another math problem :/

[Scott31 0]

but my problem is when it comes time to sit at the table, i have a hard time crunching the numbers in my head. i think its my ADD.

I have ADHD....just a hyper form of it. It's funny, most people think guys with ADD can't focus....it's quite the opposite actually. We just can't focus on one thing, but are very good at paying attention to several things. It's like this: the normal person can focus 100% on one thing, an ADD person focuses on 10 things at 10% each :-). My point? Hmmm, not sure I have one.....a terrible side affect of ADD I will say this. Multi tabling online is a gift from God for someone like me. It doesn't allow time for my mind to wonder.

[MrConceit 0]

agreed.completely valid points--NL is a different beast, and you don't want to play tiny edges in NL, you want to play big, certain ones for lots of chips. and agreed, you don't want to call all of your chips when the pot offers you a small profit in NL.i still disagree about the coin-flip busting scenario, however... here's a similar but more exxagerated scenario: i get $1 for every dice roll between 1-5, and lose $1 to a 6. theoretically, given an infinite amount of rolls, i will go bust... but that is extremely unlikely, so you can argue that eventually everyone will go bust on any kind of gamble, but that's just not the reality. admitted, coin flipping is more volatile, but to say that eventually you will go bust on $1 flips with a $100+ bankroll is unrealistic, IMHO.aseem

100 isn't a ridiculously number in this example we're talking about. Better put it at 1,000 if you want to say you have a real chance of never busting. Another thing to remember in the commonly quoted 300BB estimate for bankroll. That assumes you don't skim off the top everytime you get to 350 back down to 300 (yeah I'm sure many of you know that, but does everybody who posts/reads here?). Losing 100 coin tosses in a row (or 50 in a row twice in relatively short sucession) is very improbable, but impossible? HAH. I wish it were impossible for me to lose 100 BB in a session too. Not that that's nearly as unlikely as losing 100 coin tosses, but still.Besides, you two are arguing about a silly thing. He's talking about infinity, you're mostly talking about having it really happen.

[jogsxyz 0]

2) If I bet 1 dollar on a coin flip repeatedly, it's a mathmatical certainty that I'll go broke eventualy unless I have an unlimited bankroll. In no limit, playing marginal edges can make a bad run of cards extremely expensive. On the other hand, the ability to make very large bets makes waiting for a hand with a large edge much more palatable. By shading my odds slightly downward, I can keep my variance under control.

Not true in a fair game. Fair meaning you win $1 when you guess right and lose $1 when you guess wrong. In a house game you will go broke. Meaning if it costs one penny to play this game, eventually you will go broke. Size of bankroll doesn't matter. The larger bankroll will only ensure that it will take longer to go broke.

[Alcatraz 0]

2) If I bet 1 dollar on a coin flip repeatedly, it's a mathmatical certainty that I'll go broke eventualy unless I have an unlimited bankroll.  In no limit, playing marginal edges can make a bad run of cards extremely expensive.  On the other hand, the ability to make very large bets makes waiting for a hand with a large edge much more palatable.  By shading my odds slightly downward, I can keep my variance under control.

Not true in a fair game. Fair meaning you win $1 when you guess right and lose $1 when you guess wrong. In a house game you will go broke. Meaning if it costs one penny to play this game, eventually you will go broke. Size of bankroll doesn't matter. The larger bankroll will only ensure that it will take longer to go broke.

Ok then... prove me wrong. Give me a mathmatical proof that it's a) impossible to get a string of heads (or tails) long enough to bust me in a fair game.

