Martingale for SNGs

[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. [img]/images/graemlins/cool.gif[/img]

[GrekeHaus]

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

[/ QUOTE ]

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.

....

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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[img]/images/graemlins/smirk.gif[/img]

[FlyWf]

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