Mathematical Proof that Progressions will never Overcome a Negative Expectation Game
- Thread starter iCountNTrack
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iCountNTrack
Well-Known Member
No i am trying to prove that mathematical ignorance will unfortunately prevail.
zengrifter
Banned
That seems like a lot of calculus just to prove that progressions don't work.
On the other hand, some progressions DO work, provided there are no arbitrary limits on capital and bet-sizing. Correct? zg
iCountNTrack
Well-Known Member
If you had an unlimited bankroll why play in the first place
zengrifter
Banned
For one reason only - to prove that I can crush any casino with my martingale.
But seriously, is that your best answer? zg
aslan
Well-Known Member
On the other hand, there are no casinos where there are no arbitrary limits on capital or bet-sizing. Correct?
zengrifter
Banned
Should a "mathematical proof" include that? zg
iCountNTrack
Well-Known Member
And seriously was this your best answer, "a lot of calculus" that cracks me up, but at the same time it shows that you haven't even read the proof, there is no calculus involved just some 8th grade math that involves factorization , expansion and rearrangement.
On the other hand the very claim that progressions will work if one had unlimited funds is idiotic. Not only it is unrealistic but it is also against the fundamental objective of progressive systems: how to beat the casino and make money quick.
People mistakenly think that a martingale will succeed absent max bet limitations even in a negative EV game but it only will work when EV is positive.
Below p (prob of winning bet), doubling each bet after each loss, and prob of making each bet after n consecutive losses are considered in figuring overall expectation from a martingale progression.
Code:
Let p = probability of winning (range = 0 to 1, 0=sure loss, 1= sure win)
EV(Single trial) = 2*p - 1
[u]Progressive bet and probability of bet upon each loss[/u]
Bet 2^0, 2^1, 2^2,.......,2^(n-1)
Prob(Bet) (1-p)^0, (1-p)^1, (1-p)^2,........(1-p)^(n-1)
EV(Overall) = Sum[Bet*Prob(Bet)*EV(Single trial)] = Sum[2^(n-1)*(1-p)^(n-1)*(2*p-1)]
as n varies from 1 to infinity
[u]This is overall expectation from martingale progression:[/u]
1) EV(Overall) = (2*p-1)*Sum[2^(n-1)*(1-p)^(n-1)]
[u]let S = Sum[2^(n-1)*(1-p)^(n-1)] & do some algebra[/u]
S = 2^0*(1-p)^0+2^1*(1-p)^1+2^2*(1-p)^2+...+2^(n-1)*(1-p)^(n-1)
[u]S*2*(1-p) = 2^1*(1-p)^1+2^2*(1-p)^2+...+2^(n-1)*(1-p)^(n-1)+2^n*(1-p)^n[/u]
S*(1-2*(1-p)) = 2^0*(1-p)^0 - 2^n*(1-p)^n
S*(2*p-1) = 2^0*(1-p)^0 - 2^n*(1-p)^n
S = (2^0*(1-p)^0 - 2^n*(1-p)^n)/(2*p-1) = (1-2^n*(1-p)^n)/(2*p-1)
Plug S into equation 1) above
EV(Overall) = 1-2^n*(1-p)^n
Results of martingale as trials vary from 1 to infinity depends upon p (prob of winning)
If p = 1/2, EV(Overall) = 0, (martingale breaks even)
If p > 1/2, As n approaches infinity EV(Overall) approaches 1 (martingale succeeds)
If p < 1/2, As n approaches infinity EV(Overall) approaches -infinity (martingale loses an infinite amount)zengrifter
Banned
Mea culpa! You are correct in that I can't distinguish between junior-high math and calculus.
Still, seems like a lot of MATH just to 'prove' that 1-1=0?
Yes, but not 'idiotic' per'se - all authorities from Scarne to Griffin and Thorp point out that unlimited capital and bet limits will allow a progression to work.
Curiously, some of the 'smartest' traders of the past 20+ years have used martingalesque betting with catastrophic results.
(See Long Term Capital Growth, etc.). >> See- The Global Collapse was a MARTINGALE Play
zengrifter
Banned
I'm with k_c, with a caveat.
This problem is similar to addressing convergence of infinite, divergent (or alternating) series. The Martingale is basically the same thing at irregular intervals.
For example: http://en.wikipedia.org/wiki/Grandi's_series
Or: 1-2+3-4...
A similar treatment with Martingale would likely lead to the same result as k_c. Just as in Grandi's series, you can't really talk about the "sum" (win) in a definite sense. Probably the conflict between k_c's treatment and the others above is really semantic about what the "sum" means. Sure, if you just decide to stop somewhere you'll always be able to claim a win at some point. But the series undoubtedly "converges" to EV as k_c shows.
Of course, whether Martingale "works" or not with an infinite bankroll and no table limits is completely irrelevant to anything regarding actual play. This is to head off some moron who will use the above "working" statement to justify their nonsensical progression claims.
zengrifter
Banned
Kudos to KC for trumping Thorp, Griffin, et al ! zg
zengrifter
Banned
You need to respond to the TSA scanner dangers.
THIS thread is not for your benefit. Everyone knows that 1-1=ZERO, capish? zg
aslan
Well-Known Member
Although I usually agree with KC, this time I think he has bitten off more than he can chew. There is no way on God's green earth that a martingale can lose given no bet limits and unlimited bankroll. If a sure thing cannot beat a house edge, there's something wrong with your math.
QFIT
Well-Known Member
Can't agree.
First, unlimited bankroll is meaningless. This means that you have infinite money, the house also has infinite money, and that you both have infinite time. All three are impossible making the situation moot.
Secondly, if you have infinite money, how can you increase it? That is, how can you win?
Thirdly, the cost of money (i.e. interest rate) on an infinite bankroll is infinte. That means, to win, you have to make more than infinite money as your money is decreasing in value by inflation.
Fourthly, in an infinity of time, all things are possible. That is, it is possible that you will lose every bet. And, if it is possible, it must happen in an infinite number of trials.
I don't think that's true zg. If you have to include unlimited capital in the equation, then you also have to include the inevitable occurance of never winning another hand before you die. Apply them to a negative EV game, and I think you must die before ending every single losing streak you can have. No??