Martingale for SNGs

Has anyone here ever used the Martingale System for managing his SNG bankroll? How was it?

[microbet]

I think several posters have done something like that; right before they went broke.

p.s. In poker it's called "tilting up."

[handsome]

It's better for roulette. Never fails.

Isn't it weird that there's still poverty in the world despite the Martingale system?

Whats the Martingale system

It's funny, I go to a college that specializes in math and science. Last year a group of guys came up with the brilliant idea of using the Martingale system for roulette, and they went around getting people to join them by investing money and playing. I would hear them explaining to potential investors how it worked and they actually had at least ten people interested. They were so excited and talking about what they were going to do with the money they won. Up until that point I thought about filling them in on what would really happen, but the fact that none of them took the time to do the math made me feel like they deserved to lose. Smart people can be so dumb sometimes.

[Sciolist]

Google is your friend

[Sciolist]

Or "Winning the $5 SNG and wondering what to do next"

It's where you double your bet after every loss. So each time you make a bet it would put you 1 unit up overall when you win. For example, you bet $1 on red in roulette. If you lose, then you bet $2 next time. If you lose that you would bet $4. Let's say you win that time, then you've lost $3 and won $4 and are now up $1 overall. Obviously this system doesn't work because every single bet you make has a house edge so no matter how you bet you can't win.

[runner4life7]

the bigger reason it fails is because there is a max bet, if there was no max bet and you had an infinite bankroll it would work in theory.

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the bigger reason it fails is because there is a max bet, if there was no max bet and you had an infinite bankroll it would work in theory.

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No it wouldn't.

[splashpot]

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the bigger reason it fails is because there is a max bet, if there was no max bet and you had an infinite bankroll it would work in theory.

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No it wouldn't.

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Yes it would. If you had an infinite bankroll. And if there were no max bet. It is impossible to lose infinity times in a row.

[SuitedSixes]

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the bigger reason it fails is because there is a max bet, if there was no max bet and you had an infinite bankroll it would work in theory.

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No it wouldn't.

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Yes it would. If you had an infinite bankroll. And if there were no max bet. It is impossible to lose infinity times in a row.

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You type faster than me.

[KenProspero]

What's the difference between using the Martingale System and Card Counting in Black Jack -- I'll tell you

If they catch you counting cards, they'll kick you out of the casino. If they catch you using a Martingale, you can expect lots of comps -- they may even send an airplane to pick you up from whereever you live.

Unless you play online. Online casino software can detect people using the Martingale system so the system rigs itself and makes you lose.

[The Yugoslavian]

Perhaps the OP really means a sort of 'step' type approach to STTs.....not really martingale but more like 'parlay'.....there have been some good posts on that as well.

Yugoslav

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the bigger reason it fails is because there is a max bet, if there was no max bet and you had an infinite bankroll it would work in theory.

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No it wouldn't.

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Yes it would. If you had an infinite bankroll. And if there were no max bet. It is impossible to lose infinity times in a row.

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Yes.. it's impossible to lose infinite times in a row, but if you lose infinite times in a row, you lose infinite dollars. While this seems like a moot point, it's very significant and means that you can't ignore the possibility.

[valenzuela]

but u can lose 100000 times on a row making the martingale -ev. Even if u can bet $82192819829389389139021391283, u are still making a -EV bet.
No matter how big ure bankroll is, u cant make money on the long run making solely -EV bets.

[applejuicekid]

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the bigger reason it fails is because there is a max bet, if there was no max bet and you had an infinite bankroll it would work in theory.

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No it wouldn't.

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Yes it would. If you had an infinite bankroll. And if there were no max bet. It is impossible to lose infinity times in a row.

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This is incorrect. There was a thread about this in the SMP forum. I'll try to find it, but basically it said that at any given point the sum off all your wagers would be negative even if you were doing it with an infinite bankroll.

