SOLUTION: A popular casino game is three-card poker. One aspect of the game is the "pair plus" bet in which a player is paid a dollar amount for any hand of a pair or better. The table sho
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Question 1197105: A popular casino game is three-card poker. One aspect of the game is the "pair plus" bet in which a player is paid a dollar amount for any hand of a pair or better. The table shows the profit and probability for all of the winning hands for a player paying $5 to play this game:
Outcome Profit ($) Probability
Straight Flush 200 0.00217
Three of a kind 150 0.00235
Straight 30 0.03258
Flush 20 0.04959
Pair 5 0.16941
For any hand other than those listed in the table, the player loses the $5 bet.
Calculate the expected profit for a $5 bet in this game (answer to the nearest penny).
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Answer: -0.12 dollars
The player expects to lose about 12 cents per game.
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Explanation:
X = profit in dollars
P(X) = probability of getting that profit
Outcome | X | P(X) | X*P(X) |
Straight Flush | 200 | 0.00217 | 0.434 |
Three of a kind | 150 | 0.00235 | 0.3525 |
Straight | 30 | 0.03258 | 0.9774 |
Flush | 20 | 0.04959 | 0.9918 |
Pair | 5 | 0.16941 | 0.84705 |
Any other hand | -5 | 0.7439 | -3.7195 |
Note the last outcome "any other hand" has probability of 1 - (sum of the other probabilities)
All of the P(X) values must add to 1.
The expected value is the sum of the X*P(X) values
0.434 + 0.3525 + 0.9774 + 0.9918 + 0.84705 + (-3.7195) = -0.11675
This rounds to -0.12 when rounding to the nearest penny.
The player expects to profit about -0.12 dollars per game.
In other words, the player expects to lose about $0.12 (aka 12 cents) per game.
This is not a mathematically fair game since the expected value isn't zero.
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