Probability of exactly k runs of wins How to calculate probability of exactly k run of wins out of exactly m wins and n losses? A run of win means 1 or more successive wins.
My try: Divide the m+n matches to portions
$L_1W_1L_2W_2L_3W_3....L_kW_kL_{k+1}$
here $L_i$ denotes a run of losses, i.e. one or more consecutive loss(es). Similarly $W_i$ denotes a run of wins.
For exactly k runs of wins, $L_2,..L_k$ should have atleast $1$ loss, but $L_1, L_{k+1}$ can have $0$ loss as well. All $W_i$s should have atleast one win.
Now, number of ways to get all different $L_i$s satisfying this criteria, is same as arranging $k$ sticks between $n$ stars, so that none of the sticks are adjacent. This is same as putting (n-k+1) balls into k-1 boxes, with empty boxes allowed= $(k-1)^{n-k+1}$. But each loss is indistinguishable from one another, so total number of ways of forming $L_1,...,L_{k+1}$ is $\frac{(k-1)^{n-k+1}}{n!}$.
Similarly, number of ways to get all different $W_i$s satisfying the criteria, is same as putting m indistinguishable balls in to k boxes, with empty not allowed. Which is same as putting (m-k) indistinguishable balls into k boxes, with empty boxes allowed = $\frac{k^{m-k}}{m!}$
So, total number of ways = $\frac{(k-1)^{n-k+1}.k^{m-k}}{m!n!}$
Probability of exactly k runs of wins = $\frac{(k-1)^{n-k+1}.k^{m-k}}{m!n!(m+n)!}$
But I am not feeling confident with this logic. It will be good if anyone can confirm if it is correct, or contradict if there is any loophole.
 A: There are ${m+n \choose m}$ ways in total of ordering the equally likely outcomes given $m$ wins and $n$ losses.
Easy bits first:

*

*$\mathbb P(K=0 \mid 0,n)=1$ since there are no wins.

*$\mathbb P(K=1 \mid m,0)=1$ if $m \ge 1$ since there are only wins, so one run.

*$\mathbb P(K=0 \mid m,n)=0$ if $m \ge 1$ since there is at least one winning run.

*$\mathbb P(K=k \mid m,n)=0$ if $m \lt k$ or $n \lt k+1$ as $k$ winning runs impossible.

Otherwise with $m,n\ge1$ and $k\ge 2$, then of these:

*

*${m-1 \choose k-1}{n-1 \choose k-2}$ start with a win, end with a win, have $k$ winning runs and $k-1$ losing runs


*${m-1 \choose k-1}{n-1 \choose k-1}$ start with a win, end with a loss, have $k$ winning runs and $k$ losing runs


*${m-1 \choose k-1}{n-1 \choose k-1}$ start with a loss, end with a win, have $k$ winning runs and $k$ losing runs


*${m-1 \choose k-1}{n-1 \choose k}$ start with a loss, end with a loss, have $k$ winning runs and $k+1$ losing runs
So ${m-1 \choose k-1}\left({n-1 \choose k-2}+{n-1 \choose k-1}+{n-1 \choose k-1}+{n-1 \choose k} \right) ={m-1 \choose k-1}{n+1\choose k}$ have $k$ winning runs
making the probability  $$\mathbb P(K=k \mid m,n)=\dfrac{{m-1 \choose k-1}{n+1\choose k}}{{m+n \choose m}}$$
