Question

A casino offers the following game: you draw one card from a standard 52-card deck. If you draw a jack, you win $1.25. If you draw a queen, you win $3.25. If you draw a king, you win $4.5 dollars. If you draw any ace except the ace of spades, you win $6.5. If you draw the ace of spades, you win $7.75. The entry fee to play this game is $1.75. Compute the expected value of this gamble (include the entry fee in your expected value).

Answer 1

Let
`W` be the random variable for the winnings from playing the game once.

• There are 4 jacks in the deck, so you draw a jack with probability 4/52 = 1/13. In this case you "win" $1.25 - $1.75 = -$0.50.

• There are 4 queens, with draw probability 4/52 = 1/13 and winnings $3.25 - $1.75 = $1.50.

• There are 4 kings, with draw probability 4/52 = 1/13 and winnings $4.50 - $1.75 = $2.75.

• There are 4 aces, and 3 of these are not of the spade suit, so the probability of drawing any of these is 3/52 and you win $6.50 - $1.75 = $4.75.

• There is only 1 ace of spaces, with draw probability 1/52 and winnings $7.75 - $1.75 = $6.00.

• Adding these up, it follows that the probability of drawing any other card is 1 - (1/13 + 1/13 + 3/52 + 1/52) = 10/13, in which you have the privilege of "winning" -$1.75.

So, the probability mass function for
`W` is

`\mathrm{Pr}(W=w) = \begin{cases} \frac1{13} & \text{if } w \in \{-\$0.50, \$1.50, \$2.75\} \\\\ \frac3{52} & \text{if } w = \$4.75 \\\\ \frac1{52} & \text{if } w = \$6.00 \\\\ (10)/(13) & \text{if } w = -\$1.75 \\\\ 0 & \text{otherwise} \end{cases}`

The expected winnings from playing one round of this game are

`\Bbb E[W] = \displaystyle \sum_w w\,\mathrm{Pr}(W=w)`

`\Bbb E[W] = (-\$0.50 + \$1.50 + \$2.75)/(13) + (3\cdot\$4.75)/(52) + (\$6.00)/(52) + (10\cdot(-\$1.75))/(13)`

`\Bbb E[W] \approx \boxed{-\$0.67}`