Find the probability that an averge player wins ten or more times in 25 hands.
In the game of blackjack as played in casinos in Las Vegas, Atlantic City, Niagara falls, as well as many other cities, the dealer has the advantage. Most players do not play very well. As a result, the probability that the average player wins a hand is about 45%.
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The probability that a player wins ten or more times in 25 hands is equal to 1 - (probability that a person wins 0-9 times)
The probability that a person wins in a single hand is 45%. The probability that a person loses in a single hand is 55%. In 25 games the probability that a person wins n games is 25Cn*(0.45)^n*(0.55)^(25 - n)
`sum_(n = 0)^9 25Cn*(0.45)^n*(0.55)^(25-n) = 0.242371`
This gives the probability for winning 0 to 9 times as 0.242371
The probability that an average player wins ten hands or more is: 1 - 0.242371 = 0.757629
The required probability is 0.757629
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