Question 1194820: A game is played as follows: A player places a bet. Then, she rolls a 20-sided die with the numbers
1 - 20 on it. If the die lands on an even number, the player wins $5. If the die lands on a 1, 5, 7, or 13,
the player wins $10. Otherwise, the player loses. If it costs $3.50 to bet, what is the expected value of
this game?
Found 2 solutions by shlomitg, math_tutor2020:
Answer by shlomitg(8)   (Show Source):
You can put this solution on YOUR website! Expected value is $1.0. The expected win is 0*(6/20)+10*(4/20)+5*(10/20) = $4.5; Deducting the game cost which is $3.5, the expected value of the game is $1.0
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
Possible Outcomes:
A = Rolling an even number
B = Rolling a 1, 5, 7 or 13
C = Rolling anything else not mentioned by the previous cases
Here are the various event spaces
A = {2,4,6,8,10,12,14,16,18,20}
B = {1,5,7,13}
C = {3,9,11,15,17,19}
And here are the corresponding probabilities
P(A) = 10/20 = 1/2
P(B) = 4/20 = 1/5
P(C) = 6/20 = 3/10
Thing to notice: P(A)+P(B)+P(C) = 1
The probabilities must sum to 1 in order to have a valid probability distribution.
Also, the probabilities must be in the interval 
Let X = net winnings
This will be the result of subtracting the prize money and the cost to play the game.
For event A, we have X = 5-3.5 = 1.50
For event B, we have X = 10-3.5 = 6.50
For event C, we have X = 0-3.5 = -3.50
Let's set up a table like so
| Event | Net Winnings | Probability | | A | $1.50 | 1/2 | | B | $6.50 | 1/5 | | C | -$3.50 | 3/10 |
Now add on a third column where we multiply the net winnings with their corresponding probability
For example: 1.5*1/2 = 0.75 in the first row.
| Event | Net Winnings | Probability | Winnings*Probability | | A | $1.50 | 1/2 | 0.75 | | B | $6.50 | 1/5 | 1.30 | | C | -$3.50 | 3/10 | -1.05 |
Adding the results of the third column will get us to the final answer
0.75 + 1.30 + (-1.05) = 1.00
Answer: $1.00 is the expected winnings.
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