Question

R6.3. Keno In a game of 4-Spot Keno, the player picks 4 numbers from 1 to 80. The casino randomly selects 20 winning numbers from 1 to 80. The table below shows the possible outcomes of the game and their probabilities, along with the amount of money (Payout) that the player
wins for a \(1 bet. If X = the payout for a single \)1 bet, you can check that

${\mu }_X=\backslash (0.70$and ${\sigma }_X=\backslash )6.58$.

Matches:	0	1	2	3	4
Probability:	0.308	0.433	0.213	0.043	0.003
Payout:	\(0	\)0	\(1	\)3	\(120

(a) Interpret the values of ${\mu }_X$and ${\sigma }_X$ in context.
(b) Jerry places a single $\backslash )5$bet on $4$-Spot Keno. Find the expected value and the standard deviation of his winnings.
(c) Marla plays five games of 4-Spot Keno, betting $1 each time. Find the expected value and the standard deviation of her total winnings.
(d) Based on your answers to (b) and (c), which player would the casino prefer? Justify your answer.

Short answer

(a) The mean of $X$indicates that the predicted pay out is $\$0.70$on average, and that the payout will vary by $\$6.58$on average from the payout of $\$0.70$.
(b) The mean and standard deviation are $\$3.50$and $\$32.90$respectively.

(c) The mean and standard deviation are $\$3.50$and $\$14.71$respectively.

(d) Marla would prefer the casino.

Step-by-step solution

Part (a) Step 1: Given information

To interpret the values of ${\mu }_X$and ${\sigma }_X$ in context.

Part (a) Step 2: Explanation

Since, the given values are:
${\mu }_X=\$0.70$
${\sigma }_X=\$6.58$
Determine the mean and standard deviation as:
${\mu }_{aX+b}=a{\mu }_X+b$
${\sigma }_{aX+b}=a{\sigma }_X$

The mean of $X$indicates that the predicted pay out is $\$0.70$on average, and that the payout will vary by $\$6.58$on average from the payout of $\$0.70$.

Part (b) Step 1: Given information

To find the expected value and the standard deviation of his winnings. Let, Jerry places a single $\$5$ bet on $4$-Spot Keno.

Part (b) Step 2: Explanation

Determine the expected value and standard deviation as follows:

${\mu }_{5X}=5\times {\mu }_X$

$=5(0.70)$

$=\$3.50$

${\sigma }_{5X}=5\times {\sigma }_X$

$=5(6.58)$

$=\$32.90$

As a result, the mean and standard deviation are $\$3.50$and$\$32.90$respectively.

Part (c) Step 1: Given information

To find the expected value and the standard deviation of the total winnings. Let, Marla plays five games of $4-$Spot Keno, betting $\$1$each time.

Part (c) Step 2: Explanation

Determine the expected value and standard deviation as follows:
${\mu }_{X_1+X_2+\dots +X_5}=5\times {\mu }_X$

$=5(0.70)$

$=\$3.50$

$\sqrt{{\sigma }^2{X}_1+X_2+\dots +X_5}=\sqrt{5\times {\sigma }^{2}x}$

$=\sqrt{5(6.58{)}^2}$

$=\$14.71$

As a result, the mean and standard deviation are $\$3.50$and $\$14.71$respectively.

Part (d) Step 1: Given information

To justify the answer that based on the answers to (b) and (c), which player would the casino prefer.

Part (d) Step 2: Explanation

Since the result from part (b) as for Jerry:
${\mu }_{5X}=\$3.50$
${\sigma }_{5X}=\$32.90$
And the result from part (c) as for Marla:
$\mu =\$3.50$
$\sigma =\$14.71$
Because both players have the same mean, the standard deviation will determine the outcome.
A casino wants a player who can best forecast their wins and losses, and hence the person with the lowest standard deviation, which corresponds to the player Marla.