Hi guys, as there was a nice discussion about roulette (see here :

The best way to play roulette?) I would like to post some basic math and my notes about roulette (there is no advanced math to be found there )

What is the expected value on betting (considering standard 36 number roulette with one 0):
- Betting a colour (18/37 chance of winning double your stake):

*E=18/37*2=36/37*
- Betting a number (win 36 times your stake):

*E=1/37*36=36/37*
- Betting a column (win 3 times your stake):

*E=12/37*3=36/37*
- Betting a combination (0,5$ on red + 0,5$ odd):

*E(result red + odd) = 0.5*4*(10/37)+0.5*2*((8+8)/37) = 36/37*
**Conclusion :**
As you can see, betting red, black or any other "special" selection is exactly equal to betting each number in the group, which means, you can only get 36/37$ of every dollar bet, whatever combination you bet on

You are down 0.027$ everytime you bet a single dollar on roulette .. So the best way to play roulette is not to play (or play with play money as sydney69 suggests)

The "double each stake strategy" (aka Martingale betting)
- There is a strategy to bet 1$ on red .. if you win, OK, bet 1$ on red again, if you loose, double your stake to 2$ .. If you win, nice, you have 1$ profit, if you lose, bet 4$ (if you win you have 4-2-1=1$ profit again) and so on

This strategy of course works theoretically, but it is profitable under 2 assumptions:

1) unlimited bankroll

2) unlimited maximum stake in the casino

However these conditions are not met in reality, so this strategy cannot work in real life

A few interesting notes
- Do you know, what is the sum of all numbers on roulette

?

- Did you know, that using double betting strategy, it does not matter what colour you bet on each try ? (it does not matter if you always bet on black or you bet on random color each time)

- Americans are probably more greedy, as you lose more on american roulette (0,05263$ on every dollar), as there are 2 zeros but you still only win 36 times your bet on a number