The probability of getting ahead on a single toss of a coin is $p$

Suppose that $A$starts and continues to flip the coin until a tail shows up, at which point $B$starts flipping.

Then $B$continues to flip until a tail comes up, at which point $A$takes over, and so on.

Let $P_{n,m}$denote the probability that $A$accumulates a total of n heads before $B$accumulates $m$

Show that

$P_{n,m}=p{P}_{n-1,m}+(1-p)\left(1-P_{m,n}\right)$

Given:

$A$and $B$engage in a game. $A$is the first to leave.

With probability $p$, any of them turns head with probability $(1-p)$, any one of them flips tails with probability $(1-p)$, and the turns are lost.

$P_{n,m}$- Chances that $A$will amass n heads before $B$will accumulate m

prove:

$P_{n,m}=p\times P_{n-1,m}+(1-p)\left(1-P_{m,n}\right)$

The following are the occasions:

$A_{n,m}$-event in which $A$flips n heads prior flipping $m$

Heads were the first to flip.

Using the total probability formula,

$$\begin{gathered} P_{n,m}=P\left(A_{n,m}\right) \\ =P\left(A_{n,m}\mid H\right)P\left(H\right)+P\left(A_{n,m}\mid H^c\right)P\left(H^c\right) \end{gathered}$$

Given that $H$occurred, $A$proceeds to toss with the same probability as before - similar to a fresh game in which $A$starts first and, since $A$has already flipped one head, $A$must flip $n-1$more before $B$flip $m$of them.

$$\begin{gathered} P\left(A_{n,m}\mid H\right)=P\left(A_{n-1,m}\right) \\ =:P_{n-1,m} \end{gathered}$$

$=P\left(A_{n,m}\mid H^c\right)$

Given that $H^c$occurred, the situation is analogous to starting afresh game with $B$as the first player, both with 0 heads. The probability that $B$- the first player achieves $m$heads before the other achieves n heads is$P_{m,n}$and this is the complement of the desired occurrence.

$P\left(A_{n,m}\mid H^c\right)=1-P_{m,n}$

Substitute this $P\left(H\right)=p,P\left(H^c\right)=1-p$to the equations, the formula is proved.