You are playing a die game. If you roll 1, then the house pays you $25. If you roll 2, the house pays you $5. If you roll 3, you win nothing. If you roll a 4 or a 5, you must pay the house $10, and if you roll a 6, you must pay the house $15. What is the expected value of this game?
EV=sum (probability*outcome)
= 1/6 (25+5+0-10-10-15)
do the math.
To calculate the expected value of the game, we need to multiply each outcome by its corresponding probability and sum them up.
Let's begin by determining the probability of rolling each number on the die:
- The probability of rolling a 1 is 1/6 (since there is only one 1 on a standard six-sided die).
- Similarly, the probability of rolling a 2 is also 1/6.
- The probability of rolling a 3 is 1/6.
- The probability of rolling a 4 or 5 is 2/6 or 1/3 (since there are two numbers out of six that satisfy this condition).
- The probability of rolling a 6 is also 1/6.
Now let's determine the expected value for each outcome:
- If you roll a 1, you win $25. So the expected value for this outcome is (1/6) * $25 = $25/6.
- If you roll a 2, you win $5. So the expected value is (1/6) * $5 = $5/6.
- If you roll a 3, you win nothing. Hence the expected value is 0.
- If you roll a 4 or a 5, you lose $10. So the expected value is (1/3) * -$10 = -$10/3.
- If you roll a 6, you lose $15. So the expected value is (1/6) * -$15 = -$15/6.
Finally, let's sum up all the expected values:
Expected value = ($25/6) + ($5/6) + 0 + (-$10/3) + (-$15/6)
Expected value = $4.17 - $1.67 - $2.50
Expected value = $0
Therefore, the expected value of this game is $0. This means that, on average, you neither win nor lose money in the long run.