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The Mathenoobics of Poker The Mathenoobics of Poker

02-09-2012 , 09:27 AM
index of all the parts of this series

tl; dr

A 5000th post is traditionally used to pass on some stunning strategic insight, helping the rest of you to crush even more than you already do. And if I knew more about the game, that’s what I’d attempt, but I don’t, so this will have to do.

There's a regular - one might almost say unending - stream of questions in Beginners about the basic mathematics of poker. Guys who are struggling with pot odds, implied odds, pot equity, fold equity, how often someone needs to fold for a bluff to be profitable, do I need to multiply by 2 or 4? etc etc. All good stuff, and there are plenty of books and articles out there that will answer all these kind of questions, but for my 5000th post, I thought it might be fun to start a thread here in Beginners which outlines, in a systematic way, some of these key concepts.

This is for Beginners. It's going to be basic. We're not going to get into Game Theory, Nash Equilibria, or reverse implied fold equity (whatever that is). If you're happy working out the Expected Value of a turn semibluff shove, this isn't for you, except that some of you guys will know more about this stuff than I do, so please chip in and add your own insights and do a better job of explaining some of this stuff than I will. And if you have questions, or something isn't making sense (entirely possible when it's me doing the explaining ) then remember the First Rule of Beginners: nobody's allowed to feel awkward about not understanding something. Fire away. No such thing as a stupid question.

Right off the top of my head I'm thinking we should talk about stuff like:
- probability
- combinatorics
- expected value
- equity
- counting outs
- pot odds and implied odds
- bluffing (and semibluffing)

I’m assuming you’re comfortable with basic arithmetic, simplifying fractions, switching quickly between decimals, percentages, and fractions etc. If you’re not, just say so, and someone in this thread will help you get sorted.

We're going to need more than one post for this. Not sure how many. Let's make a start and see what happens.

PROBABILITY
We live in an uncertain world. We play a game that is centred around uncertain outcomes. In everyday language we use words like 'unlikely', 'possible', 'certain', 'favourite', 'underdog' etc to describe these events.

Probability is the mathematician's way of quantifying the likelihood of something happening. It's a tool which poker players can use to make better decisions.

Probabilities are measured on a scale which runs from 0 to 1.
An event that is impossible is assigned a probability of 0.
At the other extreme, something that is certain to happen is given a probability of 1.
Everything else is somewhere in between.

A lot of the probabilities we'll want to talk about can be determined accurately.
If I pick a card at random from a full deck, there are 52 cards, and 26 of them are red.
So the probability of picking a red card = 26/52 = 0.5 = 50%
(Note then, that a probability can be expressed as a fraction, a decimal, or as a percentage. Use whichever you prefer.)

This is sometimes written as p(red card) = 0.5

There are 13 hearts, and 39 non-hearts. in a full deck.
So p(heart) = 13/52
and p(non heart) = 39/52

Note that p(heart) + p(non-heart) = 13/52 + 39/52 = 52/52 = 1
(Hopefully, this makes sense since we must get either a heart or a non-heart. There are no other possibilities. We're therefore certain to get one or the other.)

If A is some event which may or may not happen, then
p(A) + p(not A) = 1
So p(A) = 1 - p(not A)

It's quite common, when trying to work out the probability of something happening, that's it actually easier to work out the probability of it NOT happening. We can then subtract that probability from 1, to find the probability we're actually interested in.

Note that when we said

p(heart) + p(non-heart) = 1

we were dealing with two mutually exclusive events. That is, it's not possible to select a card which is both a heart AND a non-heart.

But it's possible to have a situation where the two events we're interested in are not mutually exclusive.

For example, if we pick a card at random, what is p(Heart OR Ace)?

We could start by simply counting the number of qualifying cards.
There are 13 Hearts, and 4 Aces, but one of those Aces is already counted in the 13 Hearts.

So 13 Hearts, and 3 non-Heart Aces, makes 16 cards all together.
So p(Heart OR Ace) = 16/52 = 4/13

Our general rule for working out p(A OR B) where A,B are two events is as follows:

p(A OR B) = p(A) + p(B) - p(A AND B)

where p(A AND B) is the probability of both events happening at the same time (which will equal zero if the two events are mutually exclusive, because that's what mutually exclusive means).

