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?


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?

  • $\begingroup$ It occurred to me later that the normal approximation to the binomial is the way to go, as explained by @jay-sun. Agreed. $\endgroup$ – 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)


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