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.