Sunday, February 24, 2008

Roulette Theory

This subject has interested me for a while, mainly due to the fact that mathematically speaking, I believe the possibility of a strategy to obtain a slight statistical edge to the player exists. Well, without further or do, here goes.

The graph displays bets available for European roulette, which has a slightly lower edge for the house with the single 0, where American roulette contains a 0 and a 00.

Let us focus on the 3rd's, i.e. boxes where you can bet on "1st-12","2nd-12", and "3rd-12".

The probability of each spin landing in one of the 3rd's is 12/37, or 0.3243~ (32.43%), and 0.027~ (2.7%) for the 0 to get hit.

What this means, is that over in the long run, roughly out of every 100 spins, each 3rd would expect to get hit roughly 32 times, and the 0 roughly 3 times.

Now the payoff. With respect to betting size on a 3rd, winning on a 3rd gives a 200% profit, and a losing bet results in a loss of 100%. i.e. a $1 bet makes $2 in profit if won, otherwise the original bet amount of $1.

This presents a winner/loser size ratio of 2/1=2.

Statistical expectancy. This is how much you would expect to win or lose over in the long run. If you play like any other newbie at the casino, the following becomes expected.
$2*0.3243 - $1*(1-0.3243) = -0.0271
The negative expectancy points to a gradual loss of roughly 3% over the period of roughly 1,000 spins. So what is needed to make the expectancy positive?

The following lies the key to this theory-
The game is deterministic. i.e. we KNOW the exact statistical expectation of how it will end in the long run, and with that knowledge it then becomes entirely feasible to adopt a strategy to exploit that information.

(At this point, I know the math professors at AUT would probably disagree with me quite passionately. I welcome anyone to debate against the above notion.)

To reiterate, we can mathematically expect 32 hits out of each 3rd with each 100 spins (or 16 out of 50 spins), however we will need at least 34 winners to be profitable at the end of 100 spins (or 17 winners out of 50 spins). Now, what if in the first 50 spins, the first 3rd or the "1st-12" landed only 10 times? For the statistical expectancy of 32 hits to fulfill the next 50 spins, or the second half of 100 spins, will have a higher than 50% chance of turning out 22 hits on the "1st-12" bet spot. And viola, an expected winning session.

What is the exact probability of the statistical expectancy to be fulfilled with respect to the number of spins? I will address this issue in the near future. I hope you enjoyed this reading :)

0 Reflections: