Ask Me About Poker Math Christmas Edition

[spoonitnow]

gogogogo

[kmind]

These aren't completely math stuff but:

1. Have you ever tried making a Game Theory Optimal preflop range?

2. I completely like your combinatorics thread but are there any exercises to calculate total combos very quickly? Maybe I'm just a total noob to this but I find myself taking a bit to thinking of combos and then adding them up and trying to figure out what part folds, etc.

[Pelion]

1. Have you ever tried making a Game Theory Optimal preflop range?

Im guessing that would take some kind of super computer :P.

[spoonitnow]

These aren't completely math stuff but:

1. Have you ever tried making a Game Theory Optimal preflop range?

For certain situations with restricted betting options yes, like push/fold with small stacks or raise/3bet/4bet/5bet situations restricted for only being able to raise to certain intervals or fold preflop with no calling. It's a pretty straight-forward mechanical process, and isn't nearly as useful as people usually seem to think it is.

The process of solving poker situations and toy games for optimal strategy is covered at great length in The Mathematics of Poker by Bill Chen and Jerrod Ankenman. But I warn you, it's much more of a math textbook than it is a poker book.

2. I completely like your combinatorics thread but are there any exercises to calculate total combos very quickly? Maybe I'm just a total noob to this but I find myself taking a bit to thinking of combos and then adding them up and trying to figure out what part folds, etc.

In my opinion, you shouldn't ever be trying to break down large ranges in terms of combinations at the table. Instead you should do it a lot in your study and cultivate a feel for it. This is another skill that is probably less useful than people seem to think. It seems that whatever precision you would gain by being able to do this for large ranges within a reasonable amount of time would be mostly wasted by the lack of accuracy in the ranges you actually put people on.

1. Have you ever tried making a Game Theory Optimal preflop range?

Im guessing that would take some kind of super computer :P.

A normal computer would do, but the game tree is so large (even for heads-up fixed-limit games) that it can't be solved for within a reasonable amount of time. Some simpler games have been solved for unexploitable strategies, like heads-up push/fold no-limit hold'em, and heads-up fixed-limit single draw lowball (both 2-7 and A-5 iirc).

Also I want to note that I'm pretty sure it's not possible to just solve for the preflop play without solving for the rest of the game.

[kmind]

I thought that's what spoonitnow was?! WTF?

[elipsesjeff]

This is probably something you've done before so it shouldn't be that bad .

Calculate the total EV of calling a preflop re-raise with 55. We raise UTG 6 handed at 25 NL to $.75 and get re-raised to $2.50 by the button. Lets say he has QQ-AA and AK in his reraise range in this spot. Effective stacks were $25 to start the hand.

In a re-raised pot we will get AA to stack off 100% of the time on any board, KK 100% on any undercard boards but only 1 3/4 size pot bet on an Ace, AK 100% of the time when he flops TPTK or better and we flop our set, and QQ 50% when he flops an overpair and again only 1 3/4 size pot bet on an Ace or King. What's the total EV of calling?

Follow up question: If we include JJ/AQ in the above reraising range, and add in JJ stacking off 33% on an undercard board and only 1 1/2 pot bet on an A/K/Q. AQ you get 2 1/2 pot bets, but not his stack when he flops TPTK or better when we flop a set. Does this change the EV enough to swing the decision one way or another?

3rd Question: If villain's only reraising with AA/KK here can we call pre-flop, given that he stacks off with AA preflop 100% and KK as an overpair 100% and 1 3/4 psb when an Ace comes?

[spoonitnow]

This is probably something you've done before so it shouldn't be that bad .

Calculate the total EV of calling a preflop re-raise with 55. We raise UTG 6 handed at 25 NL to $.75 and get re-raised to $2.50 by the button. Lets say he has QQ-AA and AK in his reraise range in this spot. Effective stacks were $25 to start the hand.

