Poker bankroll management, building and strategy – online poker guide

Poker theorems; David Sklansky, Baluga Whale, Andy Morton, Zeebo and Clarkmeister

Sunday, May 22nd, 2011 | Poker Articles, Poker Mathematics, Poker Strategy | 3 Comments

The following massive piece of work on poker theorems was submitted by Conrad. Poker theorems are pieces of fundamental poker strategy and advice, usually expressed in poker literature and forums. An ‘objective’ poker strategy is hard come to come by – the generation of hyper aggressive internet hotshots have caused us to revamp our ideas as to what constitutes an ‘optimum’ strategy. Internet star Dusty ‘Leatherass’ Schmidt, who posted the world’s highest win rate for $5/$10 NL in 2007 and 2008, even released a book entitled ‘Don’t listen to Phil Hellmuth: correcting the 50 worst pieces of poker advice you’ve ever heard’. Due to the evolution of the game, advice from the ‘old guard’ of is often considered dated, and players such as Hellmuth have been heavily scrutinised for their cash game performances. That said, books such as Doyle Brunson’s Super System and the Harrington on Hold’em Series are still well respected. Although their doctrines are contested, poker theorems are good as general rules of thumb. They may not be a substitute for things like poker training, but are useful nonetheless. They are not concepts that a player should stick to religiously, but ideas that a player should
always have in mind.

The fundamental theory of poker by David Sklansky

The Fundamental Theorem of Poker is described by esteemed poker player, theorist and author, David Sklansky. Sklansky is considered to be a leading voice on gambling and poker theory in general. The theorem states:

‘Every time you play a hand differently from the way you would have played it if you
could see all your opponents’ cards, they gain; and every time you play your hand
the same way you would have played it if you could see all their cards, they lose.
Conversely, every time opponents play their hands differently from the way they
would have if they could see all your cards, you gain; and every time they play their
hands the same way they would have played if they could see all your cards, you
lose.’

This is a very basic theorem, stating that every decision we make should be in accordance with maximizing EV (expected value). In the long term, this is what counts. So even though chasing a flush on the river may be tempting, we should only call if our opponent is giving us the correct pot odds.

Morton’s addition to Sklansky’s theorem

Sklansky’s theorem is only applicable in heads up situations. Morton’s theorem, articulated in a poker newsgroup by Andy Morton, explains why Sklansky’s theorem is not applicable in a multi-way pot. It often occurs when one player has the best hand, and two players are on draws. The player with the best hand might make more money in the long run when an opponent folds to a bet, even if that opponent is making a correct fold and would be making a personal mistake to call the bet. For instance, Player A holds Ac-Qc, player B Ah-9h, and player C Js 3s on a Ad-Jh- 4h board. Player A has a made hand – top pair, and when he bets the pot Player B with the flush draw is going to call. In the long run, Player A would make profit in a heads up situation with Player B. His odds are dashed and Player B’s enhanced, however, if player C, with his mid pair, makes the call. This is because he has 6 outs to improve his hand. This concept is sometimes referred to as implicit collusion.

The Beluga Whale Theorem

Other popular theorems are documented in community site twoplustwo. The Beluga Whale Theorem states that when you are a pre-flop raiser, and your top-pair hand is raised/check-raised on the turn, it is time to re-evaluate your hand. This is because your opponent is often trying to build a pot to get paid off with his monster. If you have AK on a K-10-5-9 board, and you face a raise on the turn, it is quite conceivable your opponent has two pair or better. This theorem is reliable against weaker opposition, however shrewder players can exploit this by floating.

Zeebo’s poker theorem

Zeebo’s Poker Theorem states that nobody ever folds a full house. So, if you have any inclination that your opponent has a weaker full house, bet out. People tend to overestimate boats because in a large number of situations they tend to be good. If you have KK on a board which includes AAA, bet out even if you put your opponent on something as low as 22.

Clarkmeister’s Theorem

Clarkmeister’s Theorem argues that when you are out of position heads-up on the river, and a 4 to a flush card comes, always bet (unless you have something with realistic showdown value). This is a perfect bluff spot, and an opponent will fold something like a weak/middle flush a large percentage of the time.

To find out about more obscure poker theorems, or the mathematical explanation behind some of the ones stated in this article, be sure to browse twoplustwo along with other poker forums.