Let's use this rule to check our calculation of p(Heart OR Ace)

We know p(Heart) = 13/52 = 1/4
and p(Ace) = 4/52 = 1/13

What is p(Heart AND Ace)?
There's only one card that is both a Heart AND an Ace - the Ace of Hearts.
So p(Heart AND Ace) = 1/52

So our rule gives us:
p(Heart OR Ace) = p(Heart) + p(Ace) - p(Heart AND Ace)
= 1/4 + 1/13 - 1/52
= 13/52 + 4/52 - 1/52
= 16/52
= 4/13
which is the same result we got by counting the qualifying cards.

If you want to play along, try these:
If we pick a single card at random from a full deck, what is
(a) p(a Broadway card)
(b) p(a Broadway card OR Club)
(c) p(a black wheel card)
(d) p(a wheel card which is not a 3 or a Diamond)

Sometimes we’ll be interested in two independent events. Events are said to be independent if the occurrence (or non occurrence) of one of them has no impact at all on the likelihood of the other occurring. For example if I spin a coin and roll a die, the result of the coin (Heads or Tails) is totally independent of the result of the die (1,2,3,4,5, or 6).

In sports (here in the UK) Wolves winning the Premier League and Exeter winning League One are clearly independent events (they also happen to have vanishingly low probability but that’s another matter). However betting events like ‘Manchester United will beat Liverpool this Saturday’ and ‘United’s Wayne Rooney will score three goals’ are most certainly not independent. (If all you learned about the game afterwards was that Rooney had scored three goals, that would absolutely change your view about how likely it was that Manchester United won the match.)

We have a simple rule for dealing with the probability of two independent events both happening:
p(A AND B) = p(A) x p(B)

So if I spin the coin and roll the die,
p(Heads AND 4)
= p(Heads) x p(4)
= 1/2 x 1/6
= 1/12
(exactly the result you’d get if you were to list all possible outcomes:
(H,1),(H,2) …. (T,5),(T,6)
a list of 12 outcomes, of which (H,4) is just one.)

An obvious application of probability in poker is to calculate the likelihood of being dealt different kinds of hands preflop. But before we do that, it would be useful to know a little about combinatorics. We’ll talk about that in the next part.

Last edited by AlienSpaceBat; 03-03-2012 at 07:46 AM. Reason: added index link
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02-09-2012 , 09:36 AM
Brilliant. Look forward to reading the next part!
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02-09-2012 , 09:37 AM
Quote:
Originally Posted by SickFold89
Brilliant. Look forward to reading the next part!
+1. Nice one DD.
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02-09-2012 , 10:15 AM
Bookmarked.


Also, I'll have $5 on Wayne Rooney to score three goals and for United to beat Livepool on Saturday please. What are the odds?
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02-09-2012 , 10:26 AM
good so far - look forward to pt 2
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02-09-2012 , 10:33 AM
Quote:
Originally Posted by bumblebee99
Also, I'll have $5 on Wayne Rooney to score three goals and for United to beat Livepool on Saturday please. What are the odds?
The odds are pretty much the same as for just Wayne Rooney scoring a hat trick. The odds on Wayne Rooney scoring a hat trick and Man U losing the game would be quite long.

You'd get about 13:1 I reckon.

Last edited by gothninja; 02-09-2012 at 10:40 AM. Reason: Sorry DD - pretty bad derail I must admit (BB I'll lay you 10:1)
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02-09-2012 , 10:44 AM
Nice post DD ! I think a LOT of BQers can get value from this.
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02-09-2012 , 10:46 AM
Now why didn't my maths teacher explain it like that.
Looking forward to next installment.
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02-09-2012 , 10:51 AM
Better explained with cards than with drawing coloured balls from a sack as done in schools probability calculus.
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02-09-2012 , 11:29 AM
Quote:
Originally Posted by DiamondDog
If you want to play along, try these:
If we pick a single card at random from a full deck, what is
(a) p(a Broadway card)
(b) p(a Broadway card OR Club)
(c) p(a black wheel card)
(d) p(a wheel card which is not a 3 or a Diamond)
My attempt:
Spoiler:

(a) p(a Broadway card)
p( Broadway card) + p(non-Broadway card) = 1
p(Broadway card) = 1 - p(non-Broadway card)
p(Broadway card) = 52/52 - 32/52
p(Broadway card) = 20/52
p(Broadway card) = 5/13
(b) p(a Broadway card OR Club)
p(Broadway card OR Club) = p(Broadway card) + p(Club) - p(Broadway card AND Club)
p(Broadway card OR Club) = 20/52 + 13/52 - 5/52
p(Broadway card OR Club) = 28/52
p(Broadway card OR Club) = 7/13
For the rest of the exercises I needed to know that p(A AND B) = p(A) x p(B).

