Mathematical Proof that Progressions will never Overcome a Negative Expectation Game

[psyduck]

You've lost me... It was QFIT who introduced the concept of 'almost surely', in this thread,
but for the wrong reason. Do you disagree with the proper meaning of almost surely? zg

When your profit = -EV*(total wager), you surely will not make money. Where is the almost surely?

 

[johndoe]

When your profit = -EV*(total wager), you surely will not make money. Where is the almost surely?

It's a misapplication of the concept of EV.

 

[psyduck]

It's a misapplication of the concept of EV.

Whatever your definition for EV is, the point is profit will be minus.

 

[johndoe]

Whatever your definition for EV is, the point is profit will be minus.

Not if you have an unlimited credit line (and table limit), and control over when to stop. You will "almost surely" win, with a certainty that's much higher than anything else you'll ever encounter. Arguably definite.

(EV is a poor interpretation for a diverging series.)

I already calculated the odds of losing in a 24-hour period as absurdly small. How about for a lifetime, per k_c?

The odds of losing are 1 in 10^1806179 . How's that for a sure thing? :laugh:

 

[zengrifter]

I already calculated the odds of losing in a 24-hour period as absurdly small. How about for a lifetime, per k_c? The odds of losing are 1 in 10^1806179 . How's that for a sure thing? :laugh:

But did you factor in a housEdge of .25% ? zg

 

[k_c]

Last try for brother zg

Actually it does. You may not have really followed this thread carefully enough. zg

For example --

Tossing a coin​

Suppose that an "ideal" (edgeless) fair coin is flipped again and again. A coin has two sides, head and tail, and therefore the event that "head or tail is flipped" is a sure event. There can be no other result from such a coin.
The infinite sequence of all heads (H-H-H-H-H-H-...), ad infinitum, is possible in some sense (it does not violate any physical or mathematical laws to suppose that tails never appear), but it is very, very improbable. In fact, the probability of tail never being flipped in an infinite series is zero. Thus, though we cannot definitely say tail will be flipped at least once, we can say there will almost surely be at least one tail in an infinite sequence of flips.

It is improbable that you will lose every trial in a negative EV game. A negative EV game means chance of winning is less than 1/2 and chance of losing is greater than 1/2.

It is also improbable that you will reach the limits of a very large bankroll by starting with 1 unit and doubling the bet after each loss.

Comparing above 2 data series:

Prob(one loss) = x/2 where x > 1 and x < 2 = (more than 1/2 & < 1)
Prob(n successive losses) = (x/2)^n
Bet after losing n successive trials = 2^n

Prob(n successive losses) = (x/2)^n -> 0 as n becomes large
Bet(n successive losses) = 2^n -> infinity as n becomes latge

For a negative EV game Prob(n successive losses) -> 0 at a slower rate than Bet -> infinity
Bankroll will run out before chance of losing all reaches 0.

Alternative proof: martingale works in a negative EV game if and only if what is depicted in the image is true.

 

[psyduck]

Not if you have an unlimited credit line (and table limit), and control over when to stop. You will "almost surely" win, with a certainty that's much higher than anything else you'll ever encounter. Arguably definite.

(EV is a poor interpretation for a diverging series.)

I already calculated the odds of losing in a 24-hour period as absurdly small. How about for a lifetime, per k_c?

The odds of losing are 1 in 10^1806179 . How's that for a sure thing? :laugh:

We are arguing the theoretical aspect (or you will not have unlimited money). No matter how low the probability may be, as long as it is not zero, you cannot exclude it from happening.

Let's take a step back and assume that your losing streak will not last forever. You win a hand after wagering an infinite amount of money. At that point, the casino will have no money left to pay you. Your net profit is still minus infinity.

 

[aslan]

Actually it does. You may not have really followed this thread carefully enough. zg

For example --

Tossing a coin​

Suppose that an "ideal" (edgeless) fair coin is flipped again and again. A coin has two sides, head and tail, and therefore the event that "head or tail is flipped" is a sure event. There can be no other result from such a coin.
The infinite sequence of all heads (H-H-H-H-H-H-...), ad infinitum, is possible in some sense (it does not violate any physical or mathematical laws to suppose that tails never appear), but it is very, very improbable. In fact, the probability of tail never being flipped in an infinite series is zero. Thus, though we cannot definitely say tail will be flipped at least once, we can say there will almost surely be at least one tail in an infinite sequence of flips.

View attachment 7118 Yes, I'm almost sure of that!

 

[aslan]

Let's take a step back and assume that your losing streak will not last forever. You win a hand after wagering an infinite amount of money. At that point, the casino will have no money left to pay you. Your net profit is still minus infinity.

You couldn't have an infinite amount of money to bet in the first place. Where would you keep it? :grin::whip:

 

[aslan]

It is improbable that you will lose every trial in a negative EV game. A negative EV game means chance of winning is less than 1/2 and chance of losing is greater than 1/2.

It is also improbable that you will reach the limits of a very large bankroll by starting with 1 unit and doubling the bet after each loss.

Comparing above 2 data series:

Prob(one loss) = x/2 where x > 1 and x < 2 = (more than 1/2 & < 1)
Prob(n successive losses) = (x/2)^n
Bet after losing n successive trials = 2^n

Prob(n successive losses) = (x/2)^n -> 0 as n becomes large
Bet(n successive losses) = 2^n -> infinity as n becomes latge

For a negative EV game Prob(n successive losses) -> 0 at a slower rate than Bet -> infinity
Bankroll will run out before chance of losing all reaches 0.

Alternative proof: martingale works in a negative EV game if and only if what is depicted in the image is true.

