Can one overcome a -EV game? - Page 7

college kid · #61 10-28-2005, 05:00 AM

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I made a post on another forum in which I claimed that without betting caps on a game like Craps a double up betting system could be used to eventually ruin a Casino that has a finite bankroll.

A poster in response, called me an idiot (might be right), and said that no huge bankroll could overcome a -EV game? I believe he is wrong and I am right. I would like some help from you probability guys to decide this issue.

Vince

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First, that's what table limits are for. Second, the house's bankroll is bigger than your. Third, even with a huge bankroll, you will eventually hit a streak large enough that you can't afford the double up. And so and and so forth.

This post reminds me of all crap money management systems people talk about with the stock market and think they are such hot [censored] when the market goes up and they are temporarily playing a +EV game. It has nothing to do with the way you are betting. You must get down to the core of how the game works. Either the payoff odss must be changed, or the physical aspects of the game must be altered so the payoffs are better (read: dice control).

Now many people on this site think I'm a damn idiot, but I see so many people who just don't fundamentally understand where the expectation comes from and the uselessness of betting systems. How can they not get it!?

TomCollins · #62 10-28-2005, 10:55 AM

The probability of an infinite run of heads or tails is 0. The probability of having a run of X heads or tails where X is any number, given infinite time and infinite flips is 1.

TomCollins · #63 10-28-2005, 10:59 AM

The odd numbers are a subset of all integers. But the set of all odd numbers and the set of all integers have the same Cardinality.

jason1990 · #64 10-28-2005, 11:03 AM

The number p is the probability of winning one game. I was proving that the probability of having an infinite streak of wins is 0. Having an infinite streak of wins is the same as saying there exists k such that you win all games after the k-th game. I'm not requiring p>0. If p=0, then the probability of winning one game is zero, so the probability of an infinite streak of wins is zero.

jason1990 · #65 10-28-2005, 11:09 AM

If X is the sequence of all wins and losses, then X is a random variable taking values in the space of all sequences of 0's and 1's. Obviously, X is not a discrete random variable, so the link you provided is not relevant. In fact, the range of X is uncountable, which implies there are uncountably many sequences a such that P(X=a)=0. In fact, if 0<p<1, then P(X=a)=0 for all sequences a. For instance, the probability that every odd flip is heads and every even flip is tails is zero. This is the same phenomenon that occurs with continuous random variables. If X is a uniform or normal random variable, for example, then P(X=a)=0 for all real numbers a. In order to get a non-zero probability, you have to ask for the probability that X lies in some interval.

Vincent Lepore · #66 10-28-2005, 06:06 PM

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Since p<1, the limit as n goes to infinity of p^n is 0.

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This sounds like a self serving proof. Why does p^n = 0?

Vince

Vincent Lepore · #67 10-28-2005, 06:33 PM

O.K. Jason, I agree with your summation of infinite sequences. Since you know more about these things than I do it is my belief that I do not need to use an infinite sequence to prove that I am correct. I am confident that you can provide proof that the Casino will go broke using finite numbers for each (the casino and the opponent) given no cap on bettings. Correct?

Vince

Komodo · #68 10-28-2005, 07:25 PM

Ok, infinite losing streaks is impossible which is proven.
But its also trivial to prove that infinite money is impossible too, which kills any martingale formula both in practice and in theory.
I dont see how you can allow infinite money and finite losing streaks in the same mathematical formula. It just doesnt make any sense.

AlphaWice · #69 10-28-2005, 09:59 PM

Remember, the initial discussion was supposed to be centered around "why do casinos not have betting caps"?

Suppose you have a casino, raking in $100M a year, with a $10B roll. Investors, backed by $100B, decide to stupidly "martingale" you, in a game of craps/bacarrat. They succeed some % of the time greater than 50% (because their money is much greater than yours). So do you let them bet, knowing they ruin you - or stand pat with guaranteed income?

TomCollins · #70 10-28-2005, 10:23 PM

[ QUOTE ]
[ QUOTE ]
Since p<1, the limit as n goes to infinity of p^n is 0.

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This sounds like a self serving proof. Why does p^n = 0?

Vince

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p^n doesn't equal 0. It's limit is zero. This means that you can pick any positive number, as small as you want. Call this number x. For each x you pick, I can pick an n such that p^n < x. So if you make x 1/1 billion, I can pick an n that p^b < 1/1 billion.