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Calculating poker odds – Don’t misuse the rule of 2 and 4

Friday, June 4th, 2010 | Poker Articles, Poker Mathematics | 1 Comment

Submitted by Cory, this article belongs to the Poker Mathematics series.

One of the first things we learn about the mathematics involved in poker is figuring out our chances of winning by using the rule of four and two.  For those of you who may not be sure exactly how that works, if you have an open ended straight draw for example, you count your outs, the unseen cards that make your straight, which is 8 and multiply your number of outs by 4 on the flop and 2 after the turn.  So on the flop you’re about 32% to make your straight with the turn and the river yet to come and on the turn you’re about 16% to make your straight with just the river to come.

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Basically is what this means, is the odds of you completing your straight are just over 2:1 against it with 2 cards and just under 5:1 against it with 1 card.  This information is used to measure the likelihood of you hitting your draw against the price you’re getting on a call compared to how much is already in the pot.

If I’m playing a pot with $80 and my opponent bets $20, he creates a pot of $100 and it costs me $20 to potentially win that $100, or I’m getting 5:1 on the call.  We’re all good so far right?

The problem is, I’ve recently run in to a lot of players who are completely misusing the rule of four and two to try and justify some pretty bad calls.  You really need to be thinking about what your chances are of hitting your draw on the next card because very often, you’re going to be faced with another bet.  The rule of four and two was designed for when you’re facing an all in bet on the flop or turn and you have a draw.  If you go all in, you’re guaranteed to see all five board cards, but if you and your opponent still have chips behind, it changes the math drastically.

Let’s say you’re in a $1-$2 game.  You and your opponent each have $200 stacks to start the hand.  Your opponent opens from middle position to $8 and you call with 8C7C from the button and take a flop heads up.  The blinds have probably been taken out for rake, so we have a pot of probably $15.  The flop comes 9H, 6S, AD.  You have no clubs, but you have an open ended straight draw.  Your opponent bets $10 creating a pot of $25 and you’re getting 2.5:1.  Using your rule of four and two the odds are 2:1 against hitting your hand so this is a profitable call… if you can guarantee that you’ll see both remaining cards.

You can’t forget about how much you’re probably going to have to call on the turn.  If you call this flop bet there will be a pot of $35, so if you miss the turn, you may have to call a bet somewhere between $20 and $30 to see the river card which you counted on with your rule of four.  So really, your flop decision has to be whether or not you’ll be willing to call somewhere between $30 and $40 to see both the turn and river.

This is where implied odds come in to play.  Since you and your opponent are both playing $200 deep, it makes sense to call his bet on the flop and call a reasonable bet if you miss the turn.  Let’s say you call this $10 on the flop creating a pot of $35 and your opponent bets $20 on the turn and you missed.  Now you’re getting $55:$20, or 2.75:1 pot odds, which is not a good price, but if you call, the pot will be $75 and your opponent will have $162 to pay you off with if you hit the river.

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If your opponent should decide to bet the river when you hit, it’s going to be somewhere around $50 in to a pot of $75, in which case you’re going to be able to stack him enough of the time to make the call on the turn where you were getting slightly the worst of it profitable in the end.  Big bet poker games are more about implied odds than pot odds.  Limit games focus more on pot odds since implied odds in a limit game are reduced to maybe one or two extra bets.

Never forget when deciding whether or not to call on the flop to factor in how much that call is really going to cost you.  Always be aware of the effective stack size in the hand so you can calculate your implied odds.  Don’t make calls where you’re taking the worst of it if either you or your opponent has a shorter stack.  Stop using the rule of four and two to justify flop calls when you can’t call a bet on the turn and get that river card that you promised yourself on the flop.

Tags: Poker Articles, Poker Mathematics

The poker mathematics

Wednesday, April 28th, 2010 | Poker Articles, Poker Mathematics | No Comments

Submitted by Steve, this article belongs to the Poker Mathematics series.

In this article, Steve gives a general introduction to the poker mathematics.

Poker is a game of fine psychology-based decision, betting and bluffing. There’s no math involved in it whatsoever, is it? Wrong! While in high-level poker the above named elements do indeed substitute the poker mathematical aspect to a certain degree, every poker player knows that there is quite of bit of math involved in the game indeed. Even players who never stop to think about the mathematical odds involved use them all the time, when making decisions which ‘feel right’ to them. “The poker mathematics” may sound a little daunting, but once one gains a grasp on what these odds are about it becomes pretty simple.

