Probability Of Winning For Player 1 and Player 2

Question

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?

Answer 1

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.

Answer 2

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