Mathematical Proof that Progressions will never Overcome a Negative Expectation Game
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Is there a math proof to support your opinion? zg
*ALMOST SURELY =
*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."
zengrifter said:
Is there a math proof to support your opinion? zg
*ALMOST SURELY =
*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."
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.
g
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
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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Actually it does. You may not have really followed this thread carefully enough. zg
For example --
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.
zengrifter said:
*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.