Game with $n$ players, respective chances of winning however many games in succession A number of persons $A$, $B$, $C$, $D$, $\ldots$ play at a game, their chances of winning any particular game being $\alpha$, $\beta$, $\gamma$, $\delta$, $\ldots$ respectively. The match is won by $A$ if he gains $a$ games in succession: by $B$ if he gains $b$ games in succession; and so on. The play continues till one of these events happens. How do I see that their chances of winning the match are proportional to$${{(1 - \alpha)\alpha^a}\over{1 - \alpha^a}},{{(1 - \beta)\beta^b}\over{1 - \beta^b}}, {{(1 - \gamma)\gamma^c}\over{1 - \gamma^c}}, \text{etc.}$$and that the average number of games in the match will be$$1/\left({{(1 - \alpha)\alpha^a}\over{1 - \alpha^a}} + {{(1 - \beta)\beta^b}\over{1 - \beta^b}} + {{(1 - \gamma)\gamma^c}\over{1 - \gamma^c}} + \ldots\right)?$$
 A: I will use some different notations to make things easier. Let $n$ be the number of players and $\mathbf p =\begin{bmatrix}p_1\\p_2\\\vdots\\ p_n\end{bmatrix}$ the probability vector that players win a game and $\mathbf m = \begin{bmatrix}m_1\\m_2\\\vdots\\m_n\end{bmatrix}$ vector of number of consecutive games needed for each player to win the match. Let $T$ be the length of a match, $N_k$ the first game that the player number $k$ has won and $L_k$ the first game that the player $k$ has lost. \begin{align}\mathbb E\left[T\right] = \sum\limits_{k=1}^{n} \mathbb E\left[T\mid N_k = 1\right] p_k\end{align}
with the fact that: \begin{align}\mathbb E\left[T\mid N_k = 1\right] &= \mathbb P\left[L_k > m_k\mid N_k=1\right]\times m_k + \sum_{s=2}^{m_k}\mathbb P\left[L_k=s\mid N_k = 1\right]\mathbb E\left[T\mid N_k = 1\cap L_k=s\right]\\
&= m_k\mathbb P\left[L_k > m_k-1\right] + \sum_{s=2}^{m_k}\mathbb P\left[L_k=s-1\right]\left(s-1+\mathbb E\left[T\mid L_k=1\right]\right)\\
&= m_k\mathbb P\left[L_k > m_k\right] + \sum_{s=1}^{m_k}s\mathbb P\left[L_k=s\right] + \mathbb P\left[L_k < m_k\right] \mathbb E\left[T\mid L_k=1\right]\\
&= q_k + \mathbb P\left[L_k < m_k\right]\sum_{l\neq k}\mathbb E\left[T\mid N_l =1\right]\mathbb P\left[N_l=1\mid L_k=1\right]\\
&= q_k + \mathbb P\left[L_k < m_k\right]\sum_{l\neq k}\mathbb E\left[T\mid N_l =1\right]\frac{p_l}{1-p_k}\\
\mathbb E\left[T\mid N_k=1\right] &= q_k + u_k\sum_{l\neq k} \mathbb E\left[T\mid N_l=1\right]p_l 
\end{align}
Where \begin{align}
q_k &= m_k\mathbb P\left[L_k > m_k\right]+\sum_{s=1}^{m_k}s\mathbb P\left[L_k=s\right]\\
&= m_k\mathbb P\left[L_k > m_k\right]+\sum_{s=1}^{m_k}\sum_{t=1}^{s}\mathbb P\left[L_k=s\right]\\
&= m_k\mathbb P\left[L_k > m_k\right]+\sum_{t=1}^{m_k}\sum_{s=t}^{m_k}\mathbb P\left[L_k=s\right]\\
&= m_k\mathbb P\left[L_k > m_k\right]+\sum_{t=1}^{m_k}\left(\mathbb P\left[L_k \ge t\right] - \mathbb P\left[L_k>m_k\right]\right)\\
&= \sum_{t=1}^{m_k} \mathbb P\left[L_k\ge t\right] = \sum_{t=1}^{m_k} p_k^{t-1} = \frac{1-p_k^{m_k}}{1-p_k}
