# Related To Binomial Distribution

1. A blackjack player at a Las Vegas casino learned that the house will provide a free room if play is for four hours at an average bet of $50. The player’s strategy provides a probability of .49 of winning on any one hand, and the player knows that there are 60 hands per hour. Suppose the player plays for four hours at a bet of 50 per hand. a. What is the player’s expected payoff? b. What is the probability the player loses$1000 or more?

c. What is the probability the player wins?

d. Suppose the player starts with $1500. What is the probability of going broke? ## 2 Answers Just a few quick thoughts:$n = 240$(total hands) a. Let$Y = 50x ~ E[X] = np ~ E[Y] = 50np$b.$W + L = 240$To lose exactly 1000, you must lose 20 more than you win, so$W + 20 = L$Which sets up a binomial with$\binom{240}{110}.49^{110}\cdot .51^{130}$That's for losing exactly 1000. Losing 1000 or more... that's a lot of binomials to add up. I am wondering if poisson approximation to the binomial would work. I am just taking 5 minutes to look at this during a study break, but that's a start... huh? • It occurred to me later that the normal approximation to the binomial is the way to go, as explained by @jay-sun. Agreed. – Katyjean57 Mar 17 '13 at 6:53 a) Expected payoff = bet per hand*$(E_{profit}-E_{loss})$=$50*(240*p-240*(1-p))$where$p=0.49$b)To lose \$$1000, the player must lose 20 more hand than he wins. So he needs to win 110 hands or less in order to lose \$$1000$. Thus, you need to find $P(X\leq 110)$ where you approximate $X$ as normal by using the $\mu$ and $\sigma$ from the binomial distribution $X$. I am hoping you can find that.

c) Again as in (b), the player wins if he wins 121 hands or more, i.e. $P(X \geq 121)$

d)Use the same technique as in (b)