[akishore 0]

Besides, you two are arguing about a silly thing. He's talking about infinity, you're mostly talking about having it really happen.

we're both right.he's saying that given an infinite amount of hands, you will go bust. he applies this to poker by saying that it's much safer to wait for big edges in NL rather than risk too much with little edges, because variance will eat your bankroll that way. and he's perfectly right.i was just arguing on the side of realism given a finite amount of hands, and specifically a coin-flip scenario. theoretically, you can go bust with $1000 or even $100 on $1 flips, but realistically, it won't happen (of course it's possible, but going with the statistician's way of thinking... it won't realistically happen, because the chance of it happening is so very small).im in no way arguing his point on NL poker, he's right on. maximize little edges in limit poker (since you can't lose/win too much on a hand, and so not only is it safe to do this, you MUST do it to show any decent profit), but stay away from them in NL poker (variance will destroy you; one unlucky coin flip can cost you your stack). in limit, you can't afford to wait for the huge edges (since the amount you can win is very limited, the blinds will eat you up, etc. and you won't show a profit), while in NL, it's profitable and safer to wait for the big edges (since you can gain a significant amount on those edges).aseem

[gadjet 11]

2) If I bet 1 dollar on a coin flip repeatedly, it's a mathmatical certainty that I'll go broke eventualy unless I have an unlimited bankroll.  In no limit, playing marginal edges can make a bad run of cards extremely expensive.  On the other hand, the ability to make very large bets makes waiting for a hand with a large edge much more palatable.  By shading my odds slightly downward, I can keep my variance under control.

Not true in a fair game. Fair meaning you win $1 when you guess right and lose $1 when you guess wrong. In a house game you will go broke. Meaning if it costs one penny to play this game, eventually you will go broke. Size of bankroll doesn't matter. The larger bankroll will only ensure that it will take longer to go broke.

Ok then... prove me wrong. Give me a mathmatical proof that it's a) impossible to get a string of heads (or tails) long enough to bust me in a fair game.

This is my first post, Hi everyone…You say to prove you wrong… but you're the one who opened with the assertion that was way off base. The ball is actually in your court to prove your own statement correct. … Sort of like if I say: "There is an invisible elephant ghost floating right behind you right now."… prove me wrong… is it really in your cou… But I will address your statement anyways… you stated:" If I bet 1 dollar on a coin flip repeatedly, it's a mathmatical certainty that I'll go broke eventualy unless I have an unlimited bankroll." Mathematical certainty, eh?… ok, your bankroll is $1,000 dollars. (this is not unlimited… there is a clear limit). Now, you say it's a mathematical certainty that you WILL go broke? How so? Let's assume that somehow however unbelievable it might be, that I flip the coin 1,000 times and it lands tails, I'd lose the money… now if that was the streak from the opening toss what are the odds that the exact opposite would happen. … I'll give you a hint… it's the same odds. Any streak that you flip has exactly 50%/50% of happening the exact opposite way… there is no way that you can state that there is a mathematical certainty of going broke cause there is always a 50% chance the opposite could happen.Now what if the bankroll is $10 billion dollars… do you see now how the ball is in your court to mathematically prove your original statement?Nice to meet you all.

[akishore 0]

i'm not arguing with your actual point, but your argument is flawed in saying that any streak has a 50% chance of happening the opposite way.if a streak has a 1% chance of happening, the opposite streak has the same chance--1%. 98% of the time neither will happen.e.g. 10 coin flips, all 10 tails, probability is 1/2^10 = approx. 0.1%. the opposite streak, all 10 heads, is also 1/2^10 = approx. 0.1% again.two opposite streaks have the same chances of occuring in a binomial distribution or any distribution where the two opposing events have an equal probability of occuring.however, you are right in saying that it's not a certainty because to be a certainty you have to be 100%, and this is only 1% or less. i can see where he is coming from, because he is saying that given enough time, anything is possible, so it's possible/probable that you will go broke on any bankroll, but as for a certainty--i disagree as well.aseemp.s. welcome to the forum, nice to meet you too. :-)

[endo 0]

"If I bet 1 dollar on a coin flip repeatedly, it's a mathmatical certainty that I'll go broke eventualy unless I have an unlimited bankroll"This statement is just wrong. Period.And in a tournament, you're right: small edges aren't very desirable. But in a live cash game, if you're cringing at exploiting small edges, then your bankroll isn't big enough for the stakes you're playing, and you should find a smaller game. If I could see my opponent's cards, the pot's laying me even money, and I have 51% edge, I should take the bet. I have the best of it.And you can't beat roulette because it's not a coin flip. There are green spots.

