Mathematical Proof that Progressions will never Overcome a Negative Expectation Game
- Thread starter iCountNTrack
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I finally understand it! I'm not kidding. The way you just put it does make perfect sense, and it does explain why the opposing arguments fail. 
At last I can sleep at night without numbers racing to infinity swirling through my brain. Now, I am kidding. :joker: It was interesting, and for me it is now concluded.
At last I can sleep at night without numbers racing to infinity swirling through my brain. Now, I am kidding. :joker: It was interesting, and for me it is now concluded.
aslan said:
I finally understand it! I'm not kidding. The way you just put it does make perfect sense, and it does explain why the opposing arguments fail.
At last I can sleep at night without numbers racing to infinity swirling through my brain. Now, I am kidding. :joker: It was interesting, and for me it is now concluded.
At last I can sleep at night without numbers racing to infinity swirling through my brain. Now, I am kidding. :joker: It was interesting, and for me it is now concluded.
The game: Baccarat, min bet: $20. Table Max: $10,000
Pedro's scheme:
1) obtain several large lines of credits from casinos.
2) bet on PLAYER only (no 5%-commissions!) .
3) always start with a $20 bet, and double up the bet (up to table max of $10,000) after each loss.
Pedro had made lotsa short-term-profits from the above scheme. However, bad runs like the sample below had him in the hole of $600,000 over years of double-up plays. Unfortunately some gamblers have failed to learn from others' mistakes. You can still see the double-up-players everyday in the mini-Bacc or the EZ-Baccarat pits, sad!
According to Jose, losing 10 hands straight were not likely, but he said he was wrong. He had run into big losses (like the one below) about 20 some times over couple years. Gamblers can learn wisdom in 2 ways: (1) from his own mistakes, or (2) from Jose's mistakes
!
Code:
Baccarat
"Double-up"
bets
1 20
2 40
3 80
3b 80
4 160
5 320
6 640
7 1,280
7b 1,280
8 2,560
9 5,120
10b 5,120
10 10,000
26,700 Total losses
Note: Double-up-bets #3b, #7b, and #10b were the same amounts because the previous hands were tied, i.e., pushed in bj lingos.Jose should have studied this link first, and he might have saved himself $600,000.
Link to 70,000 hands of baccarat:
http://wizardofodds.com/baccarat/1000BaccaratShoes-8deck.txt
If you need more info in order to understand the game of baccarat, click this link:
http://wizardofvegas.com/guides/baccarat/
A simpler martingale analysis
I'm kind of a nut because I keep trying to put things in their simplest perspective.
The following shows simply what to expect from a martingale for any number of trials. Bankroll, infinity, etc. do not need to be considered.
Let W = probability of winning a trial
Let n = number of trials
I'm leaving out the math but it can be showm that probability of being up 1 unit after n trials = 1-(1-W)^n and the probability of being behind 2^n-1 units after n trials = (1-W)^n
For martingale to be expected to succeed,
probability of winning 1 unit needs to be more than (2^n-1) times the probability of being behind 2^n-1 units,
1-(1-W)^n > (1-W)^n*(2^n-1)
This simplifies to 1 > (1-W)^n*2^n and
(1/2)^n > (1-W)^n and
(1-W) < 1/2 and
W > 1/2
What this shows is that probability of winning 1 unit is greater than (2^n-1) times probability of being behind 2^n-1 units after n trials only on the condition that probability of winning a trial is greater than 1/2. It doesn't matter size of bankroll and it is true for any number of trials.
Similarly it can be shown that
Prob of winning 1 unit = (2^n-1) times Prob of being behind 2^n-1 units after n trials when W = 1/2
Prob of winning 1 unit < (2^n-1) times Prob of being behind 2^n-1 units after n trials when W < 1/2
The main contention in this thread was that somehow a martingale is discontuous for an unbounded bankroll and a 1 unit win can always be guaranteed even in a negative expectation game. Although it's possible to be up 1 unit after n trials the probability of being behind by 2^n-1 units exists no matter the number of trials. Moreover in a negative EV game the probability of being behind by 2^n-1 units outweighs the probability of winning 1 unit after n trials. An unbounded bankroll allows for continued betting but does not guarantee an eventual win.
I'm kind of a nut because I keep trying to put things in their simplest perspective.
The following shows simply what to expect from a martingale for any number of trials. Bankroll, infinity, etc. do not need to be considered.
Let W = probability of winning a trial
Let n = number of trials
I'm leaving out the math but it can be showm that probability of being up 1 unit after n trials = 1-(1-W)^n and the probability of being behind 2^n-1 units after n trials = (1-W)^n
For martingale to be expected to succeed,
probability of winning 1 unit needs to be more than (2^n-1) times the probability of being behind 2^n-1 units,
1-(1-W)^n > (1-W)^n*(2^n-1)
This simplifies to 1 > (1-W)^n*2^n and
(1/2)^n > (1-W)^n and
(1-W) < 1/2 and
W > 1/2
What this shows is that probability of winning 1 unit is greater than (2^n-1) times probability of being behind 2^n-1 units after n trials only on the condition that probability of winning a trial is greater than 1/2. It doesn't matter size of bankroll and it is true for any number of trials.
Similarly it can be shown that
Prob of winning 1 unit = (2^n-1) times Prob of being behind 2^n-1 units after n trials when W = 1/2
Prob of winning 1 unit < (2^n-1) times Prob of being behind 2^n-1 units after n trials when W < 1/2
The main contention in this thread was that somehow a martingale is discontuous for an unbounded bankroll and a 1 unit win can always be guaranteed even in a negative expectation game. Although it's possible to be up 1 unit after n trials the probability of being behind by 2^n-1 units exists no matter the number of trials. Moreover in a negative EV game the probability of being behind by 2^n-1 units outweighs the probability of winning 1 unit after n trials. An unbounded bankroll allows for continued betting but does not guarantee an eventual win.