In a re-raised pot we will get AA to stack off 100% of the time on any board, KK 100% on any undercard boards but only 1 3/4 size pot bet on an Ace, AK 100% of the time when he flops TPTK or better and we flop our set, and QQ 50% when he flops an overpair and again only 1 3/4 size pot bet on an Ace or King. What's the total EV of calling?

Follow up question: If we include JJ/AQ in the above reraising range, and add in JJ stacking off 33% on an undercard board and only 1 1/2 pot bet on an A/K/Q. AQ you get 2 1/2 pot bets, but not his stack when he flops TPTK or better when we flop a set. Does this change the EV enough to swing the decision one way or another?

3rd Question: If villain's only reraising with AA/KK here can we call pre-flop, given that he stacks off with AA preflop 100% and KK as an overpair 100% and 1 3/4 psb when an Ace comes?

These are all pretty straight-forward EV equations. Instead of answering them individually (which won't be of that much use to most people who read this thread since they'll see a wall of text with some math in it, shit themselves, and move along) I'll give some tips on making these long EV equations as painless and mistake-less as possible.

The first thing you should do is list out all of the possible outcomes. This can be done with the assistance of a diagram or whatever you need for it. The third question has the fewest outcomes so I'll do it as an example:

1) We miss the flop
2) We hit the flop, Villain has AA, Villain does not outflop us, we win showdown
3) We hit the flop, Villain has AA, Villain does not outflop us, we lose showdown
4) We hit the flop, Villain has AA, Villain does outflop us, we win showdown
5) We hit the flop, Villain has AA, Villain does outflop us, we lose showdown
6) We hit the flop, Villain has KK, Flop comes A-high, Villain does not outflop us *
7) We hit the flop, Villain has KK, Flop comes A-high, Villain does outflop us, we lose showdown
8) We hit the flop, Villain has KK, Flop comes A-high, Villain does outflop us, we win showdown
9) We hit the flop, Villain has KK, Flop does not come A-high, Villain does not outflop us, we win showdown
10) We hit the flop, Villain has KK, Flop does not come A-high, Villain does not outflop us, we lose showdown
11) We hit the flop, Villain has KK, Flop does not come A-high, Villain does outflop us, we win showdown
12) We hit the flop, Villain has KK, Flop does not come A-high, Villain does outflop us, we lose showdown

* We'll assume that when he has KK and the flop comes A-high if he misses a set that he bet/folds the flop so that we don't have to deal with the times we see a turn and he hits his set and puts more money in, etc.

The total EV of the situation will be the sum of the individual EVs of the outcomes. The EV of each individual outcome is the chance it happens times our profit for that outcome (including our call preflop). I'll work out the individual EVs of two of these outcomes, an easy one to understand the idea and a hard one to understand the detailed process.

1) We miss the flop

There are 48 cards left in the deck and two of them make our hand. The chance of none of our cards coming on the flop is (46/48)(45/47)(44/46) = 0.878 or 87.8%. That is the chance of this outcome happening (note that this also means the chance of at least one 5 coming is 1 - 0.878 = 0.122 or 12.2%, which we'll use later). Our profit for this scenario is -$1.75 since that is the amount for our call preflop. Since the EV of each individual outcome is the chance it happens times our profit for that outcome, the EV of this particular outcome is 0.878 * -1.75 = -$1.537.

9) We hit the flop, Villain has KK, Flop does not come A-high, Villain does not outflop us, we win showdown

Note: I'm going to bold a few numbers as we go down because in our final calculation for this outcome I don't want people to be thinking "Holy shit where did this number come from?"

The chance Villain holds KK is 50% since his range is {KK+} evenly distributed. Let's find the chance that at least one 5 comes on the flop, the flop does not contain an Ace, and the flop is not K5x (where x is not a 5). This is the part where peoples asses usually start churning butter milk, but it's not that bad really.

First, the chance the flop comes K5x where x is not a 5 is (2/48)(2/47)(45/46) * 6 = 0.0104 or about 1.04%.