[wmspringer]

At any given point your expectation is negative, yes (assuming a house edge)

However, due to variance there should be times when you are up. Quit during one of those times and you've made money. :-)

[runner4life7]

just think about this. if it takes 1,000,000 times to win you still are positive, if it takes infinity times you still are positive. Its just a matter of how long it takes till you win. If you have infinity time to play, infinity money to play with, and no max bet you will have to win at some point at which point you are positive and that can be repeated.

[Newt_Buggs]

this is a stupid debate because there is no such thing as an infinite bankroll.

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Unless you play online. Online casino software can detect people using the Martingale system so the system rigs itself and makes you lose.

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Untrue

[Newt_Buggs]

I think that he was being sarcastic. At least I really hope that he was being sarcastic.

[Mr_J]

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If you have infinity time to play, infinity money to play with, and no max bet you will have to win at some point at which point you are positive and that can be repeated.

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Incorrect. There is always the chance that you lose from the beginning.

[runner4life7]

no there isnt, because with infinity tries you WILL win at some point. while it doesnt matter because its not physically possible, in theory you will win. You will not win with 1,000,000 tries, but infinity, yes.

What if you used the martingale system on both red and black? Then every time the wheel stops your winning one bet.
Wouldnt you?


fwiw, it was this type of EV thinking for roulette that started my obsession for poker.

[runner4life7]

ive never heard of that before, that could be interesting.

[pergesu]

The problem with your debate is that the points you're now arguing have absolutely no substance.

"You WILL win at some point." Well...why? Can you prove it?
"Not necessarily." Can you prove it?

Someone who knows the foundation of these arguments is going to step in and say "you guys are dumb, it's ____ because of ____." I don't know what the second blank is, so I can only ask you guys to fill it in.

[Mr_J]

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because with infinity tries you WILL win at some point

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No, it's just a probability.

[runner4life7]

i agree that there is a one in a huge number chance that you will lose every time and i know that this system cannot work, but with infinity tries you will win, that or I do not understand the concept of infinity, which is quite possible.

[Mr_J]

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What if you used the martingale system on both red and black? Then every time the wheel stops your winning one bet.

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Not sure what you mean??

[runner4life7]

i think what he means is like have them be two seperate systems. So first you bet $1 on each and if its red than you bet 2 on black and 1 on red the next time and so forth. I see this just reducing variance a little maybe? but obviously has the same downfall.

Edited to apologize for bumping, and didnt want to bump again, I got side tracked on this because i was a moron and tried something similar on BJ and while i know it fails, find it interesting.


re-edited to reply without bumping:

i did not mean because you can be ahead at some point its a winning game but rather that at some point in each individual thing you will win and hence make $1. As long as you always play till you get that win you will be ahead. This is different then just quitting when you are ahead, and is all pointless anyways because it cant happen so im finally done with this thread.

[curtains]

Cmon stop bumping this retardo thread, its on 4 pages already!! I take back what I said about the forum being better.

$1 on red
$1 on black

wheel spins: lands on black =>you win $2

$2 on red
$1 on black

wheel spins: lands on black=> you win $2

$4 on red
$1 on black

wheel spins: lands on red=> you win $8

$1 on red
$2 on black

wheel spins: lands on black => you win $4

you've wagered a total of $13 and won a total of $16

Net: $3
And you spun 3 times (discounting the first one which will be breakeven.) So you're EV should be $1 per spin.

[applejuicekid]

The sum of a series of -EV wagers will never be positive no matter how you sequence them. It is really silly to think that the order in which you make your bets will make you a winner. If you took every wager you made in a Martingale system and ordered them from largest to smallest and wagered them in that order would you still be winning?

Let's say you had an infinite bankroll. Each time you want to win a $1. You win $1 then repeat. Even if you had an infinite bankroll you will be making wagers lager than all the amount you have won more often then you will be up money.

Saying that you will be up at one point and thus it is a winning system is silly. Even with infinite bankroll and infinite tries you could never be up. The same argument could be said about any -EV game. If I play blackjack for an evening I am often times up at some point. Does this mean if I quit once I get up I have found a way to beat blackjack?

