I draw a card from a standard $52$-card deck. If I draw an Ace, I win $1$ dollar. If I draw a $2$ through $10$, I win a number of dollars equal to the value of the card. If I draw a face card (Jack, Queen, or King), I win $10$ dollars. If I draw a $\clubsuit$, my winnings are subtracted, and if I draw a $\spadesuit$, then nothing happens. If I draw a diamond, I get $25$ dollars. (For example, if I draw the $8\clubsuit$, then I lose $16$ dollars.) What would be a fair price to pay to play the game? Express your answer as a dollar value rounded to the nearest cent.
Your answer should be a number with two digits after the decimal point, like $21.43$.
We find the expected value of one play of the game. We do so by calculating the expected value of each card and summing them. The probability of drawing each card is $\dfrac{1}{52}$.
The expected value for drawing an Ace is equal to the winnings, which is $1$ dollar.
The expected value for drawing a $2$ through $10$ is equal to the winnings, which can be calculated by taking the average of the numbers from $2$ to $10$, giving us $\dfrac{2+3+4+5+6+7+8+9+10}{9}=6$.
The expected value for drawing a face card is equal to the winnings, which is $10$ dollars.
The expected value for drawing a $\clubsuit$ is equal to the negative winnings, which is $-1$.
Similarly, the expected value for drawing a diamond is equal to the winnings, which is $25$ dollars.
Since drawing the $\spadesuit$ doesn't change the amount of money won, the expected value for drawing a $\spadesuit$ is equal to the winnings, which is $0$ dollars.
Summing up all these individual expected values, we get that the total expected value is
\begin{align*}
&\left(\dfrac{1}{52} \cdot 1 \right) \\
+ &\left(\dfrac{9}{52} \cdot 6 \right) \\
+ &\left(\dfrac{12}{52} \cdot 10 \right) \\
+ &\left(\dfrac{4}{52} \cdot -1 \right) \\
+ &\left(\dfrac{13}{52} \cdot 25 \right) \\
+ &\left(\dfrac{13}{52} \cdot 0 \right) \\
&= \dfrac{1}{52} \cdot 1 + \dfrac{9}{52} \cdot 6 + \dfrac{12}{52} \cdot 10 - \dfrac{4}{52} \cdot 1 + \dfrac{13}{52} \cdot 25 \\
&= \dfrac{1}{52} + \dfrac{9 \cdot 6}{52} + \dfrac{12 \cdot 10}{52} - \dfrac{4}{52} + \dfrac{13 \cdot 25}{52} \\
&= \dfrac{1 + 54 + 120 - 4 + 325}{52} \\
&= \dfrac{496}{52} \\
&= 9.54,
\end{align*}rounded to the nearest cent. Therefore, a fair price to pay to play this game is $\boxed{\text{\$}9.54}$.
To find the fair price to play the game, we need to calculate the expected value (EV) of playing the game.
The probability of drawing each type of card is as follows:
- Probability of drawing an Ace: $\dfrac{4}{52}$
- Probability of drawing a $2$ through $10$: $\dfrac{9}{52}$
- Probability of drawing a face card: $\dfrac{12}{52}$
- Probability of drawing a $\clubsuit$: $\dfrac{13}{52}$
- Probability of drawing a $\spadesuit$: $\dfrac{13}{52}$
- Probability of drawing a diamond: $\dfrac{13}{52}$
Now, let's calculate the expected value step-by-step for each card:
For an Ace, the winnings are $1$ dollar. So, the expected value for drawing an Ace would be:
$\text{EV}_\text{Ace} = 1 \cdot \dfrac{4}{52}$
For cards $2$ through $10$, the winnings are equal to the value of the card. So, the expected value for drawing a $2$ through $10$ would be:
$\text{EV}_\text{2-10} = \left(2+3+4+5+6+7+8+9+10\right) \cdot \dfrac{9}{52}$
For a face card, the winnings are $10$ dollars. So, the expected value for drawing a face card would be:
$\text{EV}_\text{Face} = 10 \cdot \dfrac{12}{52}$
For a $\clubsuit$, the winnings are subtracted. So, the expected value for drawing a $\clubsuit$ would be:
$\text{EV}_\text{Clubsuit} = -\left(\text{total winnings}\right) \cdot \dfrac{13}{52}$
For a $\spadesuit$, nothing happens, so the expected value for drawing a $\spadesuit$ is $0$.
For a diamond, the winnings are $25$ dollars. So, the expected value for drawing a diamond would be:
$\text{EV}_\text{Diamond} = 25 \cdot \dfrac{13}{52}$
Finally, we can find the total expected value (EV) by summing up the individual expected values:
$\text{EV} = \text{EV}_\text{Ace} + \text{EV}_\text{2-10} + \text{EV}_\text{Face} + \text{EV}_\text{Clubsuit} + \text{EV}_\text{Diamond} + \text{EV}_\text{Spadesuit}$
Now, let's calculate the overall expected value (EV):
\begin{align*}
\text{EV} &= 1 \cdot \dfrac{4}{52} + \left(2+3+4+5+6+7+8+9+10\right) \cdot \dfrac{9}{52} + 10 \cdot \dfrac{12}{52} + \left(-\text{total winnings}\right) \cdot \dfrac{13}{52} + 25 \cdot \dfrac{13}{52} + 0 \cdot \dfrac{13}{52}\\
&= \dfrac{4}{52} + \dfrac{54}{52} + \dfrac{120}{52} - \dfrac{\text{total winnings}}{4} + \dfrac{325}{52}\\
&= \dfrac{503}{52} - \dfrac{\text{total winnings}}{4}
\end{align*}
Now, since we want the game to be fair, the expected value (EV) should be $0$. So, we can set $\text{EV} = 0$ and solve for the total winnings:
\begin{align*}
\dfrac{503}{52} - \dfrac{\text{total winnings}}{4} &= 0\\
\implies \dfrac{\text{total winnings}}{4} &= \dfrac{503}{52}\\
\implies \text{total winnings} &= 4 \cdot \dfrac{503}{52}\\
&\approx 38.5
\end{align*}
Therefore, the fair price to play the game would be approximately $\boxed{38.5}$ dollars.