[gadjet 11]

two opposite streaks have the same chances of occuring in a binomial distribution or any distribution where the two opposing events have an equal probability of occuring.aseemp.s. welcome to the forum, nice to meet you too. :-)

Isn't this my pooint though? what I'm saying is that if hypothetically you knew a streak of ten were going to happen... it is exactly 50% heads or tails which way the streak were to actually fall... I realize that it is not 50% for a streak to happen...Aseem, I have enjoyed many of your posts, I think you make great points and look forward to discussions.

[akishore 0]

And in a tournament, you're right: small edges aren't very desirable. But in a live cash game, if you're cringing at exploiting small edges, then your bankroll isn't big enough for the stakes you're playing, and you should find a smaller game. If I could see my opponent's cards, the pot's laying me even money, and I have 51% edge, I should take the bet. I have the best of it.

I agree with you here, theoretically. To be fair, however, I think the poster who made that comment had a point as well:NL poker is exploiting big edges. you don't need to exploit little edges for too many chips because that greatly increases your variance. as an extreme example, it's the difference between going all in every time you're a 4:1 favorite vs. all-in every time you're an 11:10 favorite.in the former case, your bankroll will build much more steadily but in the latter state luck can cause a big losing streak--even though in the long run, it's a good bet and you have the best of it, there's no need for it because it'll just cause big fluctuations, and you can afford to wait for a big edge and then make the best of it.in a limit game, however, i'm 100% with you. it's the little edges that matter, and the poster realizes that, too. he's discussing NL though. so i'm saying that both of you are correct, and both of you have valid points: if you want to exploit those little edges, have a much bigger bankroll that can sustain the swings, otherwise wait for the big edges.

Isn't this my pooint though? what I'm saying is that if hypothetically you knew a streak of ten were going to happen... it is exactly 50% heads or tails which way the streak were to actually fall... I realize that it is not 50% for a streak to happen...

you're right, if a streak were to definitely happen, it can only be one of two. however, you never know if a streak will happen, so i don't think it's correct to think of it in that context. in any given number of 10-flip trials, a streak of 10 will happen 2 in 1024 times, about. one of those streaks will be all heads, the other will be all tails.so based on that, i think what you're saying is, "if it's almost a certainty that you'll suffer 1000 tails in a row and go bust, it's the same certainty that you'll suffer 1000 heads in a row and double up", which is a valid point based on an invalid hypothesis that it's almost a certainty of busting. in reality, both are almost complete impossibilities (note the word almost before any of you flame me :-) ).so you're right that one streak is just as likely to happen as the other, but both streaks are very unlikely to happen altogether. make sense? but i see what you're saying. i think you understand too, i'm just clarifying.

Aseem, I have enjoyed many of your posts, I think you make great points and look forward to discussions.

thanks, i appreciate the comment! i've come to find this forum to be a real home for my passion of poker. i used to do some rgp'ing but it was just stupid... too many flames, too much spam, too many know-it-alls, and no sense of community whatsoever, save a few names like GambleAB and CincinattiKid, etc.none of this discussion would be possible without any of the other great posters on this forum either. it's really cool to be able to post a hand and get a big thread about whether check-raising or three-betting was correct, or to post an idea like this and get so many responses, etc.so thanks again, and i look forward to discussing poker with you, too.aseem