(c) p(a black wheel card)
p(black wheel card) = p(black card) x p(wheel card)
p(black AND wheel card) = (26/52) x (20/52)
p(black AND wheel card) = 520/2704
p(black AND wheel card) = 10/52
(d) p(a wheel card which is not a 3 or a Diamond)
p(wheel card AND NOT Diamond OR 3) = p(wheel card) - [p(wheel card AND Diamond) + p(3) -p(Diamond AND 3)]
p(wheel card AND NOT Diamond OR 3) = 20/52 - [(20/52 x 13/52) + 4/52 - 1/52]
p(wheel card AND NOT Diamond OR 3) = 20/52 - (260/2704 + 4/52 - 1/52)
p(wheel card AND NOT Diamond OR 3) = 20/52 -(5/52 + 3/52)
p(wheel card AND NOT Diamond OR 3) = 20/52 - 8/52
p(wheel card AND NOT Diamond OR 3) = 12/52
Did I get it right?


Will we have to wait another 5000 posts for the next part ;D?
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02-09-2012 , 11:45 AM
Quote:
Originally Posted by bumblebee99
Bookmarked.


Also, I'll have $5 on Wayne Rooney to score three goals and for United to beat Livepool on Saturday please. What are the odds?
No chance. I already have money on suarez to score. Hopefully its in front of the stretford end.
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02-09-2012 , 12:38 PM
bigups DD
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02-09-2012 , 01:02 PM
@gothninja: nope, that's not a derail at all imo. Excellent point. And a reminder that we'll need to talk about converting probabilities into odds blah blah blah.

@nismo9:
Quote:
No chance.
Just to be a nit for a moment (he said, thrilled to get a chance to be a nit) but in the context of the thread this is clearly incorrect. There's definitely a non-zero probability of Rooney sticking in three and United winning the match. Anyone who really believes the probability of it happening is zero, should, in theory, be willing to lay any odds we request. (I did wonder whether it was a good idea, introducing words like 'Manchester United' and 'Liverpool'. Have to confess, my own team ply their trade much lower down the food chain.)

@Sworaven
Yes, those are all correct. Good stuff.

Spoiler:
Seems to me, one of the things that makes this stuff interesting - or a challenge, depending on which way you look at it - is that there’s nearly always more than one way of arriving at the answers.
In some cases, I find it easier simply to count the qualifying cards, so

(a) p(a Broadway card)
Broadway = A,K,Q,J,T = 5 cards
x 4 suits = 20 cards out of 52
20/52 = 5/13

(b) p(a Broadway card OR Club)
= p(Broadway) + p(Club) - p(Broadway AND Club)
= 5/13 + 1/4 − 5/52
= 20/52 + 13/52 − 5/52
= 28/52
= 7/13

(c) p(a black wheel card)
black wheel card = A,2,3,4,5 of Clubs or Spades
= 10 cards
10/52 = 5/26

(d) p(a wheel card which is not a 3 or a Diamond)
A,2,4,5 = 4 cards, of 3 suits = 12 cards
12/52 = 3/13

Quote:
Will we have to wait another 5000 posts for the next part ;D?
We should regard that as 'unlikely'

Last edited by DiamondDog; 02-09-2012 at 01:19 PM.
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02-09-2012 , 01:03 PM
Nice post, are u doing part 2?
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02-09-2012 , 08:22 PM
Since DD welcomed others to chime in, I will do so. Rather than Part 2, you can consider this to be an Appendix to DD’s Part 1, which I thought was excellent. It is taken directly from the beginning part of a draft Probability chapter of a book I’m writing tailored to hold’em players not afraid of a bit of math but not at the complexity of MOP, for example. It discusses four definitions of probability with the last three definitions providing insight into how to estimate probability. Here it is – unfiltered
*************************************
Probability is the underlying mathematical theory for analytical poker. Most of us have an intuitive understanding of probability, namely it measures or quantifies the likelihood of some chance event occurring. Sometimes the quantification is in relative terms, such as – two pairs are more likely or more probable than three of a kind. Sometimes, it is numerical, such as -- the probability that a pair of aces will win against nine random hands is 31.2%.

Card games in most every form involve probability, simply because the shuffling of the cards and then dealing them out to players involves a random or chance element. In fact, much of the early development of probability theory came about with reference to games of chance. For hold’em strategy purposes, the most useful application of probability is in estimating the chance that you will win the hand. However, there are many other applications such as determining how much bankroll you need to avoid going bankrupt and what opponent-fold probability you need to ensure a +EV decision.

