Question

Consider Example$5o$, which is concerned with the number of runs of wins obtained when$n$wins and $m$losses are randomly permuted. Now consider the total number of runs—that is, win runs plus loss runs—and show that

$$\begin{gathered} P(2kruns)=\frac{\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}m-1\\ k-1\end{array}\right)}{\left(\begin{array}{c}n+m\\ n\end{array}\right)} \\ P(2k+1runs)=\frac{\left(\begin{array}{c}n-1\\ k\end{array}\right)\left(\begin{array}{c}m-1\\ k-1\end{array}\right)+\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}m-1\\ k\end{array}\right)}{\left(\begin{array}{c}n+m\\ n\end{array}\right)} \end{gathered}$$

Short answer

Use the procedure from the Example$50$

Since runs of wins and losses are alternating, in$2k$runs there are$k$runs of wins and $k$runs of losses, in $2k+1$runs, there can be either$k+1$ run of wins and $k$of losses or vice versa.

Step-by-step solution

Given Information.

A number of runs of wins are obtained when $n$wins and$m$losses are randomly permuted.

Explanation.

Experiment: The distribution of$n$wins and $m$losses in runs (consecutive wins or losses).

Wanted probabilities - precisely$2k$run, $P\left(2k\right)$and precisely $2k+1$runsrole="math" localid="1649344157923" $P(2k+1)$.

As explained in the example $50$there are$\left(\begin{array}{c}m+n\\ n\end{array}\right)$equally likely outcomes (we differentiate only the order of wins and losses, choose$n$places for wins from$m+n$)

$2k$runs.

Since runs of wins and runs of losses alternate, $2k$runs are precise $k$runs of wins and$k$runs of losses.

Say that$x_1,x_2,\dots ,x_k$are the lengths of$Ist,2nd$, $\dots .k$-th run of wins.

$x_1,x_2,\dots ,x_k$can be any positive numbers such that:

$x_1+x_2+\dots +x_k=n$

Since there are $n$wins in total.

Similarly, if $y_1,y_2,\dots ,y_k$are the lengths of runs of losses, the number of possibilities $y_1,y_2,\dots ,y_k$is the number of positive solutions to:

$y_1+y_2+\dots +y_k=m$

From the chapter$1.6$, the number of possible solutions $x_1,x_2,\dots ,x_k$is$\left(\begin{array}{c}n-1\\ k-1\end{array}\right)$and the number of $y_1,y_2,\dots ,y_k$is$\left(\begin{array}{c}m-1\\ k-1\end{array}\right)$.

So the total number of dividing $n$wins and $m$losses into $2k$runs is:

$\left(\begin{array}{l}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}m-1\\ k-1\end{array}\right)$

So by the formula for probability on the sample spaces of equally likely outcomes:

$P\left(2k\right)=\frac{\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}m-1\\ k-1\end{array}\right)}{\left(\begin{array}{c}n+m\\ n\end{array}\right)}$

Explanation.

$2k+1$runs.

Since runs of wins and runs of losses alternate, localid="1649345613859" $2k$runs are either $k$runs of wins and $k+1$runs of losses or$k+1$runs of wins and $k$runs of losses.

Both of the cases are analogous to the procedure above.

Say that $x_1,x_2,\dots ,x_k$are the lengths of localid="1649345620724" $1st,2nd,$$\dots k$th run of wins.

$x_1,x_2,\dots ,x_k$can be any positive numbers such that:

$x_1+x_2+\dots +x_k=n$

Since there are $n$wins in total.

The name$y_1,y_2,\dots ,y_k,y_{k+1}$is the length of $k+1$runs of losses, the number of possibilities $y_1,y_2,\dots ,y_k,y_{k+1}$is the number of positive solutions to:

$y_1+y_2+\dots +y_k+y_{k+1}=m$

From the chapter$1.6$, the number of possible solutions $x_1,x_2,\dots ,x_k$is $\left(\begin{array}{c}n-1\\ k-1\end{array}\right)$and the number of $y_1,y_2,\dots ,y_k$is$\left(\begin{array}{c}m-1\\ k\end{array}\right)$.

So the total number of dividing$n$wins and$m$losses into $k$runs of wins and $k+1$runs of losses is$\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}m-1\\ k\end{array}\right)$.

And if there are $k+1$runs of wins and $k$runs of losses there are$\left(\begin{array}{c}n-1\\ k\end{array}\right)\left(\begin{array}{c}m-1\\ k-1\end{array}\right)$

possible distributions of wins and losses into runs.

So by the formula for probability on the sample spaces of equally likely outcomes:

$P(2k+1)=\frac{\left(\begin{array}{c}n-1\\ k\end{array}\right)\left(\begin{array}{c}m-1\\ k-1\end{array}\right)+\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}m-1\\ k\end{array}\right)}{\left(\begin{array}{c}n+m\\ n\end{array}\right)}$