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Disclaimer: I know all about the house edge, negative expectation, expected value, gambler's fallacy, etc.


In the roulette forum I frequent, there are various debates about the inevitability of loss of a roulette player. There are people who insist that "the math" says that that any roulette system will lose and they can prove it. They base their reasoning on spin independence and house edge.

I try to explain that since there are probabilities involved, one can not be certain, let alone prove, that a specific system will lose after say 10,000 spins.

Yes, the expected value is negative and the probability of winning may diminish, but there can be no proof, in the mathematical sense, that a player will definitely lose playing a specific system. Because there will always be a chance that he will win. After all we can not exclude the case that he is lucky and wins more bets that he "should".

My questions:

  1. Am I correct that there can be no mathematical proof that a specific roulette player will lose money by playing roulette for, say, 10,000 spins?
  2. How can I explain mathematically the fact that it can not be proven that a roulette player will lose?

PS: It is interesting that, although by "lose" I meant "overall monetary loss" (which would require many losing bets), most replies interpreted "lose" as "lose at least one bet". Which is even more probable, yet it can not be proved it will happen.

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    $\begingroup$ Of course you can't prove that a specific player will win or lose a specific game. That's not possible. Otherwise you would know in advance what the outcome of, say, the toss of a coin is with certainty. What you can prove is that on average a player will lose in the long run. That's exactly the "expected" values you mention. $\endgroup$ – MPW Feb 6 '18 at 15:54
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I'll answer your questions, then follow with a few comments.

  1. You are correct.
  2. Since there are only finitely many possibilities in a finite number of roulette games -- say $n$ possibilities -- and each possibility is equally likely, and there is at least one possibility where the gambler wins money, there is a probability at least $\frac1n > 0$ that the gambler wins money. So there is positive probability of winning. It's reasonable to claim that "having positive probability" is a sufficient mathematical criterion to establish that winning is "possible."

What proof you may have seen likely proves that in infinite games, the gambler will lose with probability $1$. This is often stated as "the gambler will lose almost certainly." This may fall short of "the gambler will lose inevitably." Consider a simpler case: you toss a fair coin $n$ times. The probability that you will toss no tails is $\frac1{2^n}$. As $n$ approaches $\infty$, this probability approaches $0$; thus in infinite coin tosses, with probability $1$ you will toss a tail at some point. Nevertheless, there is a possible sequence of coin tosses -- HHHHHH... -- which does not include a tail. It's just that this sequence occurs with probability $0$.

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  1. You are correct.
  2. Consider the possible case of the player always wagering on black for 10,000 spins, and each spin actually falling out black. The probability of such an event is microscopic, but strictly positive, meaning it is possible. Moreover, given enough tries, it should actually happen...
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    $\begingroup$ ... although I think current physical theories predict there can't be enough tries? I mean in the sense that things like the heat death of the universe would prevent roulette tables from existing that long. $\endgroup$ – user14972 Feb 6 '18 at 16:16
  • $\begingroup$ @Hurkyl yes. i never said it is practical $\endgroup$ – gt6989b Feb 6 '18 at 19:29
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Yes, you are right. It is thinkable that there exist some people who have been playing roulette and not lost money. If the player places en plain bets for 10,000 rounds in a row, it may even happen (with the extremely low probability of $\frac1{37^{10000}}$) that he wins every single time. Therefore, a proof that this player must lose is impossible.

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