# Odds/Probability question - Roulette

1. Jun 10, 2009

### bobbo7410

Alright, my brains a little shot right now so I need some help. I've been studying the Casino game of Roulette and I was curious how this idea would work out.

Lets say you put in $20 each time and there are 2 different systems. 1st way: 50% chance to win back what you put in. So your original$20 + $20. 2nd way: 66% chance to win back half what you put in. So your original$20 + $10. Either way the max you could lose is$20 yet the 2nd way you would be risking more for less. Over the course of lets say 10 turns doing the same way each time, which system would provide the best results??

From what I've gathered:
[edit: Revised the work below in post #3.]

Method 1 would give roughly 5 wins. 5 wins = $40 per win =$200
5 loses = $-20 per loss =$-100 In this method, you win $100 roughly. Method two, you'd expect around 6.66 wins. 6.66 wins =$30 per win = $199.8 3.34 losses =$66.8
210 - 60 = $133 Yet there has to be a flaw somewhere, that would increase the odds of winning to over 50% for method two.. Last edited: Jun 10, 2009 2. Jun 10, 2009 ### CRGreathouse Your numbers are wrong, because you're including your original money as 'winnings'. It should be clear that the first method has an expected value of$0 per turn, not $10. 3. Jun 10, 2009 ### bobbo7410 Yeah I was simply going along with a helping answer from someone else, I'm not sure why I followed it and included the bet money. So to re-do the work: Lets say you put in$20 each time and there are 2 different systems. (10 times each)

Method 1 would give 5 wins/5 losses.
5 wins at $20 per win =$100 won
5 losses at $20 per loss =$100 lost
aka 50%

Method 2, would give around 6.66 wins/3.34 losses.
6.66 wins at $10 per win =$66.66 won
3.33 losses at $20 per loss =$66.66 lost
aka 50%

So both methods are equal?? Or did I mess this up again.

4. Jun 11, 2009

### HallsofIvy

1st way: .5 chance of winning $20, .5 chance of losing$20: expected value .5(20)+ .5(-20)= 0.

2nd way: .66 chance of winning $10, .34 chance of losing$20: expected value .66(10)+ .34(-20)= -.02.

If you meant 2/3 chance of winning $10, 1/3 chance of losing$20 then: expected value (2/3)(10)- (1/3)(20)= 0.

They are both "fair bets". If you look at the standard deviation, however:
$\sqrt{.5(20)^2+ .5(-20)^2}= 20$ and $\sqrt{(2/3)(10)^2+ (1/3)(20)^2}= \sqrt{200}= 14.14$. The first is a "riskier" bet than the second.