Am I stupid? I can't fit these two concepts into any type of harmony.

[Alex/Mugaaz]

A: Infinity
B: Chance

Basic, maybe flawed question: Let's say you have a bankroll of 1 trillion BB's, and your winrate is 3bb/100. If you played an infinite amount of time, would you eventually bust out?

Since it is always possible that you could lose X hands, will this eventually happen since the time range is infinity? How does this concept mesh with the fact that you should always be up in the long run since you have an edge?

[PairTheBoard]

I think this is tricky but I don't remember exactly how to handle it. It should be easy for the practicing mathematicians around here.

It's true that an event with low probabilty will occur infinitely many times given an infinte sequence of trials. However, a sequence of events whose non-zero probabilities become smaller and smaller might not occur infinitely many times given an infinite sequence of trials. Your problem has that flavor because as time goes on and you build up more winnings your element of ruin gets smaller.

PairTheBoard

[daryn]

this one seems easy. your bankroll is finite, $1 trillion you said. so of course probability dictates that you will eventually go bust.

[pokerplayer28]

i am stupid, but here are my thoughts

since the risk of ruin is greater than 0 and will always be greater than 0 if you play for infinity it is almost certain you will go broke, but even if you play for infinity you will never be 100% certain you will go bust. So it is likely youll go bust but not certain.

This is just what popped into my head, i would really like to hear from someone qualified.

[Darryl_P]

The best way to model this is with a random walk. You start at 0 and then you take a step up to +1 with probability p>.5 and a step down to -1 with a probability q=1-p. You repeatedly take steps either up or down in this manner until you either reach + 1 trillion or - 1 trillion.

There is a formula which says you will reach + 1 trillion first with a probability of [(q/p)^1 trillion-1]/[(q/p)^2 trillion-1], which is very, very close to 1 in the example described.

The formula and its development can be found here

To answer the question of EVER reaching -1 trillion (ie. going broke), simply consider a repeated application of doubling your money before losing it, first with 1 trillion, then with 2 trillion, then 4 trillion, etc.

Your probability of winning this infinite series of trials is approximated by N*(N^1/2)*(N^1/4)*(N^1/8)*..., where N is your probability of winning your first trial, ie. a number just very slightly less than 1.

This infinite product of probabilities (all less than 1) is equal to

N^(1 + 1/2 + 1/4 + 1/8 + ...) = N^2

which is still very, very close to 1.

So while you are not guaranteed to win after an infinite amount of time, your probability of doing so is very, very high.

The key to understanding why the infinite product of numbers less than 1 can still remain very close to 1 in this case is to understand that the infinite series of fractions 1 + 1/2 + 1/4 + 1/8 + ... equals 2.

[Alex/Mugaaz]

I will admit that I am not able to fully understand what you are saying, however in trying to decipher it looks like it doesn't apply since this is not a random walk? Am I incorrect?

[Darryl_P]

Actually it is because you can define each step up as reaching a certain goal like +1000 BB, say and each step down as -1000 BB. Since you have an edge your probability of reaching +1000 BB before reaching -1000 BB is greater than 50%.

Of course if you do it this way then your ultimate goal of 1 trillion BB becomes 1 billion "steps" which is not exactly the same number, but the principle remains the same and the result will still be that you have greater than a 99.999999999 (tack on a few more if you like) % chance of winning.

[Alex/Mugaaz]

If what you're saying is that whether you bust out or not is not certain either way, even over infinity, then I agree. I think there is a chance of busting out, but that it is less than 100% and greater than 0%. Also it seems to follow that the longer this game goes on less chance of you busting out.

[fnord_too]

[ QUOTE ]
this one seems easy. your bankroll is finite, $1 trillion you said. so of course probability dictates that you will eventually go bust.

[/ QUOTE ]

I disagree. This is a random walk about a positively sloped line. If he were even money, then I would say he will go bust, since eventually he will hit a bad spot long enough to depleat any arbitrary number, but here the length of the bad streak needed is increasing, since he has positive expectation.

I have a picture in my head that I am not expressing well. Let me try it this way: Say this were flipping a coin where hero is getting 1.1:1 on his money. As time wears on, he will need a bad streak considerably longer than his longest good streak to go bust. To frame it slightly differently, and possibly sophisticly, over any set of trials, cancel the wins and losses. In this case, you cancel 11 losses with 10 wins. Then you are left with pure loss or gain. For him to go bust in this coin example, his longest loosing streak needs to be very much longer than his longest winning streak to date. (I know he can have multiple losing streaks, but you can just extend the period where you cancel wins and losses there.)

Is this making any sense? Basically, if you take a random walk whos expectation has a non zero slope, you do not have to eventually get to zero if you start sufficiently away from zero on the side that your slope is pointing.

[fnord_too]

[ QUOTE ]
If what you're saying is that whether you bust out or not is not certain either way, even over infinity, then I agree. I think there is a chance of busting out, but that it is less than 100% and greater than 0%. Also it seems to follow that the longer this game goes on less chance of you busting out.

[/ QUOTE ]

Yeah, that's what I was trying to say. And since you are starting with one trillion BB, I think that your chance of going bust is very close to 0.