\end{align}
and $u_k = \displaystyle\frac{\mathbb P\left[L_k < m_k\right]}{1-p_k} = \frac{1-p_k^{m_k-1}}{1-p_k} = q_k - p_k^{m_k-1}$
\begin{align}
\left(\begin{bmatrix}1+u_1p_1\\
& \ddots \\
& & 1+u_np_n\end{bmatrix} - \begin{bmatrix}u_1\\
\vdots\\
u_n\end{bmatrix}\begin{bmatrix}p_1 & \cdots & p_n\end{bmatrix}\right)\begin{bmatrix}\mathbb E\left[T\mid N_1=1\right]\\
\vdots\\
\mathbb E\left[T\mid N_n=1\right]\end{bmatrix} &= \begin{bmatrix}q_1\\
\vdots\\
q_n\end{bmatrix}
\end{align}
So $\left(J - \mathbf u \mathbf p^T\right) \mathbf e = \mathbf q$, where $J = \begin{bmatrix}1+u_1p_1\\
& \ddots \\
& & 1+u_np_n\end{bmatrix}$ and $\mathbf e = \begin{bmatrix}\mathbb E\left[T\mid N_1=1\right]\\
\vdots\\
\mathbb E\left[T\mid N_n=1\right]\end{bmatrix}$. This implies that \begin{align}
\mathbb E\left[T\right] &= \mathbf p^T\mathbf e\\
&= \mathbf p^T\left(J - \mathbf u\mathbf p^T\right)^{-1}\mathbf q\\
&= \mathbf p^T\left(J^{-1} + \frac{J^{-1}\mathbf u\mathbf p^TJ^{-1}}{1 - \mathbf u^TJ^{-1}\mathbf p}\right)\mathbf q\\
&= \mathbf p^TJ^{-1}\mathbf q + \mathbf p^TJ^{-1}\mathbf q \frac{\mathbf p^TJ^{-1}\mathbf u}{1-\mathbf p^TJ^{-1}\mathbf u}\\
&= \frac{\mathbf p^TJ^{-1}\mathbf q}{1-\mathbf p^TJ^{-1}\mathbf u}
\end{align}
Now let's compute $1-\mathbf p^TJ^{-1}\mathbf u$ and $\mathbf p^TJ^{-1}\mathbf q$.
\begin{align}
1-\mathbf p^TJ^{-1}\mathbf u &= 1-\sum_{k=1}^n p_k\frac{1}{1+p_ku_k}u_k\\
&= \sum_{k=1}^n p_k -\sum_{k=1}^n \frac{p_ku_k}{1+p_ku_k}\\
&= \sum_{k=1}^n p_k - \frac{p_k\left(1-p_k^{m_k-1}\right)}{1-p_k+p_k\left(1-p_k^{m_k-1}\right)}\\
&= \sum_{k=1}^n \frac{p_k \left(1-p_k^{m_k}\right) - \left(p_k - p_k^{m_k}\right)}{1-p_k^{m_k}}\\
&= \sum_{k=1}^n \frac{p_k^{m_k}\left(1-p_k\right)}{1-p_k^{m_k}}
\end{align}
and
\begin{align}
\mathbf p^TJ^{-1}\mathbf q &= \sum_{k=1}^n p_k \frac{1}{1+u_kp_k}q_k\\
&= \sum_{k=1}^n p_k \frac{1}{1+u_kp_k}\left(\frac{1-p_k^{m_k}}{1-p_k}\right)\\
&= \sum_{k=1}^n p_k \frac{1}{1+u_kp_k}\left(u_k + p_k^{m_k-1}\right)\\
&= \sum_{k=1}^n \frac{p_k^{m_k}}{1+u_kp_k} + \sum_{k=1}^n \frac{p_ku_k}{1+p_ku_k}\\
&= \sum_{k=1}^n \frac{p_k^{m_k} + p_ku_k}{1+u_kp_k}\\
&= \sum_{k=1}^n \frac{p_k^{m_k}\left(1-p_k\right) + p_k\left(1-p_k^{m_k-1}\right)}{1-p_k+p_k\left(1 - p_k^{m_k-1}\right)}\\
&= \sum_{k=1}^n p_k = 1.
\end{align}
So $$\mathbb E\left[T\right] = \frac{1}{\sum_{k=1}^n \frac{p_k^{m_k}\left(1-p_k\right)}{1-p_k^{m_k}}}$$ which is exactly what you are looking for. You can do the same thing to compute the winning probabilities.
A: Thanks for a very fun problem!  After several days, I finally found a solution via Martingales, which is IMHO more interesting and less tedious than using matrices.  In particular my solution below is modified from this paper section 5, which describes the classic Martingale solution to the ABRACADABRA problem.  If you like my solution and/or want more info, I highly recommend that paper.

