General Discussion Poker Forums

OK, I know how to figure out risk of ruin given SD, winrate, and bankroll but that assumes that you are going to play forever.

I want to calculate my RoR given that I'm only going to play say 4,000 hands.

I think the 6th equation in this link is what I need but that one seems to only work for blackjack as there is no wager size in poker. How do you convert that equation to work for poker?

http://www.bjmath.com/bjmath/Betsize/winwayz/tmpweb.htm

See section 3.5 on how to solve that problem: http://www.bjmath.com/bjmath/proport/riskpaper1.pdf

If you're going to play forever your risk of ruin would be 100%, unless you had 0SD, right?

If you're going to play forever your risk of ruin would be 100%, unless you had 0SD, right?

No, because if you're a winning player your BR will grow infinitely large.

No, you will go broke.

No, you won't. Risk of Ruin calculations are classic and well-known.

http://pokerbankrollblog.com/calculating-bankroll-requirements.htm

Note how the RoR is not bound by a goal, either profit or number of hands. Note how it is not 100%. Apply palm to face.

Note that that's a poor way to model poker

Note that that's a poor way to model poker

That might be true - because poker results are probably not normally distributed - but it's close enough, and the outcome of the formula is very valid.

This is so counter-intuitive to me.

I would understand it if "ruin" was -∞, but since the number for "ruin" is 0(or -X, where X is our original bankroll), I would think we eventually hit it.
If we start with 1 BI in a flipping game (where we would have say 60% chance of winning) and would flip an infinite amount of times, would we not eventually go bust?

It also does not intuitively make sense to me that if we run the trial an infinite amount of times that it still happens a % of times, where is the uncertainty, the variance, when we do it an infinite amount of times?

I understand that if we make a +$EV action an infinite amount of times we are going to end up with an infinite amount of $, but I just think that inevitably on our way there we are going to hit 0 at some point.

Also, if we imagine a situation where our risk of ruin would be 10% over ∞hands. If we would play ∞*∞ hands (and ofc. ∞*∞=∞) our RoR would be 100%? Or is that an "illegal" mathematical action?

Sorry if my somewhat broken English is a barrier, if there is something I wrote that doesn't make sense (grammatically not like, my logic) please let me know and I'll try to rephrase it.

If we start with 1 BI in a flipping game (where we would have say 60% chance of winning) and would flip an infinite amount of times, would we not eventually go bust?

It also does not intuitively make sense to me that if we run the trial an infinite amount of times that it still happens a % of times, where is the uncertainty, the variance, when we do it an infinite amount of times?

I understand that if we make a +$EV action an infinite amount of times we are going to end up with an infinite amount of $, but I just think that inevitably on our way there we are going to hit 0 at some point.

Also, if we imagine a situation where our risk of ruin would be 10% over ∞hands. If we would play ∞*∞ hands (and ofc. ∞*∞=∞) our RoR would be 100%? Or is that an "illegal" mathematical action?

Well theres a very high ROR if we start with only one buyin, but we won't always go bust. The key point is that every time we win, it is less likely that we bust. This is because we have a positive winrate.

So even if there is always a slight % chance that we have a -everything megadownswing, the vast majority of trials will result in winning more, which makes that chance subsequently even smaller.

I'm not sure exactly how it all works out, but my intuition is that theres a fraction involved, and the top and bottom have the same order, and it cancels out into a # between 0 and 1.

Your ∞*∞ idea is mathematically flawed, yes. ∞*∞ does not work the same way as n*n. Also, if we consider the range of possible bankrolls after ∞ trials, a 10% ROR means that 90% of our bankrolls will be of infinite order, and thus nearly impossible to bust. so after ∞*∞ or ∞*∞*∞*∞ or whatever, our ROR remains 10%.

Think about this example, maybe it will make more sense too you intuitive.

Lets say we start with a million dollar bankroll. We flip a fair coin. If we lose we pay 1$, if we win we get another million dollars. Do you really think that you will go broke in the long run playing this game? This game is fundamentally not different than one with a smaller bankroll or with a smaller edge.

Multiplying infinite with infinite doesn't make much sense by the way, unless you start with higher level mathematics. And if you have a 10% risk of ruin over a certain number of hands, you certainly can't multiply both and think you always go bust by playing 10 times as many hands. Besides the fact that your risk of ruin shrinks if you play more hands, the basic math is also wrong.

If you flip a coin you hit heads 50% of the time. So you always hit heads if you flip two times? (Obviously not)

Not to belabor the point, but you can't multiply ∞*∞. I mean, ∞ already includes everything. You can't make it any bigger. Just like you can't make 0 any less 0.

Not to belabor the point, but you can't multiply ∞*∞. I mean, ∞ already includes everything. You can't make it any bigger. Just like you can't make 0 any less 0.

Just for fun:

∞/∞ = 1
(∞^∞)/∞ = ?

DJ Sensei;

Well theres a very high ROR if we start with only one buyin, but we won't always go bust. The key point is that every time we win, it is less likely that we bust. This is because we have a positive winrate.

So even if there is always a slight % chance that we have a -everything megadownswing, the vast majority of trials will result in winning more, which makes that chance subsequently even smaller.

I'm not sure exactly how it all works out, but my intuition is that theres a fraction involved, and the top and bottom have the same order, and it cancels out into a # between 0 and 1.

Yeah, I know that if we don't eventually go bust it's because that even though we never have 0% RoR, the "added" RoR is a fraction of the RoR that was "added" before. Kinda like in the Achilles and the tortoise problem where he would catch him at 1.111..., there the added time is always 1/10 of what you added before. (like, you start with a 1BI roll. The odds of you busting are X(which is smaller than 1) and then if you win the flip it's X^2 and then X^3, so the odds of you busting at all would be (X+X^2+X^3+X^4+X^n)) But I thought that it would approach 1 if n were infinite. But it might approach some other number depending on the X. Actually, after thinking about it some more, I think it only works when the X is 0.5 and you start with a br of 1bi. But I think that only means that my math is wrong, cause if you'd be slightly -EV and started with a huge roll, you'd not go bust in 100% of instances according to that calculation. Fuck.

Anyways, if your roll is infinite, and you're a -EV player, do you not hit a profit of X in 100% of instances?
That's kinda the thing I have a problem with, it's that the scale is open to one end, but not the other, but you still don't hit the closed end in 100% of instances.



Reading this over it doesn't make much sense in English(might not in my native language anyways, since I've changed my view of this multiple times through writing it, but I think it would be better to leave it as it is then to change it) but you might to be able to decipher wtf I'm trying to say.

Hielko;

Lets say we start with a million dollar bankroll. We flip a fair coin. If we lose we pay 1$, if we win we get another million dollars. Do you really think that you will go broke in the long run playing this game?

Yeah, I do. Or well, I don't anymore, but I still don't fully understand why not.
EDIT: I don't mean that we stand to lose money long term in this game, I just mean that eventually our bankroll will hit the number X, which just so happens to be 0. I do believe this is incorrect though, now, but I don't intuitively believe it.

And if you have a 10% risk of ruin over a certain number of hands, you certainly can't multiply both and think you always go bust by playing 10 times as many hands.

No, you'd go bust 1-(0.9^10), which was what I was trying to do with infinity, but I was really skeptical it was mathematically correct to multiply infinity that way, it is okay with finite numbers. I wasn't adding them up

(∞^∞)/∞ = ?

That's a tough one.