Now what's the chance total that the flop comes with at least one 5 but without an Ace? The chance of a 5 coming on the first card is 2/48. The chance of an Ace NOT coming on the second card is 43/47. The chance of an Ace NOT coming on the third card is 42/46. Then the total chance of this type of flop is (2/48)(42/46)(43/47) * 6 = 0.2088 or 20.88%.

Then the chance of a flop coming with at least one 5, no Ace, and not K5x (where x is not a 5) is 20.88 - 1.04 = 19.84% or 0.1984.

Once we see this flop, what is the chance we win a showdown? It's going to be roughly our equity on the flop, so we'll use that, which is about 91% give or take some runner draws for KK.

Our profit for this outcome is that we win what was in the pot before we called plus the rest of our opponent's stack. The pot when it is our turn to act is $3.60 and the remainder of our opponent's stack is $22.50. This assumes 0.10/0.25 blinds (some sites use 0.15/0.25 instead) and no rake. Accounting for the rake in these types of calculations is pretty easy, you just subtract it from the profit in the outcomes where you win the hand. So our total profit for this outcome is 3.60 + 22.50 = $26.10.

Then the final calculation is easy. Recall our outcome: We hit the flop, Villain has KK, Flop does not come A-high, Villain does not outflop us, we win showdown.

- We hit the flop and it's not A-high and Villain doesn't outflop us: 19.84%
- Villain holds KK: 50%
- We win showdown: 91%
- Our profit: $26.10

So the total EV of this outcome is 0.1984 * 0.5 * 0.91 * 26.10 = $2.356.

This goes a lot faster when you use a spreadsheet and some sort of chart or program to give you the flop/turn/river and equity calculations.

[spoonitnow]

Also I think GrinderSchool needs a math coach amirite

[elipsesjeff]

hah, ty. i figured someone shoulda asked you a math quesiton

[Fielmann]

I have infinite bankroll and I have an ability to play an infinite number of hands in one day. Will there come a moment this evening where I have won $1.000.000?
(I am playing 2NL and my plan is to go all-in every hand, if this should change the answer somehow.)

[spoonitnow]

I have infinite bankroll and I have an ability to play an infinite number of hands in one day. Will there come a moment this evening where I have won $1.000.000?
(I am playing 2NL and my plan is to go all-in every hand, if this should change the answer somehow.)

Edit: See below:

[Pelion]

Suppose after H hands your chance of having a profit P at some point in that sample is n and n > 0. Then in 2*H hands, your chance of having a profit P at some point in that sample is higher than n. In 3*H hands, your chance of having a profit P over that sample is going to be even higher, etc. As your sample size approaches infinity, so does your chance of having a profit P at some point over your sample.

While I havnt fully thought about this yet, I dont think its necessarily true. For example, think about a 6 sided die. You bet on a 6 and get a 1:1 payout. You bet $1 and you want to know the chance you will have a profit of P = $1.

If H = 1.

You have a 1/6 chance of meeting P. n is therefore 1/6.

If H = 2.

If you lose your first roll you can only ever breakeven. Therefore you have a 5/6 chance of missing with your first roll. Even if your first roll hits you still have a 5/6 chance of losing your second bet and breaking even. The only case where you will have a profit of more than P = $1 after 2 rolls is if you win both rolls. This is a 1/6 * 1/6 chance, so 1/36. Since your expectation is negative with one roll, it becomes more negative the more rolls you make, and the chance n of making a profit P therefore becomes smaller (given that your standard deviation doesnt change in some unfeasably huge way).

So as H increases, n tends to 0.

Likewise, if you play an infinite number of hands at negative expectation (and this is where some pretty hardcore maths probably has to come in), there will be streaks in that number of hands where your profit is infinite, and there will be streaks where your profit is negatively infinite. However, the negative infinities will both be bigger and more frequent (and more likely to happen first) than the positive infinities, so I think if you worked out the sum you would not manage to get to a positively infinite P. Theres a whole branch of maths devoted to comparing different sized infinities which you could go into if you really wanted to :P.