The main idea of the Martingale solution is that if you place fair bets, the expected gain (or loss) is exactly zero no matter your betting "strategy".
So imagine there is a casino offering fair bets on individual games.  If you bet $1$ dollar on $A$, you get back $\frac{1}{\alpha}$ if $A$ wins, so the bet is fair: your expected gain $= \alpha \times \frac{1}{\alpha} - 1 = 0$.
So you bet using this strategy: You bet $1$ dollar on A, and as long as A keeps winning you keep re-betting the entire amount on A, until either you lose (i.e. A loses) or A wins the entire match.  At the end, either you have $\frac{1}{\alpha^a}$ (if A wins the match) or $0$.  Since the former happens with probability $\alpha^a$, the expected result is still $1$ and the expected gain is still $0$.
In fact you are a REAL big fan of A and go crazy like this: at the beginning of every single game, you make a new $1$ dollar bet, potentially starting its own sequence of exponentially growing results.  You keep doing this for as long as the match lasts.  Conceptually, consider each such $1$ dollar bet separately.  Now when the match is over (many games later), there are two possible outcomes:

*

*If anyone other than A wins the match: your final result is $0$.


*If A wins: your last bet would pay $1/\alpha$, the second-to-last bet would pay $1/\alpha^2$, etc., up to the bet you placed exactly at the beginning of the $a$-games winning streak, now worth $1/\alpha^a$.  Your final total winning result, which I will denote as $w_A$, is:
$$
w_A = \frac{1}{\alpha} + \frac{1}{\alpha^2} + \dots + \frac{1}{\alpha^a}
= \frac{\alpha^{a-1} + \alpha^{a-2} + \dots + 1}{\alpha^a}
= \frac{1 - \alpha^a}{(1-\alpha) \alpha^a}
$$
Now let $p_A$ be the probability of A winning.  So your expected final result is $p_A \times w_A$.
Here is the fancy Martingale argument: Since all bets are fair, your expected final gain must be $0$ [see footnote], which means your expected final result $p_A \times w_A$ must equal your expected total expenditure.  But you have been betting $1$ dollar per game, so your total expenditure $=$ no. of games $T$, and your expected total expenditure $= E[T]$.
There is nothing special about A, and the same argument applies to B, C, etc.  So:
$$E[T] = p_A w_A = p_B w_B = p_C w_C = \dots $$
This immediately shows $p_A, p_B, p_C, \dots$ are proportional to $\frac{1}{w_A}, \frac{1}{w_B}, \frac{1}{w_C}, \dots$ which is the first desired result.
The second desired result follows easily since sum of all probabilities $=1$:
$$
1 = p_A + p_B + p_C + \dots = E[T](\frac{1}{w_A} + \frac{1}{w_B} + \frac{1}{w_C} + \dots)
$$

Footnote: The relevant theorem here is Doob's Optional Stopping Theorem and in particular, this problem meets the precondition of $E[T]$ being finite (clear from the underlying Markov Chain) and bet sizes being bounded.
Again, I highly recommend the paper I learned from, especially section 5.
