Extended Rock-Paper-Scissors

Since the players are playing until one wins 2 games, they need to play at least 2 games.

Let $A_i$ be the event that player A wins the $i$ th game where $i \in [2,\infty)$ , then $P(\text{A wins game}) = P\left ( \bigcup_{i = 2}^{^\infty} A_i \right )$ $P(A_i) = P(B_i \cup C_i )$

Where $B_i = \{\text{A wins on the ith game and B wins one game}\}$ and

$C_i = \{\text{A wins on the ith game and B wins no games}\}$

$\text{These are the only ways for} \ A \ \text{to win on the} \ i\text{th round because if} \ A \ \text{lost twice then he or she could not possibly win.}$ $\text{ Furthermore, as} \ B_i \cap C_i = \emptyset, \ \text{for all} \ i, \ \text{that is the above events cannot occur simultaneously, we have}$

$P(B_i \cup C_i ) = P(B_i) + P(C_i)$
$P(C_i) = \binom{i-1}{1}(1-2p)^{i-2}p^2, \quad P(B_i) = \frac{(i-1)!}{1!1!(i-3)!} (1-2p)^{i - 3} p^3$ As $A_i \cap A_j = \emptyset$ , when $i \not = j$ , we apply countable additivity and get $P\left ( \bigcup_{i = 2}^{^\infty} A_i \right ) = \sum_{i = 2}^{\infty} P(A_i) = \sum_{i = 2}^{\infty} P(B_i) + P(C_i) = (i-1)(1-2p)^{i-2}p^2 + (i-1)(i-2)(1-2p)^{i-3}p^3$ Letting $i-1 = n$ , we have: $P(\text{A wins game}) = \sum_{n = 1}^{\infty} n(1-2p)^{n-1}p^2 + n(n-1)(1-2p)^{n-2}p^3$

Recall, If $X \sim geo(2p)$ ,that is if $X$ denotes a geometric random variables with parameter $2p$ , then $P(X = k) = (1-2p)^{k-1} 2p$ .

Further, $\mathbb{E}[X] = \sum_{k=1}^\infty k(1-2p)^{k-1}2p = \frac{1}{2p}$ and

$E[X^2] = \sum_{k=1}^\infty k^2(1-2p)^{k-1}2p= Var(X) + \mathbb{E}[X]^2 = \frac{1-2p}{4p^2} +\frac{1}{4p^2} = \frac{2-2p}{4p^2}$ (These formulas can be found in the references below)

By algebraic mainpulation we can obtain the following: $= \frac{p^2}{1-2p} \left (\frac{1-3p}{2p} \sum_{n=1}^\infty n(1-2p)^{n-1}2p + \frac{1}{2}\sum_{n=1}^\infty n^2(1-2p)^{n-1}2p \right )$

We can now substitute $E[X] \ \text{and} \ E[X^2]$ respectively for the 2 infinite sums above. $P(\text{A wins the game}) = \frac{p^2}{1-2p} \left ( \frac{1-3p}{4p^2} + \left(\frac{1}{2}\right) \left( \frac{2-2p}{4p^2}\right) \right ) = \left ( \frac{p^2}{1-2p} \right ) \left ( \frac{2-4p}{4p^2}\right) = \frac{1}{2}$

https://en.wikipedia.org/wiki/Geometric_distribution

https://en.wikipedia.org/wiki/Variance#Definition

Because of symmetry, no calculations are needed. The chance of A to win are the same as B's. The chance of no one winning is sufficiently small that the answer must be 0.5.