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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!

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closed as too localized by Loop Space, Gjergji Zaimi, Robin Chapman, Pete L. Clark, Anweshi Aug 4 '10 at 17:49

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

2  
The place to ask is math.stackexchange.com –  Loick Aug 4 '10 at 16:21
    
Try this on the "lower level MO" site. Better yet, get a book out of the library on probability. As a hint: imagine that instead of running the same roulette wheel 100 times, you run 100 roulette wheels once. What would you bet then? Suppose you look at them one at a time and the first three are red, what would you expect to see at the next one? –  Loop Space Aug 4 '10 at 16:23

3 Answers 3

up vote 2 down vote accepted

(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.

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

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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.

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