Mathematical Proof that Progressions will never Overcome a Negative Expectation Game

[k_c]

We removed the stop of lifetime many posts ago.
Time is unlimited as well.

So my trailing question(s) is, for the last time -

Is an unbounded martingale almost surely to win in a -EV environment?

No.

And if so, does that constitute a paradox or a suspension of normal statistical logic and/or probability? zg

No.

 

[zengrifter]

Is an unbounded martingale almost surely* to win in a -EV environment?

No.

Is there a math proof to support your opinion? zg

*ALMOST SURELY =

"While there is no difference between almost surely and surely (that is, entirely certain to happen) in many basic probability experiments, the distinction is important in more complex cases relating to some sort of infinity." ​

 

[k_c]

Is there a math proof to support your opinion? zg

*ALMOST SURELY =

"While there is no difference between almost surely and surely (that is, entirely certain to happen) in many basic probability experiments, the distinction is important in more complex cases relating to some sort of infinity." ​

Common sense dictates that what we know about reality in the case of negative EV is that ALMOST SURELY = SURELY in this case.

It was shown that for any bankroll no matter how large that a martingale fails for -EV.

Martingale proponents solution to this problem is to refuse to define a bankroll even though they can make it as large as they wish. Why? Because as soon as it's defined then it has been proven that the martingale eventually fails and bankroll is eventually depleted. Any humongous bankroll that is defined will always need to be bigger yet and there is no end to this.

If EV = -100%, meaning not even 1 win is possible, the argument can be made that an unlimited bankroll will never be depleted. Obviously a martingale cannot succeed when EV = -100% (but maybe it'll work in the zenzone) so if you lose forever should we believe there is such a thing as an undepletable bankroll? Again once any bankroll is defined it can obviously be depleted (except maybe in zenzone.)

Putting all the pieces together using common sense and what we know about reality
ALMOST SURELY FAILS for -EV = SURELY FAILS for -EV (except in zenzone)

Coming soon in the zenzone is the secret to the magic bankroll that allows a martingale to win in a negative EV game. It is not available anywhere else and is available at a low, low price. There will be numerous testimonials from dozens of satisfied customers.

 

[zengrifter]

Putting all the pieces together using common sense and what we know about reality
ALMOST SURELY FAILS for -EV = SURELY FAILS for -EV

To paraphrase Aristotle - The hole (in your argument) is greater than the sum of your "pieces".

Thank you nonetheless for your unbounded .. errr unfounded opinion.

Coming soon in the zenzone is the secret to the magic bankroll that allows a martingale to win in a negative EV game.

The extra adhominen was uncalled for. A math proof, please? zg

 

[k_c]

The hole (in your argument) is greater than the sum of your "pieces".

Thank you for your unbounded .. errr unfounded opinion.

The extra adhominen was uncalled for. A math proof, please? zg

I'm sorry but the ball is in your court to show that a bankroll exists that overcomes -EV in a martingale. :devil:

 

[zengrifter]

I'm sorry but the ball is in your court to show that a bankroll exists that overcomes -EV in a martingale. :devil:

So much for science. zg

 

[k_c]

So much for science. zg

I guess that's what the zenzone is all about. :grin:

 

[psyduck]

So my trailing question(s) is, for the last time -
Is an unbounded martingale almost surely to win in a -EV environment? zg

If you already know the game has a -EV, then your profit = -EV*(total wager). Even with infinite backroll to keep you going, your profit will still be minus.

 

[zengrifter]

If you already know the game has a -EV, then your profit = -EV*(total wager). Even with infinite backroll to keep you going, your profit will still be minus.

Almost surely? :laugh:

I really hate to see this thread locked without a math-proof, unless its already been posted? zg

 

[psyduck]

Almost surely? :laugh:

I really hate to see this thread locked without a math-proof, unless its already been posted? zg

Under what condition will profit = -EV*(total wager) be positive? What mathematical operation can you use to make it become plus?

 

[zengrifter]

Under what condition will profit = -EV*(total wager) be positive? What mathematical operation can you use to make it become plus?

Never, it will always be -EV, yet the martingale will almost surely win.
That is my hypothesis. zg

 

[psyduck]

Never, it will always be -EV,

So is the proof mathematical enough?

yet the martingale will almost surely win.
That is my hypothesis. zg

Can you prove it with math or it will remain as a dream?

 

[zengrifter]

So is the proof mathematical enough?

I hope so. Where is it? zg

 

[psyduck]

I hope so. Where is it? zg

Your profit = -EV*(total wager)

You just agreed that this equation will always be minus.

 

[zengrifter]

Your profit = -EV*(total wager)
You just agreed that this equation will always be minus.

Yes, I agreed from the start. Then I just pointed out the obvious.

And given that I will ALMOST SURELY come out ahead with another bet, in an unbounded scenario, I asked
if this was a paradox, or suspension of common probability theory, or simply an asymptotic function? zg

 

[aslan]

Your profit = -EV*(total wager)

You just agreed that this equation will always be minus.

That he did, that he did! View attachment 7116

Did you know, horse sense is the thing a horse has which keeps it from betting on people. Plagiarized by WC Aslan, left field

​

 

[psyduck]

Yes, I agreed from the start. Then I just pointed out the obvious.

And given that I will ALMOST SURELY come out ahead with another bet, in an unbounded scenario, I asked
if this was a paradox, or suspension of common probability theory, or simply an asymptotic function? zg

"Almost" is the key word. "Almost surely come out ahead" does not mean "will come out ahead".

 

[zengrifter]

"Almost" is the key word. "Almost surely come out ahead" does not mean "will come out ahead".

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

*ALMOST SURELY =

"While there is no difference between almost surely and surely (that is, entirely certain to happen) in many basic probability experiments, the distinction is important in more complex cases relating to some sort of infinity." ​

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.

 

[psyduck]

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.

I challenge you to spend the rest of your life tossing that coin and see if you will reach a point where you keep having heads and no tail. Let me know.

 

[zengrifter]

I challenge you to spend the rest of your life tossing that coin and see if you will reach a point where you keep having heads and no tail.

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