aslan
Well-Known Member
You left out Sarah Palin, Shad. :grin:
Mathematical Proof that Progressions will never Overcome a Negative Expectation Game
- Thread starter iCountNTrack
- Start date
I was referring to the weight given to Thorp et al.'s opinions on the matter.
KC proved that with any finite bankroll, it won't work.
Dyepaintball12
Well-Known Member
Nice!
Also, just take a look at the Billion Dollar casinos being built and that's pretty good proof progressions don't work.
Also, just take a look at the Billion Dollar casinos being built and that's pretty good proof progressions don't work.
aslan
Well-Known Member
While on the subject of upper limits, does anyone know how the casino calculates it RoR? It would seem after this discussion that table limits are unnecessarily low. I can see how a billionaire might close down a casino with a couple of well placed bets if there were no limits, but these $500, $1,000 $2,000 and $5,000 maximum limits seem unrealistically low. Anyone?
zengrifter
Banned
They definitely work some of the time. The successful progressionists pride
themselves in avoiding more losing sessions than could be expected statistically. zg
What the proof is showing is that with an unlimited bankroll overall EV of a martingale based on original bet is negative when EV for each individual trial is negative and that the bigger the bankroll you start with the more you can expect to lose by virtue of being able to continue.
If the game is such as a fair coin toss it shows that overall EV = 0. This simply means that for a large enough number of trials ratio of units won to units lost approaches 1.
If the game is positive EV for an individual trial it shows that you can expect to eventually win 100% of initial bet given an unlimitied bankroll. This is the definition of a successful martingale as you have won 1 unit.
Basically all I did was to sum all possible bets weighted by their probability of occurrence. This results in a geometric sequence which has limiting values as number of terms becomes large.
It's really not surprising that positive EV cannot happen in a negative EV game but there seems to be something about a martingale that makes people think it can be done with an unlimited bankroll. The proof shows that even with unlimited funds each time a martingale is started and EV is negative you can expect to lose more than you win and the greater the bankroll the greater the expected loss will be in the long run.
It's splitting hairs, but the proof shows that for any finite bankroll (regardless of how large) it's an unwinnable system. If the bankroll is infinite (what I meant by unlimited, and presumably what the others meant), you can win any arbitrary amount you want, by expanding your bets by any ratio desired (i.e. triple after a loss). You're guaranteed to be ahead at some point, after which you can stop with whatever win you choose.
This does not, however, asymptote to a +EV for infinite trials.
This does not, however, asymptote to a +EV for infinite trials.
QFIT
Well-Known Member
The flaw is that when you bring infinity into the question, then you must include an infinite number of possibilities. One possibility is that you will not be ahead at some point. In fact, in an infinity of time, this must happen.
aslan
Well-Known Member
As much as I love blackjack, I would not want to be stuck at the table for an infinite number of hands anyway.
I disagree; there is no case where you will not be ahead at some point; it requires only one win. The odds of that happening approach zero as trials increase.
QFIT
Well-Known Member
Yes, the odds approach zero per trial. But, there are an infinite number of trials. What's infinity times zero again.
Point is, when we talk about infinity, we must include every possibility. It is possible you won't ever be ahead.
iCountNTrack
Well-Known Member
Being ahead does not mean that the system mathematically works. If i played one hand of blackjack in my life and decide to hit my hard 19 i get a deuce win the hand and never played for the rest of my life does it mean it was a good strategy.
If also play an infinite number of hands, how am it going to deal with the infinite sum of accumulated expectations.
Yes, the odds approach zero per trial. But, there are an infinite number of trials. What's infinity times zero again.
Point is, when we talk about infinity, we must include every possibility. It is possible you won't ever be ahead.
Ah, but there need not be an infinite number of trials. You get to stop when you have a profit. And you always will at some point. That's why it "works".
If you're forced to play forever, or have no control of stopping point, it all asymptotes to the EV, of course.
I never said it was a good strategy! (I need to explain that?) I'm just saying what the math says. If you have no bankroll or time limit, you can play until you win whatever arbitrary amount you want, then quit. Those conditions obviously do not exist in reality, but we're talking about a mathematical proof here. Like I said, KC did a good job proving the failure for any finite bankroll, but not an infinite one, which is what others (Thorp, etc.) were talking about. They're both right.
