2. A game of chance costs $5 to play and consists of rolling a five fair...

Question

Question

2. A game of chance costs $5 to play and consists of rolling a five fair dice. If at least four of the dice shows a number (s

Answer 1

Answer #1

Probability of getting a number greater than 2 in roll of a dice = 4/6 = 2/3

Thus, probability of getting a number greater than 2 in atleast four of the five dice = P(X = 4) + P(X = 5)

= (2)*(2/3)*(1/3)+(2/3)

= 0.4609

(a) Expected net winnings from one game = $10*0.4609 - $5

= -$0.391

(b) Expected number of games played until the gambler loses a total of four games = 4/(1 - 0.4609) = 7.42

Thus, expected net loss under this strategy = $0.391*7.42

= $2.9

Answer 2

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