get $$\sum_{j=0}^N{\binom{N}{j}(q)^j(1-q)^{N-j}}\sum_{k=j+1}^M{\binom{M}{k}p^k(1-p)^{M-k}}$$

I don't know any way to make this much simpler.

Now to find the probability that $P_2$ wins, you have to do the same things in reverse. I'll leave that to you.

One word of warning, though. Don't think that the probability that $P_2$ wins is one minus the probability that $P_1$ wins. The match could very well end in a tie.

• Can you check my answer now @saulspatz?
– QHZ
Apr 14 '18 at 5:59

Is this correct for the P(Player 2 wins)? For (k > N)

• This looks right to me. Apr 14 '18 at 6:03
• Thanks for the help @saulspatz !
– QHZ
Apr 14 '18 at 6:04
• It was my pleasure. Apr 14 '18 at 6:04