Doomed to go bust?

[RocketManJames]

Talking with a couple friends...

You're in a very very good poker game. You have a significant edge over your opponents, but that edge is not 100%. Assume that collectively, your opponents have an infinite bankroll... you will play an infinite number of hands.

Is it true that you are guaranteed to go bust at some point in this infinitely long game? Isn't it true that at some point, you're guaranteed to lose some 'infinite' hands in a row? Or is there some limit equation that comes into play that says you are not guaranteed to go bust?

Or take a +EV game where a fair coin is flipped. Say that if it comes heads, you win $10, if it comes tails you lose $1. The house has an infinite bankroll... you have a finite bankroll. Are you guaranteed to go bust if you play for an infinitely long time?

-RMJ

[Cerril]

Pretty much. If you look around here or the internet in general for risk of ruin equations, you'll notice that you can never achieve 100% certainty that you won't go broke. That 100% basically means that it -is- certain that over an arbitrarily long period of time you're going to run out of money.

[BruceZ]

[ QUOTE ]
If you look around here or the internet in general for risk of ruin equations, you'll notice that you can never achieve 100% certainty that you won't go broke. That 100% basically means that it -is- certain that over an arbitrarily long period of time you're going to run out of money.

[/ QUOTE ]

ARRGH! These statements are not the same. Your first sentence is true, and your second is patently false. See this post for an explanation from when the last guy made the same mistake.

And Rocket, I thought you understood this. Yes, you will eventually lose M games in a row, for any M no matter how large. That doesn't mean you go broke. In the limit as you play to infinity, the size of your bankroll relative to these swings goes to infinity, the size of your bankroll goes to infinity, and your risk of ruin goes to zero.

[Cerril]

Hmm... I see the arguments but the math still seems a little opaque for me.

If we assume you have an EV of something like +.02BB/hand (2BB/100), theoretical risk of ruin decreases at a linear rate. That would seem to imply as the number of hands increases to infinity, the cumulative chance of actually going broke also goes to infinity (in any given hand there is a finite chance that it is the first in a slide resulting in a loss of your entire bankroll).

It's been a few years since I've done this sort of calculus and I don't actually have the formula for RoR on hand, but if someone could point out the errors and show me the math for this I'd be grateful.

[Rockatansky]

There's no way you could ever have a risk of ruin equal to zero playing poker. To use an extreme example, it is possible that you will be dealt 7-2o for the rest of your poker playing career until you finally run out of money.

[Cerril]

But it's not, strictly, possible that you would be dealt 72o for the rest of eternity. That's where things get sticky. Specifically I'm replying to the post above mine though, which states that the cumulative risk of ruin over infinite hands is a finite number. Since obviously it would have to be infinite (1.0), nonexistent (0), or a finite number in between, there has to be some math that takes the assumptions and comes up with a cumulative RoR given a SD, WR, and other factors played out over an infinite number of hands.

i.e. with a 2BB/100 winrate and a 15.0BB/100 SD, there is a x% chance that at some point in your career, never withdrawing, you will go broke; 0<=x<=100

My contention is that the RoR will always go to 100% over infinity given an SD greater than a certain point, but I've seen enough equations that derive out to finite real numbers that I am open to being proven wrong. I'd just like to see the mathematical proof/derivation or a technical english language explanation of how that works.

[Rockatansky]

[ QUOTE ]
But it's not, strictly, possible that you would be dealt 72o for the rest of eternity.

[/ QUOTE ]

It most certainly is. It is highly improbable, but it is possible. My point is that, all other concepts aside, risk of ruin will always be non-zero due to the chance element of poker.

[Cerril]

Over an infinite number of hands you'll be dealt 7c2h (and so on) precisely as often as every other combination of two cards. Once you bring infinity into play the possibility that you'll be dealt only 16 different combinations out of 2652 shrinks to 0. Infinity truly is the 'long run' where every combination appears with exactly equal frequency.

If the math matters:

16/2652 < ~.1 (it's actually close to .06, I'm just using a much larger number for the sake of the proof)

The probability of being dealt 72o is less than 10%. Of being dealt it twice in a row less than 1%.

P(72o n times) < (1/10)^n; as n goes to infinity, P(72o n times) goes to 0

[elwoodblues]

You are also guaranteed to be up 1 billion dollars at some point in infinity.

Some things just aren't worth worrying about.

[Sephus]

but you could be dealt 72o 100000000 times in a row and go bust.