[pergesu]

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And you spun 3 times (discounting the first one which will be breakeven.) So you're EV should be $1 per spin.

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rofl you can't base your EV on the results of three previous trials. You can't base your EV on any results of previous trials. That's absolutely ridiculous.

[FlyWf]

And when it lands on O or OO?

[GrekeHaus]

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Cmon stop bumping this retardo thread, its on 4 pages already!! I take back what I said about the forum being better.

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Sorry to bump this retardo thread, but I have the answer to the question.

The problem with this question is that it is a divergent series. This is why people are confused and arguing about it. Given an integer N, you will at one point be up N units and at one point be down N units with probability 1. This is because no matter how well you've done so far, the probability that you will never go on a streak that will win or lose you that much is precisely 0.

Because of this, it is impossible to say anything about what will happen over an infinite amount of time. It is simply a question of the way you choose to group the terms when you sum them.

[Guthrie]

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this is a stupid debate because there is no such thing as an infinite bankroll.

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Congress would disagree.

[Pokey]

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The sum of a series of -EV wagers will never be positive no matter how you sequence them.

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This statement that sounds so logical is actually incorrect because of two bizarre assumptions in this problem:

1. Infinite number of possible bets.
2. Ever-increasing wagers.

Here's a thought exercise to prove the point. Let's say that we play a game. We flip a fair coin. If it comes up heads, you pay me 1/2 my bet. If it comes up tails, you win my entire bet. Every individual bet is clearly +EV for you. You promise to play as many times as I want, and you promise to offer me as large a "tab" as I want. I can guarantee myself that, with enough time, I can bankrupt you, no matter how large your (finite) starting cash is. Here's how:

1. If I haven't bankrupted you, I bet $2.
2. Flip the coin. If it comes up heads, go to step #1 ahead $1.
3. If it comes up tails, triple my previous bet and go to step #2.

Run that sequence forward in your mind. Divide it into "sets" where each set ends with a flip of heads. The probability of a set continuing to N flips is (1/2)^N. KEY POINT: since N can be infinite by (nonsensical) assumption, the probability that a set never ends becomes (1/2)^infinity, which limit theory tells us equals zero. So every set WILL end, and whenever a set ends, the sum total of all my bets and wins equals exactly $1. I can play as long as I want (by assumption), and you'll extend me all the credit I want (also by assumption), so I can bankrupt you, guaranteed.

Note that along the way, there will be plenty of times when I'm *seriously* in the hole, but because you allow me to continue increasing my bet to whatever amount I want, it doesn't matter.

Note also that everybody's intuition that -EV bets can't be bad remains true in all real-world situations; since nobody has an infinite bankroll and since no casino allows infinite betting, none of this matters in the real world, and martingale betting techniques will eventually lose you money.

Final side note: if you use a martingale betting method, you don't change your expected value, but you DO change the distribution of outcomes. A martingale betting strategy will let you win far more often than you lose, but your losses will be so staggeringly large that they wind up more than offsetting all your wins in the long run.

[Big Limpin']

That was a great post, Pokey. Thank you for taking the time. /images/graemlins/cool.gif

[GrekeHaus]

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The sum of a series of -EV wagers will never be positive no matter how you sequence them.

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This statement that sounds so logical is actually incorrect because of two bizarre assumptions in this problem:

1. Infinite number of possible bets.
2. Ever-increasing wagers.

Here's a thought exercise to prove the point. Let's say that we play a game. We flip a fair coin. If it comes up heads, you pay me 1/2 my bet. If it comes up tails, you win my entire bet. Every individual bet is clearly +EV for you. You promise to play as many times as I want, and you promise to offer me as large a "tab" as I want. I can guarantee myself that, with enough time, I can bankrupt you, no matter how large your (finite) starting cash is. Here's how:

1. If I haven't bankrupted you, I bet $2.
2. Flip the coin. If it comes up heads, go to step #1 ahead $1.
3. If it comes up tails, triple my previous bet and go to step #2.

