Can one overcome a -EV game?

[TomCollins]

What you are arguing is not debated. A big stack will bust a small stack a large percent of the time even if the small stack has an edge. This percent depends on betting patterns, bankroll differences, and edge in the game. Will you beat the casino with this? No. Do you have a large proabability of "breaking the bank"? Depends on those three variables.

[pzhon]

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Alright, I thought you could have an infinite series of wins if you had infinite money and time.

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Yes you can and will.

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No, that is another false statement by you. The probability of an infinite streak of wins is 0. However, there will be streaks of wins of every finite length with probability 1.

This is really easy to prove.

[TomCollins]

Its sad how few people understand the concept of infinity [img]/images/graemlins/frown.gif[/img]

[SheetWise]

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Its sad how few people understand the concept of infinity.

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And sad how many use it in the same way a drunk uses a lamp post -- for support rather than illumination.

[Vincent Lepore]

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The probability of an infinite streak of wins is 0.
.....This is really easy to prove.

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O.K. Prove it. I'm open to learning.

Vince

[Vincent Lepore]

I believe the following applies.

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Theorem 4. The cardinality of the power set of an arbitrary set has a greater cardinality than the original arbitrary set, or |*A| > |A|.
This is called simply Cantor's Theorem. It generalizes the previous theorem, in which we proved that the power set of a particular set, N, had a greater cardinality than the original. The present theorem is trivial for finite sets, but is fundamental for infinite sets.
Proof. Let A be an arbitrary set of any cardinality, finite or infinite. Again we supply a negative proof, and assume that the members of A can be put into one-to-one correspondence with the subsets of A. Take any one of the supposed ways of pairing off the members of A with the subsets of A. Let us say that if a member of A is paired with a subset of A of which it happens to be a member, then it is happy; otherwise it is sad. Let S be the set of sad members of A. Clearly S is one of the subsets of A. Therefore S is paired off with one of the members of A, say, x. Is x happy or sad? If x is happy, then x is a member of the set to which it is paired, which is S, but that would make it sad. If x is sad, then x is not a member of the set to which it is paired, which is S, but that would make it happy. So if x is happy, then it is sad, and if it is sad, then it is happy. Our assumption implies a contradiction and is therefore false. So the members of A cannot be put into one-to-one correspondence with the subsets of A.
But if A and *A cannot be put into one-to-one correspondence, then they cannot have the same cardinality. If so, then the larger one must be *A, for A can be put into one-to-one correspondence with a proper subset *A. For example, if the members of A are A1, A2, A3..., then they can be put into one-to-one correspondence with this subset of *A: {{A1}, {A2}, {A3}...}.


Many profound consequences follow directly from Cantor's theorem. But we make the most important of them explicit in the next theorem.

Theorem 5. If S is a set of any infinite cardinality, then its power set has a greater infinite cardinality, or |*S| > |S|.
This follows directly from Cantor's Theorem (Theorem 4). Cantor's theorem applies equally to finite and infinite sets; this corollary focuses on the important consequence for infinite sets.
If we follow the notation for finite sets, and say that a set of cardinality a has a power set of cardinality 2a, then this theorem asserts that 2a > a, for each transfinite cardinal a.
This theorem asserts that for any infinite cardinality, there is a larger infinite cardinality, namely, the cardinality of its power set. Hence, there is an infinite series of infinite cardinal numbers. We will meet some of the infinite cardinals larger than N shortly.
This theorem also implies that, for every set, there is a greater set. It follows that there is no set of all sets, or no set of everything.

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I believe that an infinite set of wins is a subset of the infinite set of events of wins and losses.

Vince

[jason1990]

Let p be the probability of winning the game and assume p<1. Let P_k be the probability that we win every game after the k-th game. If we can show that P_k=0 for all k, then the result is proven.

Let k be an arbitrary integer and let x be an arbitrary positive real number. Since p<1, the limit as n goes to infinity of p^n is 0. Therefore, there exists and integer n such that p^n<x. Let P_{k,n} be the probability that we win all the games from game number k+1 to game number k+n. Then

P_k < P_{k,n} = p^n < x.

So P_k<x for all positive numbers x. Hence, P_k=0. Since k was arbitrary, P_k=0 for all k, and this completes the proof.

[jason1990]

Cardinality and measure are two different ways of describing the "size" of set. The concept which is relevant in probability is measure, not cardinality.

[Vincent Lepore]

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Let p be the probability of winning the game and assume p<1


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I take it that you mean p<1 and >0. You also mean the probability of having an infinite streak of wins and not winning ALL games. Right?

Vince

[Vincent Lepore]

If you toss a fair coin 2 times the probability distribution will be as follows.

2 heads: .25
2 tails: .25
head:tail:.25
tail:head:.25

If you toss the coin an infinite number of times the probabiliy distribution will be similar. One of the distributions will be an infinite series of heads and one an infinite series of tails, etc. No? Or are you through has mathematics able to eliminate a potential result? I'm sure common sense plays no part in this discussion but even common sense tells us that one possibility is to continuously toss a head.

http://www.netnam.vn/unescocourse/statistics/52.htm

Vince