The bottom line about the poker mathematics aspect of the game is the comparison between the pot odds and the hand equity which gives the player a mathematically correct way to proceed. Basically, what one does is that he/she calculates his/her pot odds and then compares it to his/her odds for making a potentially winning hand. The first stage of the move is to calculate your pot odds.

Suppose you’re playing for a $40 pot and the player in front of you makes a $10 bet. It now costs you $10 to tag along for the ride, and to possibly take down a $50 pot. It basically costs you $10 to gain a shot at a $50 pot, which means your pot odds are 50-10 = 5-1. In order to find out your pot odds, always divide the current size of the pot (make sure you take all bets and raises occurring in front of you into account) with the amount you’re required to call to stay in contention. The bigger the pot odds, the better off you are, because  the bigger these odds are, the weaker the hand you can afford to make the call on.

According to Sklansky, the pot odds can be exaggerated in such a way that it becomes a viable choice for players to make the call on just about any two cards. Obviously, this can actually be proven mathematically too.

The next step of your odds escapade is to find out your hand equity. Suppose you’re looking to make a flush, a hand which you’re pretty certain would win the pot for you. In order to find out your hand equity (or the odds against your draw filling up), you need to take your number of outs into account. I’ll give you a simple example to make things easier to understand: you have a 4-card flush on the flop. You have the above situation on your hand, with the 5-1 pot odds. You know that a flush would win you the pot, what do you do? First, you calculate your number of outs: there are 52 cards in the deck, of which 3 are on the board face-up and 2 are in your possession. That leaves 52-5=47 cards. There are 13 cards of the same suit as your 4-card flush, 4 of which are already revealed. That leaves you with 9 outs. Of the 47 remaining cards, 9 will help you hit your flush and 47-9=38 will not. That means the odds against your flush filling up are 38-9. That leaves you with 4.22-1 odds (remember that these odds are against your flush), which are smaller than the 5-1 pot odds, which means it is mathematically correct for you to make the call here.

When counting your outs, make sure you take ALL your outs into account (you may have cards which could hit you for a straight or maybe even a set in the above situation), and to disregard all your anti-outs (cards which would hit your opponent for a better hand than yours).

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EV in poker

Thursday, March 18th, 2010 | Poker Articles, Poker Mathematics | 2 Comments

This article belongs to the Poker Mathematics series.

Calculating EV in poker

Before going into depths with the concept of EV in poker, some math is needed. In mathematics, the EV or expected value of a variable is defined as the probability weighted sum of the possible values for the variable in question. This may sound complicated, but it really isn’t that bad. Let’s look at the simple example of rolling a dice. In this case the possible values and probabilities are as follows:

• 1 will come up in 1 out of 6 rolls on average yielding a probability of 1/6 = 16,7%
• 2; 16,7%
• 3; 16,7%
• 4; 16,7%
• 5; 16,7%
• 6; 16,7%

So the EV for rolling a dice is: (1 x 0,167) + (2 x 0,167) + (3 x 0,167) and so on. In total this adds up to 3,5.

When it comes to EV in poker, there are basically two situations to consider, namely EV when you call a bet and EV when you make a bet.

In both cases, I will define EV in poker as the amount you will be payed back on average on a 1$ bet (including your original 1$ bet):

• EV < 1 – If you always play hands with an EV of less than 1 you will lose on average
• EV = 1 – If you always play hands with an EV equal to 1 you will break even on average
• EV > 1 – If you always play hands with an EV of more than 1 you will win on average

So ALWAYS aim to play hands with an EV larger than 1!!!

EV IN POKER WHEN CALLING A BET

It is pretty straight forward to calculate your EV when calling a bet. It’s simply the product of your probability (P) of winning the hand and the decimal pot odds (DPO) offered:

• EV = P*DPO

For a review of the difference between decimal odds (European style odds) and fractional odds (American style odds) check out my article: poker pot odds; all you need to know.