Run that sequence forward in your mind. Divide it into "sets" where each set ends with a flip of heads. The probability of a set continuing to N flips is (1/2)^N. KEY POINT: since N can be infinite by (nonsensical) assumption, the probability that a set never ends becomes (1/2)^infinity, which limit theory tells us equals zero. So every set WILL end, and whenever a set ends, the sum total of all my bets and wins equals exactly $1. I can play as long as I want (by assumption), and you'll extend me all the credit I want (also by assumption), so I can bankrupt you, guaranteed.

Note that along the way, there will be plenty of times when I'm *seriously* in the hole, but because you allow me to continue increasing my bet to whatever amount I want, it doesn't matter.

Note also that everybody's intuition that -EV bets can't be bad remains true in all real-world situations; since nobody has an infinite bankroll and since no casino allows infinite betting, none of this matters in the real world, and martingale betting techniques will eventually lose you money.

Final side note: if you use a martingale betting method, you don't change your expected value, but you DO change the distribution of outcomes. A martingale betting strategy will let you win far more often than you lose, but your losses will be so staggeringly large that they wind up more than offsetting all your wins in the long run.

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Apparently, you didn't read (or understand) my previous point. I'll elaborate.

The problem again here is that you're talking about a series which doesn't converge. Here's a simple example:

1-2+3-4+...

Person A sees this series and says

1+(-2+3)+(-4+5)+...=1+1+1+...=infty

Person B sees the series and says

(1-2)+(3-4)+(5-6)=...=-1-1-1-...=-infty

What you're looking at here is an example similar to this. It's a series that doesn't converge, so summing it has no meaning. You're taking the "Person A" approach. Here's the "Person B" approach to summing the same series.

Let the person flip until they are behind. For the same reasons as stated in the previous argument, they will be behind, with probability 1. Group these flips together and call the sum a_1. Now, let the person keep flipping until the sum is less than a_1. Call this next group of flips a_2. Keep doing this for each n until a_{n+1}-a_n-...-a_1<0.

In this way, we get a_n <=-1 for every n. So taking the sum gives

a_1+a_2+...<=-1-1-...=-infty

So now your winning strategy is actually losing you an infinite amount of money.

The *key* when calculating EV is that you're actually taking a limint as n->infinity. In this case, your EV will be -1/4*(average bet size). Simply summing the numbers in a way that supports your hypothesis doesn't work since you are summing a series that doesn't converge.

[runner4life7]

i didnt want to have to come back to this but you guys are really over thinking what i said. just stop trying to be like ohh i know calculus and just think for a second.

If you just double your bet till you win and then go back to one each time tell me why that does not make you money. If there is no max bet and you have all the money in the world.

[Pokey]

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Apparently, you didn't read (or understand) my previous point. I'll elaborate.

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The *key* when calculating EV is that you're actually taking a limint as n->infinity. In this case, your EV will be -1/4*(average bet size). Simply summing the numbers in a way that supports your hypothesis doesn't work since you are summing a series that doesn't converge.

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Sorry; I wasn't trying to address your specific point, but now I will -- you're making a different assumption from the one I started with. You're assuming we'll actually *play* an infinite number of games. I've assumed, as has the original poster, that the gambler facing the -EV gets to choose when to stop. That's all that's necessary to make the original position correct and my statements hold true. So long as one of the two players has a finite amount of money and the other has an infinite amount of money, this contest is over before it begins.

At any given time, I'm likely in debt as I play along. I'll go for long swings where I'm ridiculously far in debt. However, the expected time until my next profit is always finite, and I will eventually be positive again. Then, all I have to do is quit.

You assume we keep playing forever, at which point you're absolutely right -- the series never converges. However, that's the one feature of the game that allows us to always say that we can make a profit if we get to choose when we stop.

In short, we're both right. Of course, that's the kind of crazy thing that often happens when you have infinities running around.

[Pokey]

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If you just double your bet till you win and then go back to one each time tell me why that does not make you money. If there is no max bet and you have all the money in the world.

