Let [tex]W[/tex] 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 [tex]W[/tex] is
[tex]\mathrm{Pr}(W=w) = \begin{cases} \dfrac1{13} & \text{if } w \in \{-\$0.50, \$1.50, \$2.75\} \\\\ \dfrac3{52} & \text{if } w = \$4.75 \\\\ \dfrac1{52} & \text{if } w = \$6.00 \\\\ \dfrac{10}{13} & \text{if } w = -\$1.75 \\\\ 0 & \text{otherwise} \end{cases}[/tex]
The expected winnings from playing one round of this game are
[tex]\Bbb E[W] = \displaystyle \sum_w w\,\mathrm{Pr}(W=w)[/tex]
[tex]\Bbb E[W] = \dfrac{-\$0.50 + \$1.50 + \$2.75}{13} + \dfrac{3\cdot\$4.75}{52} + \dfrac{\$6.00}{52} + \dfrac{10\cdot(-\$1.75)}{13}[/tex]
[tex]\Bbb E[W] \approx \boxed{-\$0.67}[/tex]