[PA32R 0]

one thing i wanted to point out about the coin flip argument:although your application makes sense that calling is not necessarily the right play if the pot is offering you exactly the same odds as your chance of having the best hand, your argument that we will all eventually go broke flipping coins at some time or another doesn't hold water.a coin flip is a binomial distribution--it can only be heads or tails, one of two outcomes. if you have a sufficient bankroll, say $100, and you're betting $1 per flip, the chance of you busting is the chance of the coin landing the same way 100 times in a row. the probability of that is (1/2)^100, which is in the order of 1/(10^30)--infinitesmally small. that doesn't just mean that if you flip a coin 10^30 times you might bust once, it means if you conduct a 100-flip trial 10^30 times, so in actuality the probability of busting is even smaller, 1/(10^32).in this way, if the pot is offering you any amount better than the odds of your draw, it is always a correct play to call. you should never undercount your outs or fold if the pot isn't offering you enough, because poker is all about maximizing little edges (limit poker at least), and this is one example of how a little edge should be exploited to the fullest. otherwise, you could argue that you shouldn't call a 4:1 draw when the pot is offering you 9:1, because it's possible in the long run to lose money on those draws (which it is, but it's highly improbable). poker is a gambling game, so you must forget the luck factor and play long-term results.i'm not arguing with your end statement of calling/folding if the pot is offering you no profit, i'm just pointing out a seperate application of what i believe to be a flaw in your argument.as a side note, this same logic can be applied to get the number 300 for the number of big bets your bankroll should have in order to sustain itself. if you win 1 big bet per hour with a standard deviation X, you can derive the number 300 if you ask, "how big should my bankroll be so that the chance of me busting over any given time period will be less than 5%? less than 1%? less than 0.1%? etc." of course, the tighter you are, the smaller your standard deviation will be, so the less this number will be, but the number 300 seems to suit most people well.hope this helps,aseem

argh... no, the probability of bankruptcy is NOT (1/2)^100. one hundred bad calls in a row needn't happen, only 100 more bad calls than good calls. Still somewhat unlikely but by no means infitesimal. let's say you flipped the coin 2000 times. If you pick heads each time and tails comes up 1050 times, you're busted. If you flip it 100,000 times and it comes up tails 50,050 times, you're busted. The chances of this are hardly infinitesimal. In fact, if you toss long enough, it's nearly a certainty that at some point there will be 100 more tails than heads and you're out.

[akishore 0]

argh... no, the probability of bankruptcy is NOT (1/2)^100. one hundred bad calls in a row needn't happen, only 100 more bad calls than good calls. Still somewhat unlikely but by no means infitesimal. let's say you flipped the coin 2000 times. If you pick heads each time and tails comes up 1050 times, you're busted. If you flip it 100,000 times and it comes up tails 50,050 times, you're busted. The chances of this are hardly infinitesimal. In fact, if you toss long enough, it's nearly a certainty that at some point there will be 100 more tails than heads and you're out.

that's a very good point that i never realized. i've been thinking about this for a while now and have no response yet. thanks for pointing it out, though, it brings up a really interesting idea. my wonder is:we all know that in the short run, the number of tails and heads will not be exactly half and half... e.g. in 100 flips, it could be 59 heads and 41 tails. in 1000 flips, it could be 590 heads and 410 tails. in 10000 flips, it could be 5900 heads and 4100 tails.however, i know without a doubt that standard deviation is independent of the number of trials, so the probability of having a difference of 100 between heads and tails is the same given 500 flips as it is given 500000 flips (believe it or not, i am sure about this). so what i'm trying to do is use the binomial distribution to see what sufficient bankroll you need to say "the chance of there existing a 1000 flip difference between heads and tails is less than 0.1%, so a bankroll of $1000 is enough", etc.i am looking into this, please be patient... i'll reply soon. :-) aseemp.s. i understand what you're saying, but i am still disagreeing--it is not a near certainty that you will go broke. just wait, i will come up with a valid argument.