Have you forgotten Professor Dimmwitt's first rule of evidence: View attachment 7119 I saw it on the Internet; therefore, it must be true?

 

[k_c]

Have you forgotten Professor Dimmwitt's first rule of evidence: View attachment 7119 I saw it on the Internet; therefore, it must be true?

At times I think there are no words
But these to tell what's true
And there are no truths outside the Gates of Eden.

-Bob Dylan
Gates of Eden

 

[aslan]

At times I think there are no words
But these to tell what's true
And there are no truths outside the Gates of Eden.

-Bob Dylan
Gates of Eden

Touché!

 

[johndoe]

We are arguing the theoretical aspect (or you will not have unlimited money). No matter how low the probability may be, as long as it is not zero, you cannot exclude it from happening.

Let's take a step back and assume that your losing streak will not last forever. You win a hand after wagering an infinite amount of money. At that point, the casino will have no money left to pay you. Your net profit is still minus infinity.

We've been through this. You can exclude it from happening, because, by definition, you do have an unlimited bankroll (credit). It will never end, until you win. So you'll win eventually. The odds of this not happening in a lifetime is about 1 in 10^1806179 ; I'm not sure you appreciate how small this really is.

If your losing streak doesn't last "forever" as you say, then you'll absolutely be wagering a finite, and not infinite amount of money when you win that bet. And then you'll be ahead.

If you have an unlimited bankroll (credit) can it ever run out? I say no, by its own definition. Infinity never arrives.

 

[aslan]

If you have an unlimited bankroll (credit) can it ever run out? I say no, by its own definition. Infinity never arrives.

Which is why this discussion is entirely pointless--it can't happen in the real world. But take a finite br and lift the house max, and you have an entirely different ball game. :cool2:

 

[psyduck]

We've been through this. You can exclude it from happening, because, by definition, you do have an unlimited bankroll (credit). It will never end, until you win. So you'll win eventually. The odds of this not happening in a lifetime is about 1 in 10^1806179 ; I'm not sure you appreciate how small this really is.

If your losing streak doesn't last "forever" as you say, then you'll absolutely be wagering a finite, and not infinite amount of money when you win that bet. And then you'll be ahead.

If you have an unlimited bankroll (credit) can it ever run out? I say no, by its own definition. Infinity never arrives.

To theoretically evaluate the system as we have been doing, you cannot stop playing once you win a bet. That could happen in two or three hands. You have to consider what could happen if you play infinite hands. That means you need to consider all the possibilities.

One possibility is you reach a point you just keep losing (yes, the probability is small, but not zero). Then you will never recover your previous loss. To this you always say it is so small and practically zero. I say no. Approaching zero is not zero if you really argue the theoretic aspect.

The second possibility is you win a hand after you wagered an infinite or nearly infinite amount of money. The casino has no money left to pay you and you are simply in debt. No joking, you asked for unlimited credit and you got it. It is not possible both you and the casino have unlimited money at the same time.

 

[aslan]

To theoretically evaluate the system as we have been doing, you cannot stop playing once you win a bet. That could happen in two or three hands. You have to consider what could happen if you play infinite hands. That means you need to consider all the possibilities.

One possibility is you reach a point you just keep losing (yes, the probability is small, but not zero). Then you will never recover your previous loss. To this you always say it is so small and practically zero. I say no. Approaching zero is not zero if you really argue the theoretic aspect.

The second possibility is you win a hand after you wagered an infinite or nearly infinite amount of money. The casino has no money left to pay you and you are simply in debt. No joking, you asked for unlimited credit and you got it. It is not possible both you and the casino have unlimited money at the same time.

You obviously have no idea just how big infinity is! j/k You said, "One possibility is you reach a point you just keep losing." How will you know that? The answer is, you will never know that, so theoretically, you will never have to pay up your credit line either---just keep playing.

You said, "It is not possible both you and the casino have unlimited money at the same time."

Au Contraire, mon frere. the casino might well have an infinite number of $100 bills, and you might have an unlimited number of $20 bills. If you don't like that answer, the casino might have an unlimited amount in Mexican pesos and you might have an unlimited amount in American dollars. (That should keep the U.S. Bureau of Engraving busy!)

Like Qfit cautioned in one of his posts, when you start talking about infinities, all kinds of weird things are possible (I'm paraphrasing--I don't remember his exact words).

 

[psyduck]

Like Qfit cautioned in one of his posts, when you start talking about infinities, all kinds of weird things are possible

Yes indeed sir, one of the weird things is you keep losing!

 

[aslan]

Yes indeed sir, one of the weird things is you keep losing!

Possibly, but you'll never know whether a win is just ahead. But more probably, you'll go back and forth, at least for the first few trillion years. :laugh:

 

[johndoe]

To theoretically evaluate the system as we have been doing, you cannot stop playing once you win a bet. That could happen in two or three hands. You have to consider what could happen if you play infinite hands. That means you need to consider all the possibilities.

Nope, that's not a condition of this hypothetical. You get to stop whenever you like. But it doesn't really matter which win you stop after; you'll still be a winner.

As for the point, aslan, I'm just countering the earlier claim that it was proven mathematically that a martingale wouldn't work for an unlimited/infinite bankroll (credit) and table limits. Of course there's no reality involved. In any actual conditions, it's certainly a loser, and the original proof demonstrates that succinctly.

 

[psyduck]

Nope, that's not a condition of this hypothetical. You get to stop whenever you like. But it doesn't really matter which win you stop after; you'll still be a winner.

Okay, you play martingale and win a bet after three hands. Then you stop and say it works. There you have it. It does work after all these arguments. What else can I say. You have the profit to prove it.