[Mathematics] [Casino] Roulette 07-05-2015, 06:04 PM
#1
This is the first in a new series about gambling in casinos and why you're doomed to fail forever. I'm also planning on making a thread about how you can beat the casino using tricks like card counting.
Roulette. When casinos come to mind, and you try to think about the fairest game in the casino, you probably picture a roulette wheel. (This thread covers the European wheel, but the concept is the same.)
In roulette, put simply, you can either bet on a specific pocket, or on reds or blacks. Now, say you bet on either red or black. The chances of you winning are 50% and so if you bet the same amount of money on read and black, you'd never loose a cent, right? Wrong. You also have to take into consideration the 0 / green pocket.
There are 37 pockets, numbered 0-36. 1 is green, and 18 are either red or black. If there wasn't a green pocket, it'd be a simple
You'd have an equal chance of winning as the house did, and so the game wouldn't exist, because they like to manipulate probability as to make it appealing. However, there is a green pocket and so:
If you bet on either black or red, you have a 48.6% chance of winning, and a 51.4% chance of losing. That doesn't seem like much of a difference, but look at it in a graph:
![[Image: VbE99h2.jpg]](http://i.imgur.com/VbE99h2.jpg)
Very slightly, the odds are shifted in favor of the house, with their winnings going up the more bets you place.
Gambling is bad, kids. They just want your money.
Oh, and roulette is cursed by the devil. Add up all of the numbers on the pockets and you'd have wished that you hadn't.
Roulette. When casinos come to mind, and you try to think about the fairest game in the casino, you probably picture a roulette wheel. (This thread covers the European wheel, but the concept is the same.)
In roulette, put simply, you can either bet on a specific pocket, or on reds or blacks. Now, say you bet on either red or black. The chances of you winning are 50% and so if you bet the same amount of money on read and black, you'd never loose a cent, right? Wrong. You also have to take into consideration the 0 / green pocket.
There are 37 pockets, numbered 0-36. 1 is green, and 18 are either red or black. If there wasn't a green pocket, it'd be a simple
Code:
18/36=0.5
You'd have an equal chance of winning as the house did, and so the game wouldn't exist, because they like to manipulate probability as to make it appealing. However, there is a green pocket and so:
Code:
18/37=0.4864864864864865
If you bet on either black or red, you have a 48.6% chance of winning, and a 51.4% chance of losing. That doesn't seem like much of a difference, but look at it in a graph:
![[Image: VbE99h2.jpg]](http://i.imgur.com/VbE99h2.jpg)
Very slightly, the odds are shifted in favor of the house, with their winnings going up the more bets you place.
Gambling is bad, kids. They just want your money.
Oh, and roulette is cursed by the devil. Add up all of the numbers on the pockets and you'd have wished that you hadn't.