[akishore 0]

ok, i've got it. here's my argument, i leave nothing out. note that we were both right in a way. you are right that if you flip till infinity, you will probably bust, but at the same time, you can have an ample enough bankroll where, if you can guesstimate how many flips you will do, you will almost never go bust.flipping coins is a binomial distribution; there are only two outcomes, one success (we'll say heads is success), and the other failure (tails), and at every trial, the probability of each remains the same. let n be the number of trials and p be the probability of failure. for a coin flip, p = 0.5.if we do n trials, we expect that, ideally, np is the number of successes, so in the case of coin flips, n/2 is the number of heads we would expect. however, this is rarely the case, as the number of heads and tails are rarely exactly half and half.we measure this spread through standard deviation. standard deviation is simply that: a measure of the variance in the mean of a number of trials. for a binomial distribution, s is the standard distribution, and s=sqrt(np(1-p)), so as you can see, i was wrong--the spread does increase as the number of trials increases. for a coin flip, s=sqrt(n)/2 if you simplify.how does standard deviation apply? well, it applies through a bell curve, or normal curve. basically, this is a graph where the area underneath the curve adds up to 1, but the graph runs on forever. it symbolizes the idea that anything is possible, just that some things are much more likely than others. here's an image:so the events in the middle, towards the mean, are very likely to occur, while the events further out towards the tails are less likely to occur. this graph is color-coded, and it shows this:68% of the time, the number of successes will fall within one standard deviation of the expected result (np or n/2). color code: red.95% of the time, the number of successes will fall within two standard deviations of the expected results. color code: red and green.99.98% of the time, the number of successes will fall within three standard deviations of the expected results. color code: red, green, and blue.so what this means is, no matter how many trials we have, the VAST VAST majority of times--99.98%--it doesn't get any more certain than that--the number of heads/tails will be within some reasonable range of our expected result.so let's apply this.if we flip the coin n times, we expect it to fall heads n/2 times. now three standard deviations is (3/2)*sqrt(n). so can safely say that it will almost always (remember, 99.98% of the time) fall n/2 +/- (plus or minus) (3/2)*sqrt(n).if n=10 flips:n/2 +/- (3/2)*sqrt(n) = 5 +/- 4.7.so 10 flips is very volatile, we might get only 1 head, so we could lose $9 in just 10 flips.if n=100 flips:n/2 +/- (3/2)*sqrt(n) = 50 +/- 15.this is a little safer in comparison to the number of trials. it will fall heads somewhere in between 35 and 65 times.if n=1000 flips:n/2 +/- (3/2)*sqrt(n) = 500 +/- 47.so it will fall heads somewhere between 453 and 547 almost always.if n=10000 flips:n/2 +/- (3/2)*sqrt(n) = 5000 +/- 150.it will fall heads somewhere between 4850 and 5150.if n=1000000 (one million) flips:n/2 +/- (3/2)*sqrt(n) = 500000 +/- 1500.it will fall heads somewhere between 498500 and 501500 times.so how do we apply this to bankroll? since a difference of 10 between heads and tails means a loss of $10 for us, we need to have enough to safely withstand 6 standard deviations.so if we plan to flip coins for 100 flips, 6 standard deviations is 30, so a bankroll of $30 should be very safe (only 0.02% of the time will it not be safe). if you want to be ultra-safe, just ronud up to $40 or $50.if you plan to flip for 1000 flips, 6 standard deviations is a about 94, so a bankroll of $100 should be more than ample. finally, if you round high to a million flips, you need a bankroll of $3000+. with this bankroll, you will almost never go bust, even with 1000000 flips.i hope this makes sense to everyone. the poster was right in saying that at a limited bankroll (e.g. $100), after a certain number of flips, you are more and more likely to bust. however, others were also right in saying that you can have an ample enough bankroll to almost shut off that possibility, as long as you have an idea of how many flips you'll be doing.how this applies to poker is for another thread, i'm too tired. :-) hope this helps,aseem

[DeNuts1 0]

There is an artile at cardplayer.com about all of this simplified odds stuff. I don't remember if it is the newest issue or not but it is Volume 17, No. 7. The article is by Lou Krieger and name "Pot Odds Made Easy". Check it out. Has charts and all that junk.

[PA32R 0]

i agree with aseem's analysis. i hope people don't take away the notion that, if $3,000.00 is enough for flipping coins (an even money proposition) that much less is needed for poker, where they ostensibly have an advantage. the standard deviation in poker is MUCH higher and really can only be assessed by looking at your records. mine would be different than aseem's, for example. required bankroll is based on three factors: expectation; variance (the square of standard deviation) and your personal risk tolerance. aseem worked numbers for a .02% chance of busting out. if you are comfortable with a 5% chance, your bankroll can be much smaller.

