Mathematical Proof that Progressions will never Overcome a Negative Expectation Game

[johndoe]

....

I don't agree with your use of limits here, since we are not playing infinite hands. We are playing only until a win. That's where it breaks.

It comes down to only this:

Will you ever win a hand?

If yes, then the (infinite) martingale works. If no, then it doesn't.

You say there's a chance you won't - i.e., there is a nonzero probability of never winning. I say that you will if you wait long enough, because never winning requires P(win)=0. Ironically, QFIT agrees with me per his russian roulette example, but in his earlier statement states that you will "almost surely" win.

I'm not sure we have the means to resolve this one way or another.

 

[QFIT]

An an infinite amount of time, all things that can occur will occur. So, you will win a hand. But, you will also lose an infinite number of hands. What's to say you won't lost an infinite number of hands before you win a hand? In that case, you win a dollar -- after you have lost an infinite amount. You still lose.

You cannot exclude the case where you lose and claim your system is a winner.

 

[QFIT]

I really wish you would stop saying I agree with you.

 

[johndoe]

An an infinite amount of time, all things that can occur will occur. So, you will win a hand. But, you will also lose an infinite number of hands. What's to say you won't lost an infinite number of hands before you win a hand? In that case, you win a dollar -- after you have lost an infinite amount. You still lose.

You cannot exclude the case where you lose and claim your system is a winner.

As I've said, always losing requires P(win)=0. Because the game requires P(win)>0, you'll eventually win. That's why the case is excluded.

You said yourself that a win will always happen, right in your roulette example.

 

[johndoe]

I really wish you would stop saying I agree with you.

I know it must bother you, but your russian roulette example clearly showed that you agree. The gun will fire at some time, just like you said.

 

[QFIT]

Just as you ignore the case where you lose, you also ignore parts of what I say and keep claiming that I agree with you. I do not agree with you. Obviously the system works if you include in the rules that the system works and you will ignore losses. But, it is progressionist thinking.

 

[QFIT]

Since you refuse to stop saying I agree with you when I don't, my part in this conversation is ended.

 

[iCountNTrack]

I don't agree with your use of limits here, since we are not playing infinite hands. We are playing only until a win. That's where it breaks.

It comes down to only this:

Will you ever win a hand?

If yes, then the (infinite) martingale works. If no, then it doesn't.

You say there's a chance you won't - i.e., there is a nonzero probability of never winning. I say that you will if you wait long enough, because never winning requires P(win)=0. Ironically, QFIT agrees with me per his russian roulette example, but in his earlier statement states that you will "almost surely" win.

I'm not sure we have the means to resolve this one way or another.

Again you have no control on when winning the hand will occur, you are the one who is advocating that the bankroll is infinite allowing to play indefinitely i.e an infinite number of hands, i have shown that if you do that you will lose an infinite amount of money.

 

[johndoe]

Just as you ignore the case where you lose, you also ignore parts of what I say and keep claiming that I agree with you. I do not agree with you. Obviously the system works if you include in the rules that the system works and you will ignore losses. But, it is progressionist thinking.

It's you that are doing the ignoring, and/or are stuck in denial. You attempted to fashion an argument using analogy that refuted my claim, but you inadvertently gave it direct support. Because of your bias, you will neither admit that's the case, nor provide any reasons at all for refuting it.

The only case I'm excluding is the one where you lose infinitely, or equivalently where P(win)=0. Since the game must allow for a win, no matter how small the odds, this cannot be included as a possibility.

 

[iCountNTrack]

It's you that are doing the ignoring, and/or are stuck in denial. You attempted to fashion an argument using analogy that refuted my claim, but you inadvertently gave it direct support. Because of your bias, you will neither admit that's the case, nor provide any reasons at all for refuting it.

The only case I'm excluding is the one where you lose infinitely, or equivalently where P(win)=0. Since the game must allow for a win, no matter how small the odds, this cannot be included as a possibility.

Let's keep it nice and civil JD

 

[johndoe]

Again you have no control on when winning the hand will occur, you are the one who is advocating that the bankroll is infinite allowing to play indefinitely i.e an infinite number of hands, i have shown that if you do that you will lose an infinite amount of money.

I "allow" only enough hands to establish a win - not an infinite number. Yes, if an infinite number of hands are played, you'll lose an infinite amount. I agree completely.

But if a win ever occurs, which it "almost certainly" will, the system is a winner. You don't have control over when the win occurs, but you can stop once it does.

 

[johndoe]

Let's keep it nice and civil JD

I apologize. I was sensing some hostility and reacted more strongly than I should have. Being called a progressionist is a big insult! :laugh:

 

[psyduck]

I "allow" only enough hands to establish a win - not an infinite number. Yes, if an infinite number of hands are played, you'll lose an infinite amount. I agree completely.

But if a win ever occurs, which it "almost certainly" will, the system is a winner. You don't have control over when the win occurs, but you can stop once it does.

Again, if you stop after a win, any system will "work". But that is not how you do statistics. Based on the way you call "works", there would be no such thing as negative expectation games.

 

[aslan]

You still do not understand. Unlimited bankroll means that you can keep playing until you have a single win. At that point, you have a profit.

As for the graph, I'll point you to qfit's own:

http://www.blackjackincolor.com/useless4.htm

It's disastrous for any finite bankroll of course. But there is a clear upward trend, is there not?

