It is a question about classical models of probability.
Assume two guys A and B play chess. The probability of A winning ONE GAME is a whilst B is b with $a + b = 1.$ (There is no draw)
Assume that A is a better player, $a > b$. And the rule is: if A wins three games in a row first, A wins; if B wins two games in a row first, B wins. And eventually one will win the game.
We can name the event of A winning as E and B winning as F. Now, if we want to calculate the probability of A winning, it must comprise a series of events such as EF EEF EF EEF. We may assume that A wins the first game, but my confusion starts here: we have no idea what will happen next, as EF and EEF are both legitimate choices throughout the process, and we don't know what will appear. For example, EF EEE is OK and EEF EEF EEF EEE also works.
(There are similar questions with some variations, but I don't really understand some of the answers)
Best Answer
Let $P$ be the probability of A winning the match (not game), and $Q$ be the probability of B winning. Then- $$P+Q=1$$ Let us attempt to calculate $Q$. If we represent B winning a game with F, then, if B wins at the $r^{th}$ game, it is decided that the outcome of $(r-1)^{th}$ game is F and that of $(r-2)^{th}$ game is E. So, we have to create a word made of E and F of length $r-3$, with no sub-string 'FF' or 'EEE'.
If we focus on the F's, it is clear that the number of F's in the $r-3$ length string can be, a maximum of $\left\lceil \frac {r-3}{2} \right \rceil =k$, if $r\geq 6$. Suppose we take $l$ number of F's, $2\leq l \leq k$. If we go by the stars and bars approach, the gap between two subsequent F's can be either $1$ or $2$. Also, for $l$ F's, the number of gaps is $l-1$. Thus, the problem to be solved is:
$$x_1+x_2+...+x_{l-1}=(r-3)-l$$ where $1\leq x_i \leq 2$. Thus, we have to find the coefficient of $x^{r-l-3}$ in $(x+x^2)^{l-1}$. Using the binomial theorem, this coefficient is $\binom {l-1}{r-2l-2}$. Thus, the number of ways that B can win at the $r^{th}$ game is: $$S=\sum_{l=2}^k \binom {l-1}{r-2l-2}$$ Hence, the probability of B winning at the $r^{th}$ game is: $$q=S\cdot a^{r-l-2}\cdot b^{l+2}$$ Hence, since $r\geq 2$, we have: $$Q=\sum q=\left(\sum_{r=6}^{\infty} \sum_{l=2}^k \binom {l-1}{r-2l-2}\cdot a^{r-l-2} \cdot b^{l+2}\right)+b^2+ab^2+(ab^3+a^2b^2)+2a^2b^3+a^3b^3+a^4b^3$$ Hence, $P=1-Q$. Note that the $b^2$, $ab^2$, $ab^3+a^2b^2$ and $2a^2b^3$ terms represent probabilities of B winning if $r=2$, $r=3$, $r=4$ and $r=5$ respectively, $a^3b^3$ represents the $l=1$ case for $r=6$ and $a^4b^3$ represents the $l=1$ case for $r=7$. From $r=8$ onwards, the $l=1$ case is impossible anyway.
Note: In particular, I put $a=b=\frac 12$ and evaluated the summation using Python code upto 2000 terms, it is approximately equal to $0.0375$. Hence, for that case $Q=0.5859+0.0375=0.6234$, which seems reasonable to me, since B only needs to win $2$ consecutive games to win.