Question

Suppose that two players A and B are trying to throw a basketball through a hoop. The probability that player A will succeed on any given throw is p, and he throws until he has succeeded r times. The probability that player B will succeed on any given throw is mp, where m is a given integer (m = 2, 3, . . .) such that mp < 1, and she throws until she has succeeded mr times.

a. For which player is the expected number of throws smaller?

b. For which player is the variance of the number of throws smaller?

Short answer

(a) Expected number of throws is the same.

(b) Variance of Player B is smaller.

Step-by-step solution

Given information

(a)

Two players A and B are playing basketball.

The probability of success of A at any throw is p. r success is needed for A to win the game.

The probability of success of B at any throw is mp. mr success are needed for A to win the game.

Defining the pdf of A and B

Generally, the pdf of a negative binomial distribution is

\[f\left( {x|r,p} \right) = \left\{ \begin{array}{l}{}^{r + x - 1}{C_x}{p^r}{\left( {1 - p} \right)^x}\;for\;x = 0,1,2\\0,\;otherwise\end{array} \right.\]

The above pdf is the same as the pdf of player A.

For player B, the pdf is

\[f\left( {y|mr,mp} \right) = \left\{ \begin{array}{l}{}^{mr + y - 1}{C_y}{\left( {mp} \right)^{mr}}{\left( {1 - mp} \right)^y}\;for\;y = 0,1,2\\0,\;otherwise\end{array} \right.\]

Step 3: Calculating the expectation

For a negative binomial distribution, the expectation and variance are defined as:

\(\begin{array}{l}E\left( X \right) = \frac{{r\left( {1 - p} \right)}}{p}\\V\left( X \right) = \frac{{r\left( {1 - p} \right)}}{{{p^2}}}\end{array}\)

Player A:

\[\begin{array}{c}{Y_a} = {X_a} + r\\ \Rightarrow E\left( {{Y_a}} \right) = E\left( {{X_a}} \right) + r\\ \Rightarrow E\left( {{Y_a}} \right) = \left( {\frac{r}{p} - r} \right) + r\\ \Rightarrow E\left( {{Y_a}} \right) = \frac{r}{p}\end{array}\]

Therefore, expected number of the throws is the same for both the players.

Calculating the variance

(b)

\[\begin{array}{c}{Y_a} = {X_a} + r\\ \Rightarrow V\left( {{Y_a}} \right) = V\left( {{X_a}} \right) + 0\\ \Rightarrow V\left( {{Y_a}} \right) = \frac{r}{{{p^2}}} - \frac{r}{p}\end{array}\]

The variance of Player B is smaller.