Question 994021:
5. In the game of blackjack as played in casinos, the dealer has the advantage as most of the players do not play very well. As a result, the probability that the average player wins a hand is about 40%. An average player decides to play 10 hands. (15 points)
f. What is the probability that he will win in all hands? (3 points)
g. What is the probability that he will win at least 5 hands? (3 points)
h. What is the probability that he will win at least 1 hand? (3 points)
i. Assuming that this player has already lost the first 6 hands, what is the probability that he will win all remaining (4) hands? (3 points)
j. What is the average number of hands that an average player should expect to win in a 10-hand game? (3 points)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 0.4^10=0.0001
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Probability wins 0=0.6^10=0.006
Probability wins 1=10*(0.4)*(0.6)^9=0.0403
wins 2= 10C2(0.4)^2)(0.6)^8=0.1209
wins 3=10C3(0.4)^3(0.6)^7=0.2150
wins 4=10C4(0.4)^4(0.6)^6=0.2508
These sum to 0.6330 This is what we don't want. The probability of winning at least 5 is everything else or 1-0.6330=0.3670
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This is the probability of 1-probability of winning no hands, and that is 0.994
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That would be 0.4^4, since this is considered independent=0.0256
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Expected value is 10*-0.4 or 4 hands.
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