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You're right, so long as you make those two assumptions. If either of those assumptions does NOT hold true, however, then you can lose your entire bankroll, and you'll do so more often than you'll double your money. In the long run, your expectation is negative because your gambles are all negative. It's only in the infinite situations that you can make money with this strategy.

Quick example: say you start with $255 and you use a martingale betting strategy where you're going to win 49% of your bets and lose 51% of your bets. You play until you either double your money or wipe yourself out. The $255 lets you bet $1, then $2, then $4, then $8, $16, $32, $64, and $128; at that point, you're out of cash. The odds of hitting a losing streak of 8 in a row (bankruptcy) is going to be (0.51)^8 = 0.46%. In order to double your money, you'd have to avoid that bad situation 255 times in a row. The odds of dodging a 0.46% event 255 times in a row is (1-0.0046)^255 = 31%. In other words, you'd lose $255 69% of the time and win $255 31% of the time. (The math isn't *quite* perfect because you'd still have a non-zero bankroll if your catastrophic loss came after the first time, but your bankroll would be smaller, giving you an even lower chance of rebuilding; suffice it to say it's in the ballpark.) Even with a very minor -EV betting situation, martingale strategies are losers in the long run. Of course, we should have already known that, since ALL -EV bets are losers in the long run.

[GrekeHaus]

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Sorry; I wasn't trying to address your specific point, but now I will -- you're making a different assumption from the one I started with. You're assuming we'll actually *play* an infinite number of games. I've assumed, as has the original poster, that the gambler facing the -EV gets to choose when to stop. That's all that's necessary to make the original position correct and my statements hold true. So long as one of the two players has a finite amount of money and the other has an infinite amount of money, this contest is over before it begins.

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In essence, this game amounts to a glorified version of "Give me a dollar". The problem as I see it, is that you're still counting one flip as an event when each event is not independent of the others. When you opponent agrees to enter the game with you, he's not agreeing to just flip the coin once. He's agreeing to flip the coin until he loses. A single event here is essentially waiting until your opponent loses, not an individual coin flip. In this sense, each individual event is still -EV for your opponent.

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In short, we're both right. Of course, that's the kind of crazy thing that often happens when you have infinities running around.

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I suppose...

What I meant in the original post was that how would using the Martingale be better than using the Kelly% when it comes to money managing SNG buy-ins?

To simplify things, let's take the case of a 66% ITM (aka 20% ROI)HEADS UP SNG player. He plays 5+5. Once he loses, he plays 5+5 again. Once he loses, he moves up to 11. Once he loses, he moves up to 33. Once he loses, he plays 33 once again. Once he loses, he plays 55. Once he loses, he plays 109. Once he loses, he plays 215. That's 8 loses in a row. There's no way a 66% ITM (aka 20%ROI) heads up player would lose 8 games in a row.

How would this system be better or worse than using the Kelly to make money management decisions?

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There's no way a 66% ITM (aka 20%ROI) heads up player would lose 8 games in a row.

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[FlyWf]

Uh, a 66% ITM at the $5s |= a 66% ITM at the $100s.

I am referring to heads up SNS. In a heads up SNG, 66% in the money equals 20% ROI.

But my question is regards to using the martingale for SNGs. I wonder how big or small the fluctuations would be relative to Kelly and Half Kelly. Assume positive EV not negative EV (like roulette).

[zipppy]

This would never work, because someone that has +ev at $109+ buyins is not going to want to play at the $5+1s to win some money playing a martingale system in SNGs.

Also, if someone wants to start at the 5s to win money using a martingale system, I'm willing to guess they don't have +ev at bigger buyin SNGs, or the bankroll for them.

so basically the only ones this might work for wouldn't want to use it.

>>>ZIP

[pergesu]

What he's saying is that if you're a 66% ITMer at the $5s then there's a really good chance you're not a 66% ITMer at the 100s.

okay /images/graemlins/cool.gif

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This would never work, because someone that has +ev at $109+ buyins is not going to want to play at the $5+1s to win some money playing a martingale system in SNGs.

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What if he did anyway? What would be the consequences fluctuationwise and EV wise?