An interesting aspect of probability is that it uses a very precise discipline, mathematics, to deal with a very imprecise situation, the occurrence of chance events. There are four basic ways in which probability is defined and for completeness we will list them here. We shall use the term ‘event’ to represent the chance occurrence we are dealing with, such as the outcome of the toss of a die or the rank of the river card in a hold’em game and we use the notation Pr(E) to mean the probability of event E.

Mathematical Definition

The probability of an event is a number between 0 and 1, which has certain properties; namely

1. If an event E is certain to occur, then Pr(E) = 1 (Example: E = red card or black card for dealing a single card)

2. If an event E cannot occur, then Pr(E) = 0 (Example: E = red card and black card)

3. If two events E1 and E2 cannot occur simultaneously, then Pr(E1 or E2) = Pr(E1) + Pr(E2). Example: (E1 = ace; E2 = king)

From these conditions we can deduce that probability is a number between zero and one. We therefore can express it as a decimal from 0 to 1, as a fraction in this range, or as a percentage from 0% to 100%. If you do a probability calculation and get a value outside this range you have, with probability 1.0 or probability 100%, made a mistake.

With these three conditions, one can develop all the necessary formulas applicable to working with probabilities. Note that this definition never really defines what probability is or how to estimate it. Also note that if two events cannot simultaneously occur, then the probability that one or the other does occur is the sum of their individual probabilities. For example, the probability that you select an Ace or a King from a deck of cards is 2/13, since 1/13 is the probability for each event separately (see the Classical definition below). But, the probability that you select an Ace or a Heart is not 1/13 + 1/4 because they occur together if the ace of hearts is dealt.

Statistical Definition

We define a random trial to be some action that results in a chance outcome or event , such as the toss of a coin or the selection of a sample element from a population such as the two cards dealt to you in hold’em. An independent trial is one where the outcome of one trial has no effect on the outcome of another trial. If, in a series of n independent trials an event occurs r times, then the probability of the event is estimated by the ratio r/n, with the estimate improving as n increases.

Statistical Probability of Event E:

Pr(E) = r/n,

where r = number of successes (occurrences of E) in n trials.

This definition simply states that probability reflects the fraction of times an event occurred in a series of n similar trials. The larger the number of trials, the better will be the estimate. The statistical probability definition is applied more often than one would think in hold’em. For many situations the equations for calculating probability are so difficult to develop that a computer simulation program is used to generate thousands or more likely millions of hands in order to make probability estimates. From such simulations, tables of win probability or odds are developed, some of which are included in this book. For example, if one wanted to calculate the probability that a pair of kings could win in a showdown against 5 other players pre-flop, it would be near impossible to enumerate all the possible ways for such a showdown to occur. However, a showdown calculator using simulation would find the answer in a matter of seconds – the answer is about 43%, by the way.

Classical Definition

If a random trial can result in n equally likely outcomes and if r of those outcomes is consistent with an event E, then the probability of event E is r/n. Note the similarity of this definition with that of the statistical. They both define probability as a ratio – one ratio determined by experimentation and the other determined by counting the known possibilities.

Classical Probability of Event E

Pr(E) = r/n,

where r = number of possible outcomes consistent with E when there are a total of n equally likely outcomes.

In card playing terms, the random trial is the dealing of cards, the outcomes refer to the individual cards that are dealt and the event is a grouping or subset of outcomes. For example, a pair is an event or a specific pair such as kings is an event or a flop with 2 of the 3 cards of the same suit is an event. With a fair shuffle, we assume that any card is as equally likely to be dealt as any other card, so the equally likely condition is satisfied. The probability of a given hand or result occurring is then the number of ways the hand or result can occur divided by the total number of hands or results that can occur.

Let's do an example. Suppose we are interested in finding an ace on top if we shuffle the deck thoroughly and then cut the deck at some random point. After cutting the deck there are 52 possible outcomes. Of these 52, four are aces, so the probability of cutting to an ace is 4/52 or 1/13. If we used the statistical definition, we would shuffle the cards a number of times and count how many times an ace appeared on the cut. If all were well, it would get closer and closer to 1/13 as the number of trials got larger.


Subjective Definition

For this definition, the probability of an event is a number between 0 and 1 that represents one's belief that the event will occur.

