A gambler plays 60 hands of poker, 30 hands of blackjack, and 10 hands of canast

Question

A gambler plays 60 hands of poker, 30 hands of blackjack, and 10 hands of canasta per day. He wins a hand of poker with probability 1/6, a hand of blackjack with probability 1/2, and a hand of canasta with probability 1/5.

1. What is the expected number of hands the gambler wins in a day?

2. Assume the outcomes of the card games are pairwise independent. What is the variance in the number of hands won per day?

3. What is Chebyshev bound on the probability that the gambler will win at least 50 hands on a given day?

Explanation / Answer

Expected number of poker hands win = 60*1/6 =10

Expected number of black jack win = 30*1/2 =15

Expected number of canasta win = 10*1/5 =2

total expected number of hands win = 10+15+2 = 27

b) variance of number of hands per day = (10*5/6)+(15*1/2)+(2*4/5) = 17.43

c) X- mean /sd = 1.32; P(z>1.32) = 0.09

1/k^2 = 0.09

Therefore k = 3.27

cheyshev bound on the probability that the gamber will win at least 50 hand on a given dat a =3.27

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