Two chess players $A$ and $B$ play a tournament. A tournament is a sequence of games, each of which is independent from the rest. There are three different results from a game: player $A$ wins (1), the game is a draw (1/2), player $B$ wins (0). Based on some rating of the players known before the start of the tournament, it is known that
- player $A$ wins with probability $a$,
- the result is a draw with probability $b$, and
- player $B$ wins with probability $c$.
These probabilities are not affected by the games. As usual, $0\le a,b,c\le 1$ and $a+b+c=1$.
Now, the first player to win two games (not necessarily in a row), wins the tournament; draws do not count. A tournament does not have a limit on the number of games (notice that if $b=1$ the tournament will go on forever, all games ending in a draw). For example, in the sequence of games
- $(1,1)$ player $A$ wins the tournament since the only two games played were won by $A$,
- $(1,0,1)$ (and in any permutation of this sequence of games) player $A$ wins the tournament, and player $B$ has won only one game.
- $(1,1/2,\dots,1/2,1)$ player $A$ wins the tournament because $A$ won the first and last games and the other were draws,
- $(0,0,1)$ it is player $B$ who wins this tournament.
QUESTION What is the probability that player $A$ wins a tournament (as a function of $a$, $b$ and $c$)?
I first tried tackling this problem by calculating the probability that a given sequence of results can happen. So, for player $A$ to win the tournament, only the three following scenarios can happen,
- ($i$ draws) (B wins once) ($j$ draws) (A wins once) ($k$ draws) (A wins once)
- ($k$ draws) (A wins once) ($i$ draws) (B wins once) ($j$ draws) (A wins once)
- ($i$ draws) (A wins once) ($j$ draws) (A wins once)
where $i,j,k\ge 0$ are integers (not necessarily equal among all scenarios). My limitation here is that I don't know what the values of $i$, $j$ and $k$ can be. Even if I knew, I'm not sure what to do with the probabilities of each scenario. I'm not good at probabilities, and I really don't know where to even begin. I thought formalizing things with Bernoulli or Geometric random variables would help, but I haven't had any result that matches the computer simulations. Any help will be appreciated.
This is my answer. I hope it's correct. The goal is to compute the value
$$ \begin{equation*} \mathbb{P}(\text{A wins tournament}) = \frac{\sum \text{probabilities events in which A wins tournament}}{\sum \text{probabilities all events in which one of A or B wins tournament}} \end{equation*} $$
To ease this task, it was suggested in the comments of OP to change the definition of winning a game to exclude draws as a possibility. For this, we calculate the conditional probability that $A$ (resp. $B$) wins assuming that the game is not a draw using Bayes' theorem $$ \begin{align*} P_A &= \mathbb{P}( \text{A wins} | \text{the game is not a draw} ) \\ &= \frac{\text{Probability of A winning and the game is not a draw}}{\text{The game is not a draw}}\\ &= \frac{a}{1 - b} \\ &= \frac{a}{a + c} \end{align*} $$ For $B$, $$ \begin{align*} P_B &= \mathbb{P}( \text{B wins} | \text{the game is not a draw} ) \\ &= \frac{c}{1 - b} \\ &= \frac{c}{a + c} \end{align*} $$ Under this new definition of game (where draws are excluded), consider the events in which one of the two players wins the tournament:
The probabilities of the events where $A$ or $B$ win are, respectively, $$ \begin{align*} \text{(1) } e_A &= 2P_BP_A^2 + P_A^2 \\ \text{(2) } e_B &= 2P_AP_B^2 + P_B^2 \end{align*} $$ Finally, $$ \mathbb{P}(\text{A wins tournament}) = \frac{e_A}{e_A + e_B} = \frac{a^2(b - 2c + 1)}{b(a^2 + c^2) - 2a^2c - 2ac^2 - a^2 - c^2} = \frac{a^2(3c + a)}{(a + c)^3} $$