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RocketManJames
12-06-2004, 04:24 AM
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
12-06-2004, 05:10 AM
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
12-06-2004, 07:03 AM
[ 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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=1041529&amp;page=&amp;view=&amp;s b=5&amp;o=&amp;vc=1) 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
12-06-2004, 07:37 AM
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
12-06-2004, 07:48 AM
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
12-06-2004, 08:04 AM
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&lt;=x&lt;=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
12-06-2004, 08:19 AM
[ 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
12-06-2004, 09:13 AM
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 &lt; ~.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) &lt; (1/10)^n; as n goes to infinity, P(72o n times) goes to 0

elwoodblues
12-06-2004, 11:58 AM
You are also guaranteed to be up 1 billion dollars at some point in infinity.

Some things just aren't worth worrying about.

Sephus
12-06-2004, 12:08 PM
but you could be dealt 72o 100000000 times in a row and go bust.

RocketManJames
12-06-2004, 01:51 PM
I thought I did as well... until I discussed this with a few friends, and then I saw myself flip-flopping based on new arguments presented. And that was when I realized, I didn't have as strong a grasp on the subject as I had hoped,
which led to my post. Haha.

-RMJ

Rockatansky
12-06-2004, 02:27 PM
[ QUOTE ]
but you could be dealt 72o 100000000 times in a row and go bust.

[/ QUOTE ]

If you play in infinite number of hands, at some point you WILL be dealt 7-2o 10,000,000 times in a row. However, you probably will have amassed a sufficient bankroll to withstand the hit.

I know I'm just restating what BruceZ already said and what the risk of ruin formula tells us. What I don't know is why I'm being taught high school algebra and being subjected to your withering sarcasm for posting something that is unequivocally correct.

EliteNinja
12-06-2004, 04:39 PM
Engineers like to thnk about what is practical and we don't always trust all that is proven mathematically.
The chances of busting out with a huge roll in your lifetime if you follow good bankroll management procedure is in a practical sense much less than the chance of winning the lottery. Just chillax and everything will be all good.

Cerril
12-06-2004, 06:17 PM
Honestly, I'm not trying to be sarcastic, and my plea for proof is completely honest here. The thread Bruce referred me to had a concensus that it was not guaranteed that you would go broke at least once over an infinite number of hands (possible, certainly, and maybe likely, but not guaranteed). I can't figure out the math for how that's true, so I'm interested in someone showing me. I've been trying to be as detailed as possible with my reasoning and the math I have.

I am -not- trying to be patronizing, I only want to make sure that whoever is going to correct me sees my thought process here so to more easily point out where I'm wrong.

Rockatansky
12-06-2004, 06:55 PM
I was mostly annoyed with Sephus for mocking my earlier post and kind of lumped you in. I think we were saying the same thing, but you actually know math and I don't.

Intuitively, it seems like over an infinite number of hands you will have to go broke eventually so long as there is any variance whatsoever. Given a large enough sample size, it seems inevitable that you will eventually hit a losing streak that is long enough to deplete your bankroll.

As for proving this mathematically, I can't offer any help there.

I think some of the reasons people automatically reject the idea of inevitably busting out is 1) it seems contradictory to say that, over time, the risk of ruin for a +EV game approaches zero, yet you will inevitably lose (so long as there is variance); and 2) what the idea implicitly says about our vulnerability at the poker table.

Sephus
12-06-2004, 07:31 PM
[ QUOTE ]
I was mostly annoyed with Sephus for mocking my earlier post and kind of lumped you in.

[/ QUOTE ]

i dont know why you thought i was mocking your post. i thought i was clarifying your point for cerril.

AngryCola
12-06-2004, 09:02 PM
I think you asked for the ROR formula. The link BruceZ gave you took you to a thread where he provided ANOTHER link to the ROR formula.

Risk of Ruin Formula (http://archiveserver.twoplustwo.com/showflat.php?Cat=&amp;Number=207100&amp;page=&amp;view=&amp;sb=5&amp;o =&amp;fpart=all&amp;vc=1)

AngryCola
12-06-2004, 09:03 PM
[ QUOTE ]
If you play in infinite number of hands, at some point you WILL be dealt 7-2o 10,000,000 times in a row.

[/ QUOTE ]

This is simply not true.

EDIT - As for as I know.