[Alcatraz 0]

Fun thread :)For those that say my assertion is ridiculous or flat out wrong, I'm still waiting for proof of such.With a sufficient bankroll and a finite number of coin flips the chances of busting out are so small as to be virtualy non existent. If we're discussing practical application, then you are correct to say you'll never bust. However, I was discussing what would happen given an infinite number of coin flips, which immediately takes us out of the realm of practicality. In fact we're dangerously close to putting a cat in a lead shielded box and bombarding it with radioactive particles at this point.Yes, I'm aware that roulette is not a fair game the way a coin flip is. My reference to roulette involves one of many flawed systems that charlatans try to sell as a surefire money making scheme. Here's how it works:Bet 1 unit on red (or black, or even, or odd). If you lose, double the size of your bet and roll again. Continue doubling your bet until you win, at which point you will have a profit of 1 unit. This is a foolproof scheme, in theory. After all, sooner or later you will win, returning all the money you've lost plus 1 unit profit. The problem with this scheme is that you must double the size of the bet each time you lose. Eventualy you will hit a losing streak long enough to make it impossible for you to double the previous bet - either because you don't have the bankroll for it, or because the casino max bet will prevent it.With the coin flip example, you don't have to worry about the casino max bet problem. However the laws of probability still tell us that over an infinite number of tosses, at some point you will hit a losing streak long enough to bust you.Now... what practical purpose does this have for poker?Ok, let's say you buy in to a NL game for 100 units. After an hour or two you've built it up a bit when the following hand comes up: Holding K 9:heart: on the button, the board is A:heart: 10 3:heart: 7:club: And there's 900 units in the pot with 3 people still in. Player 1 bets out 200, and player 2 calls. You're certain that player 1 has a pair of aces at minimum. You have 200 units in front of you. Should you call the bet? With 1300 units in the pot, the pot is laying you 6.5 to 1 odds and you're drawing to the stone cold nuts.In a limit game calling with a flush draw to the nuts when the pot is giving you 6.5 to 1 is a no brainer. (you're 4.1 to 1 against making the flush) In a limit game you can miss the draw and go on, confident that by making this call every time we'll have a winning session.However in a no limit game making this call will cost you your entire stack if you miss the draw and you're buying back in. It doesn't take an improbably bad streak of hands like this to eat your entire bankroll.The point is that it's irrelevant what the long term expectation is for the call if your bankroll can't withstand the short term variability. The closer the pot odds are to the odds of winning, the more variance you can expect and the larger bankroll you need to make the call.Another way of saying it is that pot odds only matter in the long run but are totaly irrelevant if your bankroll can't handle short term fluctuations, and in no limit playing small edges will break even healthy bankrolls if you're pushing small edges.

[SumZero 0]

Hi all, first time poster, and this seemed like a fun thread to get involved in. The fact is that on a 50/50 fair bet on a coin flip with a finite bankroll for you and an infinite bankroll (for the person you are betting against) and infinite number of flips you will go bust. It is a certainty. Alcatraz wanted a proof that he was wrong, but since he was right, I'll offer the proof below:Here is a link that may be of interest to people http://mathworld.wolfram.com/GamblersRuin.html.Basically, this is a random walk problem (follow the links there for more info) and it simplifies that if you are p1 and have a bankroll of n1 and the person you are playing against is p2 and has a bankroll of n2 then the odds of you going bust is first if you play until one of you is broke is:n2/(n1+n2). In our case as n2 approaches infinity and n1 is finite, the limit is 1. Similarly the player we are up against's probability is n1/(n1+n2) and as n1 is finite and n2 approaches infinity the probability here has a limit value of 0.Now even more interestingly, iirc, the other inifinite game described here that isn't a fair game but is instead stacked in your favor (5/6 you win $1, 1/6 you lose $1) has a non-0, non-1 probability that you go broke that is dependent on your bankroll. In an infinite random walk that is biased in one direction (rather than being fair and neutral) you are not guarenteed to go broke!The other really cool coin game that is analyzed at mathworld is one that was proposed by Bernoulli originally. The idea is the player is paid some fixed finite amout to play the game. The player wins this amount. The player then flips a coin until they get a head. If it took them 1 flip they owe $1. If it took them 2 flips they owe $2. If it took them 3 flips they owe $4. ... If it took them n flips they owe $2^(n-1). The surprising thing is that the expected value of the flipping part is infinity so if this was a repeated process there is no fixed finite value that makes this a fair or profitable game!