As for handling infinite sums on divergent series, take a look at my first post in the thread for other examples; there are ways of handling them mathematically that aren't so simple, but are interesting.
iCountNTrack
Well-Known Member
Again if you decide to stop at a profit, it does not mean that "it worked" because you have no control nor knowledge on when this is going to happen.
I could also play 10 hands using a Martingale (with a finite bankroll) and i could be ahead on the 10th hand
That said the system DOES NOT work whether i have a finite bankroll or an infinite one.
Sonny
Well-Known Member
If there is an infinite number of outcomes then you want to consider each of them. On any given hand you can push, win (between 1-8 units) or lose (1-8 units). Each trial is a sequence of hands that ends with the player being ahead by 1 unit (but not having been ahead before). How many such sequences of 10 hands are there? How many sequences of 1,000 hands? How many sequences of an infinite number of hands?
But what if you are never ahead? As Qfit said, it must happen to some people. Those poor souls will have a huge effect on the results. Sure, many of the trials will show an eventual profit but the amount will never be as much as an infinite loss. The game is still -EV. You are more likely to lose infinitely than to win infinitely. That's how negative progression systems work. They increase the probability of winning a session but the overall EV is still negative because of the inevitable huge losses.
-Sonny-
What do you define as "working"? I'm defining it only as having the ability to stop with a win. I see no reason why one can't do so. You keep playing until you win, and since your bets are unrestricted, you're guaranteed to come out ahead. I don't see what is so controversial about this.
You do get to have control and knowledge of when to stop, because you're stopping after the win. It works just fine with an infinite bankroll.
(Again, this isn't useful or practical. We're purely in mathematics here.)
There aren't an infinite amount of outcomes. You're always going to be ahead at some point, and since you can control the stopping point, you get to stop when you're ahead. As I said, the probability of all hands being lost as the number of trials increases approaches zero.
You'll always win, eventually.
aslan
Well-Known Member
I am still trying to figure out if I had an infinite bankroll how I could add to it by winning a finite sum from the house??? Can you add to an infinite bankroll??? Would it then be infinite plus the amount won??? How does one understand a greater than infinite amount??? How can anything be greater than infinity???
If I lose one hand, do I then have less than infinity???
Do you think any of us really know what we're talking about??? Do you think the math systems we are using to prove/disprove these propositions are equal to the task??? Isn't infinity an impossible "quantity" in the real world?
I'd say the proposition itself is unsolvable as stated, like, "What happens when the irresistible force meets the immovable object (and both are indestructible)?" Therefore, I have concluded that the proposition is not only impossible, it is unsolvable. It cannot exist.
Whoever said it was pointless should have stopped there. The discussion is entirely pointless.
If I lose one hand, do I then have less than infinity???
Do you think any of us really know what we're talking about??? Do you think the math systems we are using to prove/disprove these propositions are equal to the task??? Isn't infinity an impossible "quantity" in the real world?
I'd say the proposition itself is unsolvable as stated, like, "What happens when the irresistible force meets the immovable object (and both are indestructible)?" Therefore, I have concluded that the proposition is not only impossible, it is unsolvable. It cannot exist.
Whoever said it was pointless should have stopped there. The discussion is entirely pointless.
psyduck
Well-Known Member
Pointless in practice, but theoretically interesting
I have been thinking about it in another way.
Let's set a group of infinite number of players all using the Martingale strategy. Whoever wins one unit stops playing. Among this group of infinite number of players, there must be one player who never wins as he approaches playing infinite number of hands. The amount of money he is losing is infinite, more than the total win from all winners combined. As a result of this unluckiest player in the world (hopefully not me), the whole group will end up in the red.
I have been thinking about it in another way.
Let's set a group of infinite number of players all using the Martingale strategy. Whoever wins one unit stops playing. Among this group of infinite number of players, there must be one player who never wins as he approaches playing infinite number of hands. The amount of money he is losing is infinite, more than the total win from all winners combined. As a result of this unluckiest player in the world (hopefully not me), the whole group will end up in the red.