Exactly every possibility will happen... infinity is a long time. That just means 1 in 33 times you will have to replenish your BR from an outside source despite being a poker player with a decent W/R.

So, does this mean that every poker player will go busto, if they play an infinite amount of sessions, no matter their winrate/sd/br, hypothetically speaking?

I know its stupid because infinite is not attainable, but if its true that you are assured of going broke playing to infinity.... Its kind of depressing to think about lol

i think that's the point of adding your wins to your bankroll ... as your bankroll grows, your chances of going broke decrease.

if you just started with X bankroll and used your wins for hookers and blow and never added $1 to your bankroll, you would eventually bust.

Yes, but given an infinite # of sessions, we would eventually hit a downswing so hard, that we would lose our entire roll. If we played to infinity. Its kinda like my dice example, we would eventually lose it all because eventually we would hit a streak of rolling 1 that lasted so long, it would wipe our roll.

so then no player would ever make any money playing poker, so we should all just quit while we're ahead? i don't buy it.

No, its fine to play poker, because no human being can play an infinite amount of sessions. Even if you add up every human being's sessions from the history of poker till now, including all online sessions, it still would not reach infinity.

Infinity is by definition: every possible situation ocurring. Meaning eventually you will flip a coin and it will come up heads 999999999999999999999999 times in a row.

So if a poker player could play infinite sessions, eventually he would lose his entire roll because he would run his KK into AA a million times consecutively, or whatever bad beat scenario you can imagine.
Its next to impossible to occur, but if you play *infinitely* it almost surely will happen.



Think of it this way, imagine we live in an *infinite* universe.

Well that means infinite planets, solar systems, galaxies, stars, etc

Now if there are infinite planets, then Somewhere out there, there MUST exist another earth where there is an identical clone of you and me, and another earth where dinosaurs still roam the earth, and another earth where a poker deck has 51 cards instead of 52, etc etc etc

Infinity means every possibility can occur. The human mind doesnt grasp infinity quite well, i actually was confused at first myself

Dont worry though, you'll never reach infinity, so its pretty unlikely you'll get dealt KK vs AA a million times in a row

grim in the house watch your mouth

This was an exceptionally good listen, thanks for sharing. I've seen the Benford's used in some inventive ways (a friend used it to try and spot fraud in elections), makes total sense here.

Really the lesson is trying to dodge taxes is a bad idea.

HappyLuckBox, you're thinking about this the wrong way imo.

Think about it like this. You flip a fair coin - 50% chance of heads, 50% chance of tails. Every time you get heads, you are paid $101. Every time you get tails, you pay $1. Your starting bankroll is $1,000. Your win-rate is (.5)($101) + (.5)(-$1) = $50/flip.

Notice that unless your very first 1,000 flips are tails, it becomes incredibly difficult for you to bust, because the amount you have to lose consecutively to bust your bankroll becomes absolutely massive. So the probability you bust (this is your R-o-R) is tiny.

This is obviously an extreme and contrived example, but it's the EXACT same idea as for someone with a positive win-rate in poker. Your way of thinking about "every possibly situation occurring" is kind of correct, in that any ordering of heads and tails will eventually occur. Even 10,000+ tails will occur at some point. But the question is, what is the probability that that sequence of 10,000 tails in a row occurs at a point when it could threaten your bankroll?

How does ROR factor in your opponents? Let's pretend you have a game with Awful Andy. Andy is so bad, he always loses. Seriously, the guy will get in $198 of his $200 stack with a royal flush and then fold for some odd reason the last $2 bet. What's with this guy? He's such an awful Andy.

On the other hand, sometimes you play against Awesome Eric. You think Phil Ivey is good? Compared to Awesome Eric, he's like an Awful Andy. Eric is like some Jedi Knight. I'm telling you, the man knows your holecards and the cards coming on the flop, turn, and river. The guy never loses. If only I could be so awesome as Awesome Eric.

In reality, we never play against either Awful Andy or Awesome Eric because no one is ever that bad or good, but we do play against ranges of people in between and I would think that should have a profound effect on winrate and ROR.

Of course it does. But you're just talking about table selection and giving the players extreme tendencies. Table selection absolutely affects win-rate and standard deviation, which are what define R-o-R (along with your bankroll) to begin with.

Edit: Re-reading your question, I'm not sure I follow actually. But basically, who you play against will affect your win-rate (how good you are relatively to the competition) and standard dev. (how volatile the game is), and these are what define R-o-R. So asking "how does R-o-R factor in your opponents" is kind of a roundabout question. It doesn't "factor it in" except by the nature of the fact that it depends on win-rate and s.d. (which, at their core, are "average" types of values over many different opponents (unless you play in the same lineup home game and you all sit in the same seats every time)).

But Given a large enough sample size (infinity) wont that sequence almost surely occur? Obviously its an extreme hypothetical haha

Im probably just arguing semantics here.