My (latest) conclusion is that martingale will NOT succeed in an infinite/unbounded scenario... nor will it work in a real-world tightly bounded environ...
BUT it WILL work in an artificial no-limit - unlimited BR - high #trials scenario, as Thorp, et al have agreed.
For example, 100M hands - no limit on bet size or BR.
Will that do it, or was Thorp wrong? zg
BUT it WILL work in an artificial no-limit - unlimited BR - high #trials scenario, as Thorp, et al have agreed.
For example, 100M hands - no limit on bet size or BR.
Will that do it, or was Thorp wrong? zg
Thorp basically said it would work ("quite reasonably") absent table limits and credit/BR limits. I quoted the book directly and even gave the specific page number in an earlier post.
And I pretty much agree with you (and Thorp), for reasons I've made abundantly clear over the course of the thread.
(To nitpick your post though: stating 100M hands specifically isn't quite right - that 100M-th hand might well keep you deep in the hole. Perhaps "up to" 100M hands, you're "virtually certain" to win any amount of your choosing.)
And I pretty much agree with you (and Thorp), for reasons I've made abundantly clear over the course of the thread.
(To nitpick your post though: stating 100M hands specifically isn't quite right - that 100M-th hand might well keep you deep in the hole. Perhaps "up to" 100M hands, you're "virtually certain" to win any amount of your choosing.)
Ahhh when i thought this thread was over, this is again not true, you WILL NOT win any amount of your choosing because if you encounter one win you will only be ahead by one unit.
QFIT, K_C, and Myself have shown using many different approaches will fail whether your bankroll is unlimited or limited , you are playing an unlimited number of hands.
Revisit those posts:
http://www.blackjackinfo.com/bb/showpost.php?p=205537&postcount=111
http://www.blackjackinfo.com/bb/showpost.php?p=206226&postcount=206
http://www.blackjackinfo.com/bb/showpost.php?p=206575&postcount=259
http://www.blackjackinfo.com/bb/showpost.php?p=206602&postcount=276
http://www.blackjackinfo.com/bb/showpost.php?p=206896&postcount=298
http://www.blackjackinfo.com/bb/showpost.php?p=207091&postcount=304
QFIT, K_C, and Myself have shown using many different approaches will fail whether your bankroll is unlimited or limited , you are playing an unlimited number of hands.
Revisit those posts:
http://www.blackjackinfo.com/bb/showpost.php?p=205537&postcount=111
http://www.blackjackinfo.com/bb/showpost.php?p=206226&postcount=206
http://www.blackjackinfo.com/bb/showpost.php?p=206575&postcount=259
http://www.blackjackinfo.com/bb/showpost.php?p=206602&postcount=276
http://www.blackjackinfo.com/bb/showpost.php?p=206896&postcount=298
http://www.blackjackinfo.com/bb/showpost.php?p=207091&postcount=304
Yeah I thought so too. Leave it to ZG to stir it up again! :devil:
Since credit and bets are unlimited, the unit is arbitrary. So you get to win whatever you like. Or you can keep going until you're up again. It doesn't matter, mathematically.
[Again, we're well within the realm of mathematics and proofs only, not what might actually work or not in real-life. Table and credit limits kill this, and the numbers do get absurd.]
I've countered the previous posts already; primarily the issue is whether or not the calculation should be taken after an infinite number of hands or not. Sure, mathematically if there is an arbitrary (or no) stopping point, your expectation is a loss. That I agree with.
But the resulting BR curve in this case is unusual - in typical -EV situations the BR asymptotes to the -EV slope, but when the bets grow accordingly, as in this case, it oscillates to positive immediately after every win, and does not asymptote. (Nor does it asymptote positively.)
Therefore, a single win puts you in positive territory, and you're free to quit with a profit. This particular aspect is not addressed or included in any of the aforementioned limit analysis.
As pointed out previously, the only way this won't work is if you never win a hand. Whether that's possible or not after infinite hands is something we wasted a lot of time arguing about, but it doesn't really matter - thus the language "virtually certain".
But I think we'd all be hard-pressed to find any AP technique that guaranteed a win even if you happen to lose every hand forever.
Leave it to ZG to stir it up again! :devil:Since credit and bets are unlimited, the unit is arbitrary. So you get to win whatever you like. Or you can keep going until you're up again. It doesn't matter, mathematically.
[Again, we're well within the realm of mathematics and proofs only, not what might actually work or not in real-life. Table and credit limits kill this, and the numbers do get absurd.]
I've countered the previous posts already; primarily the issue is whether or not the calculation should be taken after an infinite number of hands or not. Sure, mathematically if there is an arbitrary (or no) stopping point, your expectation is a loss. That I agree with.
But the resulting BR curve in this case is unusual - in typical -EV situations the BR asymptotes to the -EV slope, but when the bets grow accordingly, as in this case, it oscillates to positive immediately after every win, and does not asymptote. (Nor does it asymptote positively.)
Therefore, a single win puts you in positive territory, and you're free to quit with a profit. This particular aspect is not addressed or included in any of the aforementioned limit analysis.
As pointed out previously, the only way this won't work is if you never win a hand. Whether that's possible or not after infinite hands is something we wasted a lot of time arguing about, but it doesn't really matter - thus the language "virtually certain".
But I think we'd all be hard-pressed to find any AP technique that guaranteed a win even if you happen to lose every hand forever.