For an easy rule of thumb to calculate poker probabilities check out my article: poker probabilities; all you need to know

At this point I think an example is called for (warning, there is some math content ahead):

Let’s say you are playing an online tournament and the pot on the turn is 3000. Your only remaining opponent bets 500, what is your EV for a call? With your opponent’s bet of 500 the pot is now 3500 and you have to call 500 to stay in the pot.

In order to calculate your decimal pot odds, simply add the amount you have to call (in this case 500) to the pot (3500+500) and divide by the amount you have to call (4000/500 = 8). So your decimal pot odds is 8 in this case.

Let’s also assume that you have a total of 12 outs to win the hand on the river. Using the easy rule of thumb to calculate poker probabilities, this means that your probability of winning the hand come showdown is (2*12+1)% = 25% or alternatively 0,25.

For simplicity we are assuming that the river is checked by both you and your opponent so no extra chips enter the pot.

Now all that’s left to calculate your EV in the situation above is to multiply the probability with the pot odds, yielding an EV of 8*0,25 = 2.

This means that for every chip you call in the situation above, you will on average get 2 chips back. Or as in the case above with your 500 chip call, you can expect to get 1000 chips back, meaning you win 500 chips in total.

This result fits very well with the mathematical definition I gave of the expected value in the beginning of this article. Say you make the 500 chip call 100 times. In 75% of the cases you will lose your 500 chips, corresponding to 37500 chips. In 25% of the cases you will win 3500 chips, corresponding to 87500 chips. This means you will win on average 50000 chips in total over 100 plays, or 500 chips per play which fits perfectly with our previous result.

EV IN POKER WHEN MAKING A BET

EV in poker when making a bet is all about pot and pot odds control. You want to be choosing your bet sizes so your opponent’s EV for calling is less than 1.

Say you have a high pair and the flop comes with 2 hearts. The pot has 2000 chips in it and you’re up against 1 opponent as the first to act. In order to make sure your opponent’s expected value for calling is less than 1, you need to pay attention to the size of the bet you make. If your opponent is on a flush draw he has 9 outs to make his flush on the turn providing you do not have any hearts in your hand. 9 outs correspond to a probability of roughly 20%. What size bet should you make?

In order to answer this we need to revisit the formula for calculating EV in poker:

• EV = P*DPO

An EV equal to 1 means that you will break even on average. Inserting this into the formula above we get the following relationships between the probability P and the decimal pot odds DPO:

• DPO = 1/P
• P = 1/DPO

So for a probability of 20% for your opponent to hit his hand on the turn, you need to be offering him pot odds 5 (1/0,2) for him to break even on average and less than pot odds 5 to ensure he has an EV less than 1.

If you bet half the 2000 chip pot in the example above your opponent will be getting pot odds 4 to call, corresponding to an EV of 0,8 meaning that he on average will lose 0,2 chips for every chip he bets.

I hope this article has made you realize that understanding the concept of EV in poker is key factor of becoming a winning player in the long run.

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Tags: Poker Articles, Poker Mathematics

AA, where are you?

Monday, April 27th, 2009 | Poker Articles, Poker Mathematics | 1 Comment

Submitted by Mctap, this article belongs to the Poker Mathematics series.

This blog consists of 3 separate posts on my site that I have consolidated into 1 longer, more detailed post here.

About a month and a half ago I decided to check out my Poker Tracker stats for Poker Stars and noticed something interesting. At that point in time I had only played about 656 tournament hands and still hadn’t seen AA. No big deal as I was still up a few bucks since I started. I just found it interesting that with the probability of AA coming up being 1 in every 221 hands, I should have seen them almost 3 times by then. What did this mean, well, since the cards are generated through a Random Number Generator (RNG) program and many people complain that online poker is fixed, I *should* eventually hit a stretch of hands where I get AA more times than expected. Now keep in mind, I don’t think that online poker is fixed/rigged in any way, shape, or form, but if you spend time reading forums/blogs, you may get the impression that with so many bad beats given out, something fishy could be awry.

Several weeks went by before I looked at my Poker Stars tournament numbers again on Poker Tracker and discovered that I had played another 500+ hands (I know, I was just tearing through those hands) and during that time, I managed to see AA 3 more times. Much better, as this worked out to be 1 in approximately every 166 hands, which is more than the probability of getting a pocket pair (1:221). During those 3 times I had AA, I managed to take down all the pots! Another positive. What made all of this interesting, to me at least, is that over 1100+ hands I had seen AA 3 times and won with them each time (everybody expects to win with AA, but that does not always happen). So even though their frequency was worse (1:366) than their probability (1:221), their win rate was better (100%) than expected (~80%).