[MrConceit 0]

Very good posts from all, I didn't read for like 24-48 hours and you guys went crazy. All fun/informative stuff to read.Aseem even went and put in a std dev graph, talk about effort! :wink:

[akishore 0]

to alcatraz and sumzero: great posts. both are totally correct, and i just want to dwell a little more on the applications of this to poker, specifically to flush draws as in alcatraz's example.he's right. in NL poker, NEVER stack off on a flush draw if your remaining stack is anything significant, say half a buy-in or a full buy-in (if it's more, don't even consider it), and even then, only if the pot is offering you much better odds than 4:1. here's why:we can model a flush draw as a binomial distribution. after the flop, you have a 1/3 chance to hit, so p=0.33. this means that you'll either hit success, or you'll hit failure. we said standard deviation was s=sqrt(np(1-p)). so here, this simplifies to about s=0.47sqrt(n).if you model a flush draw over 100 hands, s=4.7. this means that 95% of the time, you will hit somewhere in the range of np +/- 0.47sqrt(n), or 33 +/- 4.7. thus, you can hit anywhere from 28 times to 38 times. that's a difference of 10 times where you might miss.so if each time you stack off it costs you 50 units, luck over 100 hands can leave you down 500 units.now let's introduce pot odds. let's say each time you do this on the flop, you do it when the pot offers you 2.5:1, a little better than your 2:1 odds of making it. in limit hold'em, it makes sense to call a bet of 50 units every time because twice you'll make it, earning you 125, and once you'll miss it, losing you 50, for a 75 profit.so, theoretically, in NL, over 100 such flush draws, if you make the expected 33 flushes, you'll earn 125*33=4125 and you'll lose 50*66=3300, for a good profit of 825. now if we combine this with the worst case scenario which is that you'll hit only 28 times, we see that you earn 125*28=3500, and lose 50*72=3600, so you're stuck 100 units!if 100 units is significant to your bankroll, you shouldn't call off 50 units with that draw because the pot isn't offering you enough to offset the luck factor.so if we take a safe approach, and want to figure out how big the pot should be to ever call off your stack in NL poker, and we have a 5% risk tolerance (so we'll use two standard deviations--if you want a 0.02% risk tolerance, you can use three standard deviations, but that's being far too cautious IMHO), here are my figures:let x be the amount in the pot compared to what we're calling with, i.e. the pot odds. if the pot is 500 and we're calling 100, x=5.let c be the amount you're calling.over n such scenarios, we'll earn cx * (0.33n +/- 0.47sqrt(n)) - c * (0.67n +/- 0.47sqrt(n)). since we're interested in having a risk for the worst scenarios given our risk tolerance, we'll change both +/-'s to - and + respectively.so, profit = cx * (0.33n - 0.47sqrt(n)) - c * (0.67n + 0.47sqrt(n)).over 100 scenarios, profit = 28cx - 72c.over 1000 scenarios, profit = 318cx - 682c.over 10000 scenarios, profit = 3286cx - 6714c.over 1000000 scenarios, profit = 286333cx - 713667c.if we plug in an arbitrary number like c=1, you'll earn 28x-72 over 100 scenarios in the worst case, so we see that x should be 3 or better. over 1000 scenarios, x should be 2.5 or better. same with 10000 scenarios. over 1000000 scenarios, the number rises up to 3 again (2.5 will break us even).this curve shows us that the in the short run, you can lose money by calling off big numbers of chips when the pot is doing just a little better than breaking you even, but in the long run, bad luck can cost you a lot. so in general, on the flop, call big parts of your stack only if the pot is offering you significantly better than your odds, say 3:1 or 4:1 if your draw is offering you 2:1.you can change the numbers for different draws or for turn draws (as opposed to flop draws), but you'll still see that the pot should be offering you tremendously more the worse your draw is.to reiterate, let me make a few points clear:- it doesn't matter if the pot is offering you a lot, never call off a half buy-in or more, because a run of 5 misses can cost you a big part of your bankroll.- these calculations only apply if you've already invested a lot in the hand, so your stack is much smaller--say a quarter of the buy-in--and even then, the pot is offering you a lot.- implied odds don't matter in these situations since you're calling off your stack and have lost all ability to earn more if you hit.- this doesn't apply to limit poker except maybe to just point out that in even-money situations it's better to save your money (unless you're a gambler and hope that luck works in your favor rather than against).tell me any corrections or anything.hope this helps,aseem

