Q. 3.13 - A First Course in Probability

Question

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

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

Step-by-Step Solution

Answer

$P_{n,m}$ is the probability that the first player flips $n$heads in total before the second one flip$m$.

Concentrate on the first flip, the rest of the game is equivalent to a new game with possibly changed order of the players.

Step1: Probability of getting heads

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:

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

Step2: Using the probability formula.

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,

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

Step3: Prove the statement.

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.