We do understand. You can't exclude what you do not want to happen. That's like saying, "Heads I win, tails your lose." If you never get to the END of your INFINITE losing streak (which you won't by definition) View attachment 7125 , you will NEVER get to your single win. :1st: :sad:

 

[aslan]

Again, if you stop after a win, any system will "work". But that is not how you do statistics. Based on the way you call "works", there would be no such thing as negative expectation games.

Right. If I only bet in <-4 situations, I will still win one hand occasionally. Aha! Another winning system!!!!

 

[k_c]

I "allow" only enough hands to establish a win - not an infinite number. Yes, if an infinite number of hands are played, you'll lose an infinite amount. I agree completely.

But if a win ever occurs, which it "almost certainly" will, the system is a winner. You don't have control over when the win occurs, but you can stop once it does.

I tried to show the conflict in logic simply in post #246 of this thread.
I'll try for a little more detai.

1) A negative EV game is a given
2) Let L = Prob(loss)
3) L is greater than 1/2 but less than 1
4) L can be as close to 1 as you wish but cannot equal 1
5) You say the probability of never winning a trial in a martingale sequence is virtually zero because L^n approaches 0 as n approaches infinity. This means that at the very start of each martingale sequence the chance of never being 1 unit ahead at the conclusion of the sequence is virtually zero. Fair enough but note virtual zero and absolute zero are not the same.
6) You include any case where there is any possibility of a win at all, This means L can be almost 1 but never equal 1. Agreed that L^n approaches 0 as n approaches infinity for any value of L < 1.
7) Martingale sequence can be repeated an infinite number of times over an infinite amount of time.
8) Let D = Prob(disaster)
9) A disaster occurs when any martingale sequence never experiences even 1 win until the end of time.
10) D = L^n as n approaches infinity as above.
11) D is virtually zero but not absolute zero, This is the important point.
12) The probability of going s martingale sequences without a disaster is (1-D)^s
13) The limit of (1-D)^s approaches 0 as s approaches infinity as long as D is greater than zero.

What we are left with are 2 conflicting statements each of which uses the same logic. If the logic applies to one it must apply to both.

One statement says that sooner or later you must experience a win in a martingale sequence.
The other statement says sooner or later you must experience a disaster where a single martingale sequence loses forever or until the end of time if that occurs first.

 

[johndoe]

Again, if you stop after a win, any system will "work". But that is not how you do statistics. Based on the way you call "works", there would be no such thing as negative expectation games.

Incorrect; you misunderstand. With the martingale, and no limits, you can stop after whatever profit you decide to take. It will arrive eventually.

I even gave the contrary example - once table limits are introduced, the system is a true loser, because the bankroll will asymptote to a negative number over time. This does not occur when bets are unbounded; any win brings you back into positive territory.

 

[aslan]

I tried to show the conflict in logic simply in post #246 of this thread.
I'll try for a little more detai.

1) A negative EV game is a given
2) Let L = Prob(loss)
3) L is greater than 1/2 but less than 1
4) L can be as close to 1 as you wish but cannot equal 1
5) You say the probability of never winning a trial in a martingale sequence is virtually zero because L^n approaches 0 as n approaches infinity. This means that at the very start of each martingale sequence the chance of never being 1 unit ahead at the conclusion of the sequence is virtually zero. Fair enough but note virtual zero and absolute zero are not the same.
6) You include any case where there is any possibility of a win at all, This means L can be almost 1 but never equal 1. Agreed that L^n approaches 0 as n approaches infinity for any value of L < 1.
7) Martingale sequence can be repeated an infinite number of times over an infinite amount of time.
8) Let D = Prob(disaster)
9) A disaster occurs when any martingale sequence never experiences even 1 win until the end of time.
10) D = L^n as n approaches infinity as above.
11) D is virtually zero but not absolute zero, This is the important point.
12) The probability of going s martingale sequences without a disaster is (1-D)^s
13) The limit of (1-D)^s approaches 0 as s approaches infinity as long as D is greater than zero.

What we are left with are 2 conflicting statements each of which uses the same logic. If the logic applies to one it must apply to both.

One statement says that sooner or later you must experience a win.
The other statement says sooner or later you must experience a disaster where a single martingale sequence loses forever or until the end of time if that occurs first.

Funny how we return to the age old questions. Now I ask you:

Can an Immovable Object and an Irresistible Force exist in the same universe? If they meet will they destroy each other, or should one assume they can never meet.

And where does that leave out proverbial martingaler-- locked in a perpetual game of 21? View attachment 7130-- losing all the money in the universe (like there's any left over for him to repay his debt!)? View attachment 7128-- or comfortably sipping bacardi and cokes somewhere on some Caribbean isle while spending his one unit win? View attachment 7129View attachment 7127

 

[psyduck]

Incorrect; you misunderstand. With the martingale, and no limits, you can stop after whatever profit you decide to take. It will arrive eventually.

That means the more you play the more profit you will make. No matter how much you want to win, as long as you keep playing you will reach the goal. Right? That contradicts your previous agreement that if you play infinite hands you will lose infinite amount (#271).

any win brings you back into positive territory.

But if you do not stop there, you will lose at a much faster rate in a losing streak.

 

[psyduck]

Yes, if an infinite number of hands are played, you'll lose an infinite amount. I agree completely.

If a system fails approaching infinite hands, how can you call that "works"? With a real winning system, the more you play the more you win (positive expectation). Infinite hands lead to infinite profit.