Question 698017: Roll a die. If you roll a 6 then you win $100, if you do not roll a 6 then you get to roll again. If you roll a 6 then you win,If you do not roll a 6 on the second try you loose and game stops.
What is the expected value of this game?
Answer by Positive_EV(69)   (Show Source):
You can put this solution on YOUR website! The expected value of the game is equal to the sum of the products of each outcome and its respective probability. You don't specify how much you lose if you do not roll a 6 at any point - this is necessary for the calculation. For the purposes of this problem I will assume that a loss has a result of $0, and a win has a result of $100.
The probability that you roll a 6 on either roll is equal to the probability that you roll a 6 on the first roll + the probability that you roll a 6 on the second roll. If the die has 6 sides, there is 1 side out of 6 that wins, so the probability of rolling a six on the first roll is 1/6.
The probability that you roll a 6 on the second roll is also 1/6; however, if you roll a 6 on the first roll you do not need to roll twice. The probability of getting a 6 on the second roll is thus the probability that you needed a second roll times the probability that you roll a six on the second roll; that is, the probability that the first roll is not a 6 and the second roll is. For the first roll, there are 5 non-6 numbers out of 6, so the probability that you need a second roll is 5/6. Therefore, the probability that you win on the second roll is (5/6)*(1/6) = 5/36.
The probability that you win is the sum of these probabilities, 1/6 + 5/36 = 11/36. The event that you lose is the compliment of the event that you win, so its probability is 1 - P(win) = 1 - 11/36 = 25/36.
Last, the expected value is the sum of the products of the results and their probabilities, which is (11/36)*100 + (25/36)*0 = ~$30.56, assume you lose nothing if you lose. If you lose some dollar amount $X when you lose, this formula changes to (11/36)*100 + (25/36)*(-X).
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