Inactive Tutor answered 05/19/21
Player win Probability Carnival gain
10 1/10 -7 (3-10)
5 2/10 -2 (3-5)
1 7/10 2 (3-1)
So expected gain for carnival = (-7) (1/10) + (-2) (2/10) + 2 (7/10) = [ -7 -4 + 14] / 10 = 3/10 = $ 0.30
Ma F.
asked 05/19/21A carnival game involves a box which contains one $10 coin, two $5 coins, and seven $1 coins. The player is allowed to remove one coin at random from the box and keep it. If it costs $3 to play this game, how much (on the average) does the carnival gain on each play?
Inactive Tutor answered 05/19/21
Player win Probability Carnival gain
10 1/10 -7 (3-10)
5 2/10 -2 (3-5)
1 7/10 2 (3-1)
So expected gain for carnival = (-7) (1/10) + (-2) (2/10) + 2 (7/10) = [ -7 -4 + 14] / 10 = 3/10 = $ 0.30
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