# Probability - Roulette

1. Aug 25, 2009

### nicholasch

1. The problem statement, all variables and given/known data

This particular roulette wheel has 37 slots - 0,1,2...36. A gambler can bet on different combinations of numbers. Louise loves to bet on a block of 4 numbers, called a corner. The payout on a corner is 8 to 1.

(a) Let X be a gambler's winning from a $1 bet on a corner. What is the probablity distribution for X? (Hint X can be negative) (b) What is the expected value of a$1 bet on a corner.

2. The attempt at a solution

(a) Okay i figured that the chance of a corner would be.

Pr(corner) = (1/37)X4 = 4/37.

However, i dont think this counts as the probablity distribution?

Not quite sure how to approach this.

(b) Excepted value for a $1 bet. Need to take account both winning and losing? Pr (corner) = 4/37 Pay out is 8 to 1. therefore is excepted value is 4/37 X$8 = 32/37

Pr (no corner) = 33/37 ((1-(4/37))
Shouldnt the EXPECTED VALUE of the $1 bet be$1 minus $0.02 = 98 cents? Got it from http://en.wikipedia.org/wiki/Expected_Value#Examples Not very sure because the definition is E(X)= xP(x) well from my book... 7. Aug 26, 2009 ### HallsofIvy I suggest you look at your book again. The definition of expected value is the sum of xP(x) where x ranges over all possible values. Here, the two possible values of x are 8 (if you win) and -1 (if you lose). The expected value is 8Pr(win)- 1Pr(lose)= $8\frac{1}{37}- \frac{33}{37}$. You don't think you are going to have a positive expected value gambling do you? 8. Aug 26, 2009 ### HallsofIvy You mean "Pr (Corner) = 4/37, Pr(NOT Corner) = 33/37", of course. 9. Aug 26, 2009 ### nicholasch Thanks for the prompt reply. Oh my book is correct. I mistyped that. It is the sum of xP(X), therefore it is (32/37) + (-33/37) = -1/37 Therefore, the expected value of a$1 bet is \$(-1/37), which is approx -2 cents.

It was wikipedia confusing me.