How to identify if the problem is a sample deviation problem or a population deviation problem?
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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
- 1/2 = 13/26
- 2/13 = 4/26
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
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
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.