If BruceZ was to tell me otherwise, I would believe him. /images/graemlins/smile.gif The guy knows his stuff. I'm not sure, but there are proabably only 3-7 people on these forums on par with Bruce's probability(etc) knowledge.

Cerril
12-06-2004, 09:06 PM
Again, and this may demonstrate the same lack of understanding of probability and statistics, but that seems true.

There is a finite probability of x occuring... it's very, very small for x= (72o 10,000,000 times in a row) but it is positive and it is measurable. Over an infinite stretch x should occur as often as any other occurence with that probability.

Cerril
12-06-2004, 09:07 PM
Thank you, I may try to do some of the math I'm asking about myself, but I'm not feeling like I'm necessarily qualified to do the computations, especially if any of my basic assumptions are flawed.

gaming_mouse
12-06-2004, 10:17 PM
[ QUOTE ]
[ QUOTE ]
If you play in infinite number of hands, at some point you WILL be dealt 7-2o 10,000,000 times in a row.

[/ QUOTE ]

This is simply not true.

EDIT - As for as I know.

[/ QUOTE ]

This is true. But it certainly does not prove that you guaranteed to go broke.

Cerril, the math you are looking for is in the derivation of the risk of ruin formula. What are looking for that you can't find there?

gm

AngryCola
12-06-2004, 11:56 PM
I'll take your word for it, gm.

I guess I'm wrong, but I just don't understand why you MUST be dealt it that many times in a row at some point.

I'm admittedly not very advanced in the probability department, but something about this concept seems really off to me. Of course, many aspects of probability are counter-intuitive, so I'm not surprised that I'm probably wrong. /images/graemlins/grin.gif

I should know better to accuse someone of being wrong in the probability forum, as I'm certainly not smart enough to know better than someone with a solid understanding of these concepts.

Cerril
12-06-2004, 11:59 PM
I found the error in my thinking. RoR calculations -are- the chance that you'll go broke over infinite hands. I was thinking that they had some other use. I think that was a little dense of me, so thanks for bludgeoning it in.

Incidentally, if you withdraw along any sort of schedule, that will tend to send your chance of going broke at least once up through the roof, likely to 100%

AngryCola
12-07-2004, 12:08 AM
[ QUOTE ]
Incidentally, if you withdraw along any sort of schedule, that will tend to send your chance of going broke at least once up through the roof, likely to 100%

[/ QUOTE ]

I don't think this is right, but I was wrong last time.

In fact, why am I even commenting on this now?

Silly me. Please continue this thread in my absence. I really want to hear the answer to the the above statement.

Cerril
12-07-2004, 12:33 AM
Well I'm certain that you would need to withdraw a % of your bankroll at regular intervals, over a certain point. I would imagine that it would increase your RoR drastically if you did so, and would come closer to the situation described in the more general case of withdrawing all your profits every month.

No matter how small your RoR, if you start over at regular intervals in the long run (infinity again), you're guaranteed to bust out. It takes awhile, but with a 1% RoR (fairly high by some standards) by your 70th withdrawl you'll have dodged a 50% chance to lose all your money by this point. 70 more withdrawls (which, monthly, is over 22 years and probably substantially more since you don't win every month) and you're down to a 25% to never have gone broke.

So obviously with a small enough RoR and reasonable (small enough) withdrawls, you can have a minimal risk of busting out in a normal lifetime. But over infinity I think that withdrawls may drive the RoR to 100%

gaming_mouse
12-07-2004, 12:56 AM
[ QUOTE ]
I guess I'm wrong, but I just don't understand why you MUST be dealt it that many times in a row at some point.

[/ QUOTE ]

Cola,

One of the other posters explained it. It's simply a result of the fact that it will occur with some finite probability.

If you do something that has a finite probability of occuring an infinite number of times, it will eventually occur. Nothing fancy, Just the law of large numbers.

gm

AngryCola
12-07-2004, 01:00 AM
I'm pretty sure I can grasp that.
... I think. /images/graemlins/smirk.gif

Any way you could clear up the other little bit Cerril and I were just discussing?

Probability really fascinates me, but I don't think I have the knowledge or the mind for some of the concepts yet.

gaming_mouse
12-07-2004, 01:18 AM
[ QUOTE ]
Any way you could clear up the other little bit Cerril and I were just discussing?

