Question 1099170: Suppose you play a game of chance in which you toss a coin 3 times. You pay a nor refundable $2 to play the game. You win $10 in case your coin lands with a tail on top during the three tosses. You win nothing otherwise.
a) Fill out the following probability distribution table.
b) Over the long term, what is your expected value of playing the game?
c) If you played this game 100 times, how much would you expect win/loose?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
What does "lands with a tail on top during the three tosses" mean?
It sounds as if it means tails at least once on the three tosses. If that is the case, I want to know where I can play this game, because I'm going to win 7 out of 8 times. The only way not to win is to get heads on all three tosses; the probability of that is (1/2)(1/2)(1/2) = 1/8; so the probability of winning is 7/8.
If that is indeed the rule, then when you play the game 8 times you pay $2 each time (-$16); you win 7 times (+$70), and you lose 1 time (no change). So for each 8 games your expected net would be +$54. So the expected value of each game is $54/8 = $6.75.
I don't think it is likely that that is the way the problem is supposed to go. I suspect the rule is you win if you get tails on ALL THREE tosses. Then the probability of winning is 1/8 and the probability of losing is 7/8.
Then for 8 games you would again be paying $16, but you would only be winning $10 once. So you would lose $6 every 8 games; that means the expected value for each game would be -$6/8 = -$0.75.
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