So yeah, in an infinite number of hands you will certainly have streaks of positive infinity. You just wont have enough of them and they wont be big enough to bring your overall profit to a positive infinity. Youll actually tend to negative infinity pretty quickly.

[Ragnar4]

If you had to take all of the different math concepts and put them on a piece of paper, and draw circles around the math concepts that are manditory to be understood in order to beat the micro's, what concepts would you have circled?

[LawDude]

What percentage of the pot do you generally have to bet to give someone an incorrect price on their flush draw? On their open-ended straight draw?

[spoonitnow]

While I havnt fully thought about this yet, I dont think its necessarily true. For example, think about a 6 sided die. You bet on a 6 and get a 1:1 payout. You bet $1 and you want to know the chance you will have a profit of P = $1.

If H = 1.

You have a 1/6 chance of meeting P. n is therefore 1/6.

If H = 2.

If you lose your first roll you can only ever breakeven. Therefore you have a 5/6 chance of missing with your first roll. Even if your first roll hits you still have a 5/6 chance of losing your second bet and breaking even. The only case where you will have a profit of more than P = $1 after 2 rolls is if you win both rolls. This is a 1/6 * 1/6 chance, so 1/36. Since your expectation is negative with one roll, it becomes more negative the more rolls you make, and the chance n of making a profit P therefore becomes smaller (given that your standard deviation doesnt change in some unfeasably huge way).

Edit: I'm waiting on Michelle to get out of the bathroom and have a few minutes to kill so I'll extrapolate on this. For the die game you described:

When H = 1, n = 1/6
When H = 3, n = 1/6 + (5/6)(1/6)(1/6)
When H = 5, n = 1/6 + (5/6)(1/6)(1/6) + (5/6)(5/6)(1/6)(1/6)(1/6)

Let (H+1)/2 = a to make this a little more simple to write out. Then we have the following equivalent to the above:

When a = 1, n = 1/6
When a = 2, n = 1/6 + (5/6)(1/6)(1/6)
When a = 3, n = 1/6 + (5/6)(1/6)(1/6) + (5/6)(5/6)(1/6)(1/6)(1/6)

And so on. Then the chance of it happening over an infinite number of hands is:



And n quickly converges to about 0.19354838710, or 19.354838710%.

Now back to the main point: your problem is not the same as the original problem referenced for a number of reasons. First, we're not always getting it in each hand. Second, we're not always getting it in as an underdog. The original problem heavily depends on the players and the nature of the game. I chose some really terrible wording in my original answer to the question, but the basic idea is that there is a non-zero chance that you reach any level of profit.

What that exact chance is depends on too many variables that we don't have, but if you had them, the above is exactly how you would figure it out.

[spoonitnow]

If you had to take all of the different math concepts and put them on a piece of paper, and draw circles around the math concepts that are manditory to be understood in order to beat the micro's, what concepts would you have circled?

Implied odds and some level of understanding of the EV of bets.

Edit: I guess I should be slightly more detailed. By "some level of understanding of the EV of bets" I mean that you know what goes into value betting profitably and [semi]bluffing profitably.

[spoonitnow]

What percentage of the pot do you generally have to bet to give someone an incorrect price on their flush draw? On their open-ended straight draw?

It depends on what the play is going to be like after they call, if you have redraws, if some of their outs aren't good, etc.

But for a simple example answer, let's say you're going all-in on the turn against someone who has exactly 9 outs to win and no outs to tie and you've seen their cards which is how you have this information. Also assume no rake. The chance they hit on the river is 9/44 which is 3.89:1 against, so any amount that gave them worse than 3.89:1 against would force their call to be -EV.

If you weren't all-in on the turn, then you could make a bet that gave them worse than 3.89:1 if they always stack you when they hit, and then always stack off on the river too.