[/ QUOTE ]

You mean the bit about withdrawing at a schedule. Of course, from a practical perspective, when you take out 500 everytime you reach 1500, say, then your ROR just gets reset to whatever it is for 1000.

If you do this and keep playing forever, and keep on withdrawing 500 whenever you reach 1000, then yes, eventually you will go broke, because you have some finite ROR everytime you go back to 1000. It now becomes the same situation we were just discussing.

gm

EliteNinja
12-07-2004, 01:28 AM
Mathematically, using the risk of ruin formula, you need an INFINITE number of hands to be 100% sure that you will become broke. Practically, that number is unachievable and therefore you cannot reach certainty that you will go broke over a huge number of hands less than infinity.

The new question should be:
"If God played an infinite amount of hands would He go broke?" (kidding)

AngryCola
12-07-2004, 01:36 AM
Thanks guys. I appreciate your patience. /images/graemlins/smile.gif

gaming_mouse
12-07-2004, 01:41 AM
Ninja,

Are you discussing the case where you keep pulling money out, or just the normal case of playing forever?

Because, as has already been pointed out, playing an infinite number of hands DOES NOT ensure that you will go broke. The ROR forumula calculates the chance that will go broke after an infinite number of hands, and for a positive winrate and bankroll that number IS NOT 100%.

I'm guessing you know this, but I just wanted to clarify.

gm

BruceZ
12-07-2004, 12:22 PM
Here are the bankroll and risk of ruin formulas (http://archiveserver.twoplustwo.com/showflat.php?Cat=&amp;Number=207100&amp;page=&amp;view=&amp;sb=5&amp;o =&amp;fpart=all&amp;vc=1). As you realized, these give the risk of ruin for playing to infinity, without removing money from the bankroll, and that this risk of ruin can be made as small as desired as long as the EV is positive.

You wrote:

[ QUOTE ]
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).

[/ QUOTE ]

I can't make sense out of much of this paragraph. Chance of going broke is a probability, and a probability can only take on values from 0 to 1, so it cannot "go to infinity". Actually, as your bankroll increases as a linear function of hands played, your risk of ruin decreases exponentially. For example, if your bankroll doubles from B to 2B, your risk of ruin becomes squared (making it smaller since it is less than 1) because going broke now requires that you lose B twice, so your risk of ruin goes from ror to ror*ror = ror^2(smaller).

Now you are correct that you can compute the ror as the cumulative sum of chances that each future hand may be the first of a slide. It is true that each of these chances are non-zero, and that there are an infinite number of them. But the sum of an infinite number of non-zero probabilities certainly need not diverge to infinity, (or even 1!) and of course it must be &lt;= 1. In fact, this series will converge to the same risk of ruin given by our formulas based on our bankroll at the beginning of the interval. The successive terms in the sum decrease so that their contribution becomes negligible once we have passed beyond the critical period at the beginning of our trials. Hence this concern about effects of playing to infinity are misplaced since play to infinity poses negligible risk, with virtually our entire risk occurring in the first few hundred hours.

Now it is certainly true that as we play to infinity that we will encounter arbitrarily large negative swings. The likelihood of such swings depends on the number of standard deviations below the mean that they represent. The square of the standard deviation, or the variance, will increase linearly with the number of hands played, so the standard deviation will increase only as the square root of hands played. Meanwhile, our bankroll will increase linearly with hands played, so these swings, no matter how large, will represent only a fraction of our bankroll by the time they occur, so they are safely absorbed and do not cause us to go broke.

The above is an explanation of how the risk of ruin behaves. It is not a proof. For a proof, see this post where I rule out the ror = 100% solution from the derivation of the ruin formula for coin flips and finite bankrolls (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=791989&amp;page=&amp;view=&amp;sb =5&amp;o=&amp;vc=1). Then this solution holds in the limit as our opponents bankroll -&gt; infinity, which is the same as number of hands -&gt; infinity. This derivation for the coin flip problem is used in the derivation of the more general risk of ruin formula (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=683150&amp;page=0&amp;view=ex panded&amp;sb=5&amp;o=14&amp;fpart=2#Post682045683150) which use EV and SD. Also in this same thread, Pzhon has also outlined another proof.

gaming_mouse
12-07-2004, 02:43 PM
Bruce,

Well said, as always.

gm