Subjective probability is just that - a guess about an event happening that may be based on experience or on logic, or a combination of the two. However, a guess cannot be made in such a way as to violate basic probability rules. For example, if one says subjectively that the probability of a 1 on tossing a die is greater than the probability of tossing either a 1 or a 2, then that will lead to nonsensical or inconsistent results.

In this book, we actually use all four forms of probability. The mathematical definition actually forms the basis for manipulations of all probability numbers, however these numbers are determined. The statistical definition is used when we calculate the winning probability of a hand using simulation or we cite probability numbers based on actual data. The classical definition, the one most often applied, is used in many ways such as when we count outs to determine the chances of winning. And, the subjective definition is used, for example, when we make reasoned guesses about what an opponent may have based on his prior betting habits and other clues.
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02-09-2012 , 09:06 PM
Quote:
Originally Posted by statmanhal
It is taken directly from the beginning part of a draft Probability chapter of a book I’m writing tailored to hold’em players not afraid of a bit of math but not at the complexity of MOP
Cool, look forward to reading this.

Spoiler:
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02-09-2012 , 11:15 PM
5 star thread!

I'm about to have my 3000th post and I think I'll struggle to write anything as useful as this. If anyone has any great ideas, please PM me.
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02-10-2012 , 12:17 AM
Nice start. This is really useful to anyone that wants to be serious about the game.
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02-10-2012 , 03:46 AM
Quote:
Originally Posted by statmanhal
...tailored to hold’em players not afraid of a bit of math but not at the complexity of MOP, for example...
Very good, and I look forward to seeing the final book! But it must be said DD's writing is much more beginner-friendly and perhaps better suited for people who are a little afraid of the maths!
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02-10-2012 , 04:52 AM
@maxostoch: the next couple of bits are pretty much written, and I'm planning a couple more after that.

@gothninja: Thanks for your words of encouragement. I'm definitely making it as simple as I can. Aimed at the math-phobics without wanting to scare off those who are afraid of being considered math-phobics. It's a tricky line to tread but bah, we'll give it a go.

@statmanhal: Thanks for your input. Awesome stuff. I most definitely want others to chip in and add to/correct/expand on my own offering.

Your reference to card games forming one of the early motives for studying probability is very interesting. I've no idea if it's a true story, but I once read that back in the day (in France, I believe) it was not uncommon for a bunch of noble guys to be playing cards and for one of them to get called away to fight a duel (or maybe it was two guys at the same table who fell out and decided to try and shoot each other).

So in the middle of the hand one or two of them would leave the table. It's the nature of duelling of course, that they didn't always return.

Then a few days later the widow of the loser would sometimes approach the guy running the card game and enquire about whether there was any money owing to her dead husband, giving rise to some of the earliest discussions of equity.

You can imagine the kind of conversations that might have ensued. "Well, let's see, Madam - and by the way, my condolences on the death of your husband, fine man, etc etc - he had pocket Aces, while his opponent in the hand held JT and the flop, which had just been dealt when the argument broke out, was 382." Maybe there was a lowly geek, with a ledger, who hired himself out, running a manual version of Le Stove, who, three weeks later, would declare that the widow was owed x francs.
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02-10-2012 , 04:56 AM
DD the OP was like soft porn for geeks. Drop me a pm when the book is done i will put a copy on the desk
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02-10-2012 , 06:37 AM
DD - it's people such as yourself that make 2p2 a great place.

WTF IS CHEN itt?
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02-10-2012 , 07:21 PM
Excellent poster makes excellent post

This should be really useful for beginners, good work DD.

Looking forward to the rest.
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02-17-2012 , 08:06 AM
Very helpful information!!!! Thanks
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07-26-2012 , 01:41 PM
Quote:
I’m assuming you’re comfortable with basic arithmetic, simplifying fractions, switching quickly between decimals, percentages, and fractions etc. If you’re not, just say so, and someone in this thread will help you get sorted.
Unfortunately, I'm that guy who needs help getting sorted. I'm an artist type, not a left brain type.

I haven't dealt with fractions since school. So simplifying fractions gets me lost, I'm embarrassed to say. I mean I get that 4/8 = 1/2 and such. But that is obvious, I tend to get lost on the more non-obvious stuff.

So any tips or links to help me sort that would be great. Also, switching from decimals to percentages to fractions is another thing I find difficult. I've been trying to read a lot of these odds articles around the net, and having a hard time. This series of posts looks great. I do feel like I got the basic idea in this first one.
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