At this point I really started to expect that, according to odds and statistics, I *could* see them again soon. This time, I *could* also expect to possibly lose with them in order to balance out the numbers. Now some may say that this is Gambler’s Fallacy. For those who don’t exactly know what Gambler’s Fallacy is, it is to assume that one random event is somehow connected to another random event. Wikipedia has a great write up on Gambler’s Fallacy if you want to know more. Keep in mind, that when I say I *could* expect to see AA again, doesn’t mean I will to see them again. I am a numbers person and fully understand that each and every hand is randomly generated and has no correlation to any other hand played or to be played. Anyway I look at it, the odds of getting a pocket pair (in this case AA) is always 1:221, for each and every hand played, and it is very possible that I get the same exact hand dealt to me multiple times in a row. Although with a good RNG, the odds of this happening is quick small. Bottom line is that with the presumptions that I *could* lose with them, or that I may get AA more often, my thoughts are to play them strong and IF I lose, that is the nature of the game and really can’t complain about it (no matter how much it hurts to get your Aces Cracked).

Finally after playing another 1000 hands over the next 3 weeks (I’m just a machine, lol), my numbers stated that I was lucky enough to get AA only once more. Although that doesn’t fall within the realms of probability, it does show that the RNG is exactly that, random. Also, I did manage to win with them, which demonstrates that AA is a dominant hand against most random hands. So much for the Gambler’s fallacy.

Now to this point on Poker Stars, I’ve played a little over 2000 hands and have seen AA 4 times in my hand. That is roughly once every 500 hands. Nowhere near the 1:221 probability of getting a PP, but I have managed to make the most chips with them while playing (according to PT3). Again, at some point I *should* expect to start seeing AA more often and *might* lose with them (no hand can run 100% forever). Even if this becomes true, I will still play them aggressively to maximize my chip return.

Another interesting point is that during this stretch of hands on Poker Stars I did managed to see 77 18 times (1:111), which is much greater than the probability of seeing a pocket pair (1:221), but with 7′s I only won 56% of the time and actually was down chips overall. If you are confused, I won 10 out of 18 pots, but the total amount of chips won is less than the chips I lost during the other 8 times I had 77.

Bottom line is that I’m not complaining about not getting AA, nor any other pocket pair, I’m just monitoring this trend to see how it plays out for me. Always try to remember, that just because someone was lucky enough to get AA, or any PP for that matter, doesn’t mean they can’t get it again the very next hand.

See you on the felt.

McTap03

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Tags: Poker Articles, Poker Mathematics

Online poker tournament statistics : probability of winning a poker tournament

Tuesday, April 7th, 2009 | Poker Articles, Poker Mathematics, Poker Tournament | 12 Comments

This article is a part of the Poker Mathematics series.

Recently I have wondered if it is possible to calculate the probability of winning a poker tournament based on which strategy you use and how your all in moves are distributed. In this article I will share with you my findings. Hope you have the patience to read it all the way through:-). The table below (the culmination of way too many hours of research) is a teaser of what is ahead if you continue reading.

Check out the previous post in this series:

• Online poker tournament statistics: player exit percentage
• Online poker tournament statistics: cummulative blinds

Just to remind you, in my first post in this series I discovered an exponential relationship between the time an online freezeout tournament has been running and the percentage of players exited from the tournament. This enabled me to estimate both the time needed to reach the final table of a tournament given the number of players registered and what size tournament you should choose given the time you have available to play.

I needed this relationship to be able to estimate the number of players entering a tournament of a given duration and thus the total amount of chips in play. Once I know the total amount of chips in play is known, I can calculate the number of successive all in wins needed to win all the chips in the tournament. Finally, the last step in my calculations will  be to set up different all in probability distributions to end up with the probability of winning an online poker tournament. Confused? Don’t worry:-) I will guide you through my calculations step by step in the remainder of this article. If you don’t like math, simply scroll down to the final table where I summarize my most important findings.