[spoonitnow]

I agree

Wow I bet you're not just trying to get posts so you can get into the freerolls.

[LawDude]

I agree

Wow I bet you're not just trying to get posts so you can get into the freerolls.

I don't think you are getting very good pot or implied odds on that bet, spoon.

[JKDS]

spoon do you have a sort of general equation or system of equations that could be used as a substitute for pokerstove? I understand that ps does it better and faster from a practical sense, but i like seeing the actual mathematics going on.

[spoonitnow]

spoon do you have a sort of general equation or system of equations that could be used as a substitute for pokerstove? I understand that ps does it better and faster from a practical sense, but i like seeing the actual mathematics going on.

It can be done exactly, but it's not very fun and involves counting out every board that the losing hand wins on then taking that as a fraction of the total number of boards that can come. Then when you have a range against a range you have to weight the equities against each other and all that. Not very difficult, but very long and tedious and easy to make mistakes.

However, when you're without PokerStove for some reason and a close estimate will do, you can do something like this. Say we have AhAs v 9c9d all-in preflop and we'd want to calculate the equity for each hand. The equity 99 has here is going to be pretty close to just the % of time a 9 hits.

So how often does a 9 hit? It's easier to answer how often does a 9 *not* hit then subtract that from 100%. There are 48 cards left in the deck and 2 nines are in there, so the chance no nines coming on the board is (46/48) * (45/47) * (44/46) * (43/45) * (42/44) = 0.8005 or about 80.05%, so the chance that a nine does come is about 19.95%. PokerStove puts it at 19.806%, so we're close enough.

[JR9477]

spoon do you have a sort of general equation or system of equations that could be used as a substitute for pokerstove? I understand that ps does it better and faster from a practical sense, but i like seeing the actual mathematics going on.

I remember you posting something like this on your website, but it dissapeared when you tore your page down a while ago. I take it you couldn't get it to work well enough?

[spoonitnow]

spoon do you have a sort of general equation or system of equations that could be used as a substitute for pokerstove? I understand that ps does it better and faster from a practical sense, but i like seeing the actual mathematics going on.

I remember you posting something like this on your website, but it dissapeared when you tore your page down a while ago. I take it you couldn't get it to work well enough?

There are a number of ways to approximate equity in your head. After you do so many calculations though you just develop a feel for it. You also tend to memorize a lot of common match-ups too.

[EasyPoker]

Do the likes of Phil Ivey analyse the game to this super-mathematical level?

I'd be interested to know.

[spoonitnow]

Do the likes of Phil Ivey analyse the game to this super-mathematical level?

I'd be interested to know.

While everything in poker can be expressed in terms of math, it doesn't mean that it's a game strictly based on math or learned best by using math exclusively. The math needed to get decent at poker isn't very intensive at all.

[EasyPoker]

Cool, my bad.

[revolvingiris]

Its always at times when great opportunities arise that I have nothing to say or ask. So until I can think of something I just want to say thanks for the offer

[Sasquach991]

Math Question:

If I deposit $600 on FTP, how many hands will it take me to earn my 100% bonus playing at

10NL
25NL
50NL

At FTP you get $0.06 per point and one point per dollar of rake.

This is being discussed here http://www.flopturnriver.com/phpBB2/...it-t91104.html

[Jason]

Yeah, let's get some good math going on FT bonus clearing.

I played about a 1000 hands last night @ $25NL on FullTilt and I believe I have cleared $6. It's only a one day sample, but at that rate, I would clear it in 100 days or 100,000 hands ... hrm, that seems doable but a lot. I don't know if you can say that $50NL will clear it twice as fast as $25NL or $100Nl will clear it 4 times as fast as $25NL - could be more or less. But, if you could, that would be 50,000 hands in 50 days @ $50NL or 25,000 hands in 25 days @ $100NL. Or, said another way, you'd need to play 834 hands per day for 120 days each @ $25NL; 417 hands per day for 120 days each @ $50NL; 208 hands per day for 120 days each @ $100NL.

