Answer:
Expected value = 40/26 = 1.54 approximately
The player expects to win on average about $1.54 per game.
The positive expected value means it's a good idea to play the game.
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Further Explanation:
Let's label the three scenarios like so
- scenario A: selecting a black card
- scenario B: selecting a red card that is less than 5
- scenario C: selecting anything that doesn't fit with the previous scenarios
The probability of scenario A happening is 1/2 because half the cards are black. Or you can notice that there are 26 black cards (13 spade + 13 club) out of 52 total, so 26/52 = 1/2. The net pay off for scenario A is 2-1 = 1 dollar because we have to account for the price to play the game.
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Now onto scenario B.
The cards that are less than five are: {A, 2, 3, 4}. I'm considering aces to be smaller than 2. There are 2 sets of these values to account for the two red suits (hearts and diamonds), meaning there are 4*2 = 8 such cards out of 52 total. Then note that 8/52 = 2/13. The probability of winning $10 is 2/13. Though the net pay off here is 10-1 = 9 dollars to account for the cost to play the game.
So far the fractions we found for scenarios A and B were: 1/2 and 2/13
Let's get each fraction to the same denominator
Then add them up
13/26 + 4/26 = 17/26
Next, subtract the value from 1
1 - (17/26) = 26/26 - 17/26 = 9/26
The fraction 9/26 represents the chances of getting anything other than scenario A or scenario B. The net pay off here is -1 to indicate you lose one dollar.
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Here's a table to organize everything so far
![\begin{array}{|c|c|c|}\cline{1-3}\text{Scenario} & \text{Probability} & \text{Net Payoff}\\ \cline{1-3}\text{A} & 1/2 & 1\\ \cline{1-3}\text{B} & 2/13 & 9\\ \cline{1-3}\text{C} & 9/26 & -1\\ \cline{1-3}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Cc%7Cc%7C%7D%5Ccline%7B1-3%7D%5Ctext%7BScenario%7D%20%26%20%5Ctext%7BProbability%7D%20%26%20%5Ctext%7BNet%20Payoff%7D%5C%5C%20%5Ccline%7B1-3%7D%5Ctext%7BA%7D%20%26%201%2F2%20%26%201%5C%5C%20%5Ccline%7B1-3%7D%5Ctext%7BB%7D%20%26%202%2F13%20%26%209%5C%5C%20%5Ccline%7B1-3%7D%5Ctext%7BC%7D%20%26%209%2F26%20%26%20-1%5C%5C%20%5Ccline%7B1-3%7D%5Cend%7Barray%7D)
What we do from here is multiply each probability with the corresponding net payoff. I'll write the results in the fourth column as shown below
![\begin{array}{|c|c|c|c|}\cline{1-4}\text{Scenario} & \text{Probability} & \text{Net Payoff} & \text{Probability * Payoff}\\ \cline{1-4}\text{A} & 1/2 & 1 & 1/2\\ \cline{1-4}\text{B} & 2/13 & 9 & 18/13\\ \cline{1-4}\text{C} & 9/26 & -1 & -9/26\\ \cline{1-4}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cc%7Cc%7Cc%7Cc%7C%7D%5Ccline%7B1-4%7D%5Ctext%7BScenario%7D%20%26%20%5Ctext%7BProbability%7D%20%26%20%5Ctext%7BNet%20Payoff%7D%20%26%20%5Ctext%7BProbability%20%2A%20Payoff%7D%5C%5C%20%5Ccline%7B1-4%7D%5Ctext%7BA%7D%20%26%201%2F2%20%26%201%20%26%201%2F2%5C%5C%20%5Ccline%7B1-4%7D%5Ctext%7BB%7D%20%26%202%2F13%20%26%209%20%26%2018%2F13%5C%5C%20%5Ccline%7B1-4%7D%5Ctext%7BC%7D%20%26%209%2F26%20%26%20-1%20%26%20-9%2F26%5C%5C%20%5Ccline%7B1-4%7D%5Cend%7Barray%7D)
Then we add up the results of that fourth column to compute the expected value.
(1/2) + (18/13) + (-9/26)
13/26 + 36/26 - 9/26
(13+36-9)/26
40/26
1.538 approximately
This value rounds to 1.54
The expected value for the player is 1.54 which means they expect to win, on average, about $1.54 per game.
Therefore, this game is tilted in favor of the player and it's a good decision to play the game.
If the expected value was negative, then the player would lose money on average and the game wouldn't be a good idea to play (though the card dealer would be happy).
Having an expected value of 0 would indicate a mathematically fair game, as no side gains money nor do they lose money on average.