# Roulette problem

Tags:
1. Mar 9, 2015

### dthmnstr

Hey guys long time since i've posted on here, got this random question a friend of mine and myself were wondering over.

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

A player bets 100 dollars continuously on any number besides 0 (european roulette so 1/37 chance to win). He plays until he wins once and stops playing.

a) Z is the amount of times he plays. What is the distribution of Z?
b) What is the probability of the player winning at least 1000 dollars and the probability of the player losing 1000 dollars?

2. The attempt at a solution
I'm not so sure as how to go about this as i don't have any background in statistics and am currently studying calculus. But statistics is on my curriculum after the summer so it would be nice to have a little heads up on whats to come.

2. Mar 9, 2015

### LCKurtz

Start by figuring out the probability that he loses on the first $n-1$ plays and wins on the $n$th play. That will give you the distribution.

3. Mar 9, 2015

6. Mar 9, 2015

### dthmnstr

Aha thats true for b), didnt see that....as for losing 1000, he has to lose 10 times in a row before quitting, how likely is losing 10 times in a row?

7. Mar 9, 2015

### LCKurtz

Respond to post #2 and you will have your answer. I suppose you realize that, in the real world, your "roulette" game is completely absurd. You bet $100 with only 1/37 probability of winning, and in the unlikely event you are lucky, you just get$100 back??? So you lose $100 virtually every time you spin and win$100 once every blue moon. Some game.

8. Mar 9, 2015

### Ray Vickson

Your description does not match the actual way roulette is played. See, eg, http://en.wikipedia.org/wiki/Roulette .

In particluar, the section entitled "Bet Odds Table" states that the payoff is $\frac{1}{n}(36-n) = (36/n) - 1$, where $n$ is the number of squares the player is betting on. In your description you say the player bets on ONE square, so the payoff is 35 to 1; that is, they would receive a payoff of $3500 for their$100 investment, in the lucky event that they win. (Just to be clear, the $3500 is a net gain, so they paid$100 and got back \$3600.) Betting on more than 1 square would be needed if they wanted to increase the probability of winning while lowering the possible payoff.

BTW: If we assume they bet on single squares each time, the probability of losing 10 times in a row would be $(36/37)^{10} \doteq 0.76034$.

Last edited: Mar 10, 2015