The table below summarizes the total amount of chips in play for different tournament durations and the successive all ins you will need to win to win the tournament:

Again, in order to keep things simple I assumed that each all in would double the  1500 chip starting stack. According to this assumption, 2 successive all in wins will increase your stack from 1500 to 6000, 3 successive all inn wins from 1500 to 12000 and so on.

To make things a bit more realistic, let’s assume that you win half the chips you need to win the tournament by making your opponent fold (i.e no show down). In this case the number of  successive all ins needed to win is reduced by 1 for each of the tournament durations shown above.

I have chosen the following scenarios that IMO cover the typical all in situations you will experience during a tournament.

The simple reference calculation

Setting probability calculations aside and assuming all the players in the tournament (including yourself) have an equal chance of winning it, you will win a tournament with x registered players 1 out of x times. This means you will win a 100 player tournament 1 out of 100 times, a 200 player tournament 1 out of 200 times and so on. Obviously you should aim higher than this otherwise your bankroll will hit zero in no time.

1st scenario: On a roll

• You have an 80% probability of winning all your hands.
• If for example you need to win 3 successive all ins to win the tournament, the probability of winning it is 0,8*0,8*0,8 = 0,5 corresponding to 1 out of 2 tournaments.

2nd scenario: Coinflip

• You have a 50% probability of winning all your hands.

3rd scenario: Underdog

• You have a 30% probability of winning all your hands.

4th scenario: Realistic?

• 20% of your hands you are underdog (30% probability)
• 40% of your hands you have a coinflip (50% probability)
• 40% of your hands you are favorite (80% probability)

The table below summarizes the probability calculations for the 4 scenarios:

Interestingly, it turns out that in all but one scenario (the underdog) your tournament win rate is significantly larger than the simple reference calculation. In conclusion I think I have successfully managed to give a qualified estimate of the probabilities of winning online poker tournament. In my next article I will try to make some use of the numbers I have come up with. An obvious approach would be to calculate the expected return on investment (ROI) for each of the scenarios listed above.

I would greatly appreciate any comments on the math and final numbers.

You could be posting your articles on the Poker Bankroll Blog. Read all about it here.

Tags: Poker Articles, Poker Mathematics, Poker Tournament

Calculating Bankroll Requirements

Wednesday, April 1st, 2009 | Poker Articles, Poker Bankroll Management, Poker Mathematics, Poker Tools | 2 Comments

Submitted by Scott McIntosh, this article belongs to the Poker Bankroll Management series

In gambling games such as poker, blackjack or video poker where the player is trying to make money not only is it important to be playing with a positive expectation but also much care must be taken not to lose all of one’s money. As the saying goes – “It takes money to make money”. This requires winning players to maintain a certain amount of money in reserve to handle losing streaks and is commonly known as the player’s “bankroll”. This article will explore how to estimate the amount of money that is required to be kept as a poker bankroll in order to play so as reduce the probability of going broke to an acceptable level.

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Risk of Ruin Formula (warning serious math content ahead)

The risk of ruin formula is given below. The formula is based on a random walk with upwards drift and has some assumptions that although not exactly correct are good enough that it still gives a good approximation of a poker player’s risk of ruin. These assumptions include: a) normal distribution of results b) constant win rate and standard deviation. If the actual parameters vary too far from these assumptions care must be taken as the formula is likely to underestimate the actual risk of ruin.

Risk of Ruin = e ^ (-2WB / (S ^ 2))

e = Constant (2.718281828)
W = Win Rate, measured in $ pr. hour
S = Variance (has a unit that doesn’t really make any logical sense)
B = Bankroll, measured in $
^ = Power symbol (e.g. 3^2 = 9; 3^4 = 81)
/ = Division Symbol (e.g. 24/6 = 4; 56/7 = 8 )

Reference : D. Schlesinger, Blackjack Attack, RGE publ., Oakland 1997 (2nd ed. 2000)

Note: A player’s win rate and standard deviation is usually obtained from a Poker Database program such as “Poker Tracker”.

Example

Given a winning poker player with win rate of $30/hour, standard deviation of $600 and bankroll of $15000:

Risk of ruin = e ^ (-2*30*15000/(600^2))
Risk of ruin = 8.208500%

Although this is a precise answer given to 6 decimal places it is only an approximation and probably would be rounded up to give a 10% risk of ruin in practice.

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What Risk of Ruin should you accept?