[gadjet 11]

The funny thing about this thread to me is that I think we all agree on the actual poker issue, but as analyticals we all love a good argument so we're discussing one side point…

Alcatraz wanted a proof that he was wrong, but since he was right, I'll offer the proof below:Here is a link that may be of interest to people http://mathworld.wolfram.com/GamblersRuin.html.Basically, this is a random walk problem (follow the links there for more info) and it simplifies that if you are p1 and have a bankroll of n1 and the person you are playing against is p2 and has a bankroll of n2 then the odds of you going bust is first if you play until one of you is broke is:n2/(n1+n2). In our case as n2 approaches infinity and n1 is finite, the limit is 1. Similarly the player we are up against's probability is n1/(n1+n2) and as n1 is finite and n2 approaches infinity the probability here has a limit value of 0.

Okay using this formula… as n2 approaches infinity so might n1 while still remaining a finite number. For mathematics sake let's say infinity minus 1… which leaves us with:n2/((n2-1)+n2) which leaves us with a probability that approaches 50/50. I realize that a bankroll of n2 minus 1 is in all practical terms an infinite bankroll, mathematically speaking it is not. Great posts all around, I think I've found a poker forum home… and as I said before I feel we are all on the same page on the main issue, just a little side debate going, that's all.

[akishore 0]

The funny thing about this thread to me is that I think we all agree on the actual poker issue, but as analyticals we all love a good argument so we're discussing one side point…

haha damn! you figured us out. :-) aseem

[Fooney 0]

In all the talk about math and probability you guys missed the simple, yet correct answer.In a $1 to $1 coin flip you will lose eventually regardless of the size of your bankroll because.......of the rake. with the house taking about 5% of every win you will eventually be bankrupt. I realize this doesn't hold true in a home game but in that case you will eventually go bankrupt buying beer and munchies while playing for an infinite amount of time.

[akishore 0]

yes, rake will screw you. so in poker, don't call even money situations with even-money pot odds because of the rake.but for the example of the coin flip, the reason there was no rake was to specifically show a situation where no result is a favorite, it is an exactly 50-50 gamble. if there was a rake, it would be like any other casino game, where it's no longer a 50-50 gamble, and that's not what we were interested in.aseem

[Fooney 0]

I understand. I just thought it was funny that there was this long string of discussion regarding 50-50 calls without a mention of rake.I'll be quiet and crawl back in my hole and let you discuss your coin flips.Damn, thought I was on a poker site.

[akishore 0]

I understand. I just thought it was funny that there was this long string of discussion regarding 50-50 calls without a mention of rake.I'll be quiet and crawl back in my hole and let you discuss your coin flips.Damn, thought I was on a poker site.

hahahaha, nice post. :-) i really have no idea how a pot odds question evolved into a discussion of standard deviation of coin flips. i should ask the forum mod to make a new forum, "Coin Flip Forum (Off-Topic)".aseem