Roulette probability - MathOverflow [closed]

Roulette probability

2010-08-04T16:18:50Z

I'm looking for some knowledge on probability, I've scoured the net but I can't really grasp the answer.

I was having a discussion with a co-worker about roulette probability. He says that at any given spin the probability that the outcome being red or black is equal (not taking into account the 0, which is neither).

My understanding of probability is that you should take into account the whole set of past outcomes. So if the outcome is red three times in a row, the probability that the next outcome is black will get bigger.

So to get a definitive answer I've created a roulette simulator and an artificial player. The player only bets when the outcome was the same three times in a row, then he bets the opposite. So if the outcome was red three times, he bets black.
To my surprise, the win/loss ratio was practically equal given a large enough simulation.

To finalize, my question is: how come that past outcomes have exactly zero influence on the probability of any given outcome?
I get the feeling (seeing some other (related) questions) that this may not be the place to ask, but would you then be so kind to at least get me in the right direction or point me to some resources explaining this? Thanks!

Answer by Andrea Mori for Roulette probability

2010-08-04T16:41:06Z

How come that past outcomes have zero influence? Simply because a roulette is just wood, plastic and brass and got no memory.

Answer by Dennis Haarbrink for Roulette probability

2010-08-04T16:46:10Z

Now I've asked the question I came upon a resource which states:

If you were to count all the occurrences of eight blacks in a row FOLLOWED BY A RED, you will find an equal number of occurrences of eight blacks in a row FOLLOWED BY A BLACK (9 blacks in a row).

Now, this makes sense in my wee little head :)

Anyway, thanks for the other comments and answers.

Answer by Tracy Hall for Roulette probability

2010-08-04T16:58:06Z

(It's true that this question will probably be closed soon.)

Ask yourself this question: Does the roulette ball or table have a memory? If not, then past events cannot possibly affect the next probability. "No memory", or more technically independent outcomes, is a standard hypothesis in probability problems like the one you are interested in.

The deeper question (but still not at the research level) is how statistics always seem to even out in the long term, even if there is no memory forcing them to do so. This is the content of the Central Limit Theorem of probability theory, and is closely related to the Second Law of Thermodynamics, whose rigorous treatment comes from statistical mechanics. The short answer is this: Statistics always even out in practice because in the long term there are many, many, many combinations with nearly even statistics, compared to just a handful of combinations that are greatly skewed. To be more concrete: (black, red, black, black, red, black, red, red, red) has precisely the same probability as (black, black, black, black, black, black, black, black, black), but no one ever asks about that precise first sequence; instead it gets lumped together with the 125 other sequences that have the same overall statistics, whereas all-black has no statistical compadres to share the burden of occurring more than one time in 512.