One can never eliminate the risk of going broke, that is reducing the risk of ruin to 0%. Whilst 5% might be acceptable to a part time player, a professional whose only source of income is from poker should use something like a conservative 1% risk of ruin. Also given the error likely in measuring the win rate and standard deviation and also changing table conditions one should maintain a higher bankroll than calculated.

This article was written by Scott McIntosh of ReviewPoker Rooms which has a Bankroll Calculator based on the formula given above.

You could be posting your articles on the Poker Bankroll Blog. Read all about it here.

Tags: Poker Articles, Poker Bankroll Management, Poker Mathematics, Poker Tools

Online poker tournament statistics: cummulative blinds

Sunday, March 8th, 2009 | Poker Articles, Poker Mathematics, Poker Tools | 4 Comments

This article is a part of the Poker Mathematics series.

Recently I have wondered if it is possible to calculate the probability of winning a poker tournament based on which strategy you use and how your all in moves are distributed. During the process of collecting data to solve this problem I have run into some interesting observations which I would like to share with you.

Check out the previous post in this series: Online poker tournament statistics: player exit percentage

Having discovered a relationship between player exit percentage and the time any given tournament has been running for, the next step in calculating the probability of winning a poker tournament was to estimate the cumulative amount of blinds payed for any given period of time a poker tournament has been running for.

Just to remind you, the calculations shown below are based on the blind structure of FTP (Full Tilt Poker) regular tournaments with 10 minute blind levels (I’ve only included the first 10 levels )

In order to calculate the amount of blinds payed pr. level shown in the table above, I estimated a total of 10 rounds played pr. level. Because FTP tables are 9 handed the assumption of 10 rounds pr. level means that on average a player will pay the blinds a little more than once pr. level; 1,11 times to be exact.

Summing the “Blinds pr. level” over the “Tournament duration” up to blind level 34 led to the following table:

I myself was quite amazed when I saw these numbers. 866000 chips payed in total blinds for a 6 hour tournament. That’s a lot of chips!!!!

To have some sort of reference I wanted to calculate how many successive all ins you would need to win to pay these blinds. To keep things as simple as possible I assumed that each all in would double the  1500 chip starting stack. According to this assumption, 2 successive all in wins  increase your stack from 1500 to 6000, 3 successive all inn wins from 1500 to 12000 and so on.

The rounded necessary all in wins are shown in the table above (the precice numbers in brackets). Luckily it only takes roughly 9 successive all ins to win 866000 chips….aren’t you relieved:-)

I am now ready to calculate the probability of winning an online poker tournament, but you will have to wait until my next article to find out what it is.

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Tags: Poker Articles, Poker Mathematics, Poker Tools

Online poker tournament statistics: player exit percentage

Sunday, March 1st, 2009 | Poker Articles, Poker Mathematics, Poker Tools | 7 Comments

This article is a part of the Poker Mathematics series.

Recently I have wondered if it is possible to calculate the probability of winning a poker tournament based on which strategy you use and how your all in moves are distributed. During the process of collecting data to solve this problem I have run into some interesting observations which I would like to share with you (warning: math content ahead). First a little teaser….my findings indicate that it is possible to make good mathematical estimations on how many hours it will take before the final table in a tournament is reached based on the number of players registered for the tournament:

In the table below you will get a feel for how many hours you will have to play to reach the final table based on the number of players registered.

In the following table you will get a feel for the size of tournament you should choose given the time you have available.

Here’s a recollection of how I calculated the numbers above:

My tournament statistics project is based on online poker tournaments at Fulltilt poker and more precisely freezeout tournaments (i.e no rebuys and add ons). First of all, I came to realize that I would need to be able to estimate the number of players in an MTT given the time it takes for the tournament to finish. For example if a tournament lasts 3 hours, how many players were registered to play in it from the beginning?

What I did was to note down the following information for 25 online poker freezeout tournament on Fulltilt poker:

• The time the tournament had been running
• The number of players registered
• The number of players remaining

From this data set I was able to calculate the exit percentages, that is the relative number in percent of player exits, and plot them against the time the tournaments had been running. For example, in one tournament running for 3 hours and 28 minutes, 611 players registered and of them 25 remained yielding an exit percentage 95,9%.

