In the game of blackjack as played in casinos in Las Vegas, Atlantic City, Niaga

Question

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 0.37. Find the probability that an average player wins twice in 5 hands. Probability = 10 or more times in 25 hands. Probability = .4517 There are several books that teach blackjack players the "basic strategy" which increases the probability of winning any hand to 0.49. Assuming that the player plays the basic strategy, find the probability that he or she wins twice in 5 hands. Probability = .3185 10 or more times in 25 hands. Probability = .8646

Explanation / Answer

Solution

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and

p = probability of one success, then

probability mass function (pmf) of X is given by

p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n ……………..(1)

Now, to work out solution, [all probabilities are obtained using Excel Function]

Let X = number of wins in n hands. Then, X ~ B(n, p), p being the probability of winning a hand.

Part (a) n = 5, p = 0.37 (given), x = 2.

P(X = 2) = 0.3423 ANSWER

Part (b) n = 25, p = 0.37 (given), x 10.

P(X 10) = 1 - P(X 9) = 1 – 0.5483 = 0.4517 ANSWER

Part (c) n = 5, p = 0.49 (given), x = 2.

P(X = 2) = 0.3185 ANSWER

Part (d) n = 25, p = 0.49 (given), x 10.

P(X 10) = 1 - P(X 9) = 1 – 0.1355 = 0.8645 ANSWER

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