Mathematical Proof that Progressions will never Overcome a Negative Expectation Game
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zengrifter
Banned
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."
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
Banned
I'm sorry but the ball is in your court to show that a bankroll exists that overcomes -EV in a martingale. :devil:
zengrifter
Banned
So much for science. zg
I guess that's what the zenzone is all about. :grin:
psyduck
Well-Known Member
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
Banned
Almost surely? :laugh:
I really hate to see this thread locked without a math-proof, unless its already been posted? zg
psyduck
Well-Known Member
Under what condition will profit = -EV*(total wager) be positive? What mathematical operation can you use to make it become plus?
zengrifter
Banned
Never, it will always be -EV, yet the martingale will almost surely win.
That is my hypothesis. zg
psyduck
Well-Known Member
So is the proof mathematical enough?
Can you prove it with math or it will remain as a dream?
zengrifter
Banned
I hope so. Where is it? zg
psyduck
Well-Known Member
Your profit = -EV*(total wager)
You just agreed that this equation will always be minus.
zengrifter
Banned
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
Well-Known Member
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
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psyduck
Well-Known Member
"Almost" is the key word. "Almost surely come out ahead" does not mean "will come out ahead".
zengrifter
Banned
Actually it does. You may not have really followed this thread carefully enough. zg
For example --*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."
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
Well-Known Member
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
Banned
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? z
g