Player exit percentage raw data

I was quite surprised to see the raw data plot shown above because it indicates an exponential mathematical relationship between the exit percentage and the time an online freezeout tournament has been running for. This relationship seems to be independent of the buyin of the tournament and the number of players entering.

I transferred the raw data to Origin and did a peak fit analysis to determine the mathematical relationship between the exit percentage (EP) and the hours played (x). Based on my original data set I imposed the following boundary conditions:

• EP(x=0)=0% : No players exit the tournament before it starts
• EP(x=6 hours) = 100% : All the freezeout tournaments I sampled had ended within 6 hours

Player exit percentage fitted data

Turned out the relationship was exponential as expected:

• EP(x) = A(1-exp(-Bx)), in this case A was 99,5 and B was 0,96

Now here comes Now here comes the interesting part. Given the equation above, relating the exit percentage with the number of hours played, it is now possible to estimate both the hours it takes a tournament to finish and the inverse, namely the number of players entering into a tournament given the hours it takes to finish.

Since I’m not able to paste Excel formulas into WordPress blog posts, I have made some tables with some sample data just to give you an idea of how you can use the formula.

In the table below you will get a feel for how many hours you will have to play to reach the final table based on the number of players registered.

In the following table you will get a feel for the size of tournament you should choose given the time you have available.

Hope you found the information in this article useful. If you have any comments or want a copy of the Excel spreadsheet I used to calculate the data in the tables, please let me know.

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Tags: Poker Articles, Poker Mathematics, Poker Tools

Implied odds; all you need to know

Thursday, December 25th, 2008 | Poker Articles, Poker Mathematics | 2 Comments

This article is a part of the Poker Mathematics series.

Implied odds is a more advanced poker concept than pot odds, poker probabilities and EV. You should therefore familiarize yourself with these basic concepts first.

Implied pot odds take into account future bets made to the pot and are therefore not as straightforward to calculate as regular pot odds. Implied pot odds are most relevant when you are on a draw and stand to gain additional bets if you make the draw but on the other hand won’t make additional bets yourself if you miss the draw in question. The following examples will give you an idea of how to calculate your implied pot odds in different situations:

You are playing in a large online poker tournament. Blinds are 200/400 and you have 20000 chips. In middle position you are dealt QK of diamonds and raise 3xBB. The action is folded to BB who calls your raise leaving him with 18000 chips. The pot is 2600. The flop is A(d)3(d)8(c). Your opponent bets the pot signalling he has an Ace which leaves him with 15400 chips. From a regular pot odds viewpoint you are getting pot odds 3 to call and with 9 outs you have a 19% chance of hitting your flush on the turn. From a regular pot odds viewpoint you are not getting the right odds to call. However let’s assume that you call your opponents pot bet and your opponent will make another pot sized bet (of 7800) on the turn. If you hit your flush your call on the flop made you an additional bet of 7800. Therefore your implied pot odds on the flop were 7800 (the pot after the flop) + 7800 (your opponents bet on the turn)/2600 (your call on the flop) = 6. Now with odds 6 to call you are making a +EV play by calling the potbet on the flop; you need a 17% probability for winning the hand to make a breakeven play with pot odds 6 and you in fact have a 19% probability of hitting one of your 9 outs to complete your flush.

Please be aware that justifying a call due to implied odds is a tricky business. You have to be pretty sure that you will be able to gain additional bets from your opponent on the following streets. Drawing to straights is therefore usually better than drawing to flushes since straights are better concealed and are thus more likely to extract additional bets from your opponents.

The example from above demonstrates that the stack size of your opponent is also important. Calling drawing hands due to implied odds is done against deep stacks where you have the possibility of doubling up or at least making a large contribution to your chip stack. To illustrate this, let’s use the sample example from above, but now your opponent has a chip stack of 3000 after making his pot bet. Your implied odds for calling have now dropped to (7800+3000)/2600 = 4,2, which is not enough to call.

There is one situation in particular where you will almost always call due to implied odds, namely calling preflop raises with small pocket pairs. Most of the times you will be calling with the wrong regular pot odds, but your implied odds are great because you will usually earn big pots when you hit your set. When playing small pocket pairs remember the two rules: “no set, no bet” and “yes set, yes bet”.

Merry X-Mas and happy New Year to all!!!

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Tags: Poker Articles, Poker Mathematics