If i could take 9trillion flips of that coin i obviously would do it in a heartbeat

Yes, given enough time, a monkey sitting at a typewriter will eventually, yet randomly, type out the complete works of Shakespeare.

You guys argue about the silliest things

I don't think it's all that hypothetical, and it's really nothing to do with semantics. If the way you're thinking about it were correct, no one would ever make money - that can't be, in a zero-sum game, right? Yes, that particular sequence will occur. The question is how big is your bankroll at that point?

P(E) = P(flip 1,000 tails in a row in 1,000 flips) = (0.5)^1,000 = 9.33x10^(-302)

Now how many ways can that happen within the first 101,000 flips, let's say: # Ways = 101,000 - 1,000 = 100,000 ways.

P(flip 1,000 tails in a row in first 100,000 total flips) = P(E)x(# Ways) = (9.33x10^-302)x(10^5) = 9.33x10^-298

Hope this kind of illustrates the point. Like the odds that you flip 1,000 tails in a row within trillions of trillions of flips is pretty damn good. But how ****ing enormous is your bankroll by the time it does? Like we're dealing with expectations and probabilities here, you don't automatically bust if you lose a bunch in a row.

It would take approximately 10^301 flips to get to 90% probability of having had a sequence of 1,000 tails in a row at any given point. Your bankroll after just 1% of that time had passed is 10^299[($101)(.5) + (-$1)(.5)] = $5x10^300... So probably not too concerned at that point that we've lost 1,000 times in a row. Even if a I fudged a factor of 10 or something in there, it's still pretty clearly not a concern.


Another way to think about it would be: H = number of heads, T = number of tails, SB = starting bankroll. (Remember we win $100 on heads, and lose $1 on tails) So to bust our bankroll, we'd have to reach the point where:

H($100) + T(-$1) < -$SB.
T(-$1) < -$SB - H($100)
T > H(100) + SB

Just because we reach any sequence of heads and tails at some point doesn't mean we ever satisfy the above relation. There is some probability for it (the risk-of-ruin), but it's not a guarantee.

Sorry for thread derail, btw.

Great stuff thanks sir

Something similar has already happened. It's called Fox News.

Good post scourrge, I was just responding to Happyluckbox's post #6388 where he mentions if your bankroll never grows for example you have 50 BI and you spend everything you win over 50 BI you will eventually (it might take many years) have a bad run where you lose 50BI, the scientific term for this is called being "doomswitched".

A more practical way to look at ROR is simply subtracting the amount of money you plan to spend out of your winnings. Say you win $50/hr and plan to spend $20/hr, just plug in $30/hr into the ROR formula. The $30/hr is being put back into your bankroll.

Discussions like this really make me wish I had my standard deviation stat. =[ would be way easier (and by easier I mean possible) to calculate my risk of ruin.

I'm slowly transitioning from $1/2 to $2/5. I've lost my first few $2/5 sessions so far, but have not made any horrible mistakes. Lost some big pots where the EV was slightly positive.

In terms of BBs, how should I expect my win rate to change in terms of %? In other words, say I my $1/2 win rate is 7BBs/hr. What can I expect at $2/5?

This might not be as simple an answer as it seems, due to the fact that the rake at $2/5 is will be less of a drain as in $1/2. At the room I play at (and most it seems), the rake at both stakes are identical.

This must be balanced against the fact that the $2/5 player pool will be better on average.

Thoughts?

I'm really not sure if there is an answerable question here.

Obviously no one knows how your win-rate will change. Maybe you'll win 7BB/hr at $2/5. Maybe you'll win 1BB/hr. Maybe you'll lose 3BB/hr. Maybe you'll go completely busto.

The 2/5 pool is usually not that much better.

You're WR should go up because the rake is a lower % of the pot and the stacks are deeper

This depends on too many factors that we don't know for us to be able to give any kind of valid answer beyond "it depends."

On the one hand, in larger casinos that spread a lot of mid-stakes and higher games, $2/$5 may not have a substantially tougher player field than $1/$2. But if it does, then you're going to be balancing the rake difference with your supposed baseline win-rate decrease from moving up. If there's no decrease, but the relative rake is decreased (same absolute amount), then obviously your relative win-rate will improve.

Speaking as someone who hasn't moved up to $2/$5, I can't speak from personal experience, but even if I could, my message would be the same:

• Your sample at $2/$5 is likely tiny.
• Your sample at $1/$2 is likely not even close to large enough to have a reasonable estimate of your true win-rate there.
• Don't worry much about win-rates - worry more about winning plays.
• If you're beating $1/$2, are rolled for $2/$5, and feel you are not being destroyed at $2/$5, then play $2/$5 for a while and see what happens.

i'm beginning to believe that i can lose 50 straight sessions ... 4 in a row and 7 of 11 this year.

u-g-l-y

Only 11 sessions since the new year? Need more sample