[spoonitnow]

Math Question:

If I deposit $600 on FTP, how many hands will it take me to earn my 100% bonus playing at

10NL
25NL
50NL

At FTP you get $0.06 per point and one point per dollar of rake.

This is being discussed here http://www.flopturnriver.com/phpBB2/...it-t91104.html

From bonuswhores, how much of it you clear per hour per table:

No Limit $.10/$.25: $0.60/hour at FR, $0.72/hour at 6m
No Limit $.25/$.50: $1.14/hour at FR, $1.29/hour at 6m

I don't know how accurate these figures are. If anyone has a sample of hands for a limit it would take a very simple filter in PT3 or HEM to quickly determine how long they would take, etc.

[spoonitnow]

Yeah, let's get some good math going on FT bonus clearing.

I played about a 1000 hands last night @ $25NL on FullTilt and I believe I have cleared $6. It's only a one day sample, but at that rate, I would clear it in 100 days or 100,000 hands ... hrm, that seems doable but a lot. I don't know if you can say that $50NL will clear it twice as fast as $25NL or $100Nl will clear it 4 times as fast as $25NL - could be more or less. But, if you could, that would be 50,000 hands in 50 days @ $50NL or 25,000 hands in 25 days @ $100NL. Or, said another way, you'd need to play 834 hands per day for 120 days each @ $25NL; 417 hands per day for 120 days each @ $50NL; 208 hands per day for 120 days each @ $100NL.

I think 25nl has a worse rake structure than 50nl, so you'll pay a higher rake. Additionally, the rake will cap more often at 50nl than 25nl (and also more often at 100nl than 50nl, etc.) so that will also affect it.

[Sasquach991]

Math Question:

If I deposit $600 on FTP, how many hands will it take me to earn my 100% bonus playing at

10NL
25NL
50NL

At FTP you get $0.06 per point and one point per dollar of rake.

This is being discussed here http://www.flopturnriver.com/phpBB2/...it-t91104.html

From bonuswhores, how much of it you clear per hour per table:

No Limit $.10/$.25: $0.60/hour at FR, $0.72/hour at 6m
No Limit $.25/$.50: $1.14/hour at FR, $1.29/hour at 6m

I don't know how accurate these figures are. If anyone has a sample of hands for a limit it would take a very simple filter in PT3 or HEM to quickly determine how long they would take, etc.

$/ hour doesn't really help that much as you don't know how many tables/hour the $/hr is based on.

I guess if I can't get half my bonus I'm still doing good.

[spoonitnow]

Math Question:

If I deposit $600 on FTP, how many hands will it take me to earn my 100% bonus playing at

10NL
25NL
50NL

At FTP you get $0.06 per point and one point per dollar of rake.

This is being discussed here http://www.flopturnriver.com/phpBB2/...it-t91104.html

From bonuswhores, how much of it you clear per hour per table:

No Limit $.10/$.25: $0.60/hour at FR, $0.72/hour at 6m
No Limit $.25/$.50: $1.14/hour at FR, $1.29/hour at 6m

I don't know how accurate these figures are. If anyone has a sample of hands for a limit it would take a very simple filter in PT3 or HEM to quickly determine how long they would take, etc.

$/ hour doesn't really help that much as you don't know how many tables/hour the $/hr is based on.

I guess if I can't get half my bonus I'm still doing good.

Wat.

[Sasquach991]

Is it per hour per table?

thx spoon

[spoonitnow]

I had a cute example come up earlier of a math concept that's related to a number of things in poker like multiway bluffing or figuring out the chance of flopping different holdings, and some other stuff. But anyway, here's the example.

Suppose my genius girlfriend has a puppy that she puts on her bed sometimes, and the chance of the puppy pissing on her bed is 30%. Then what is the chance of her bed getting pissed on if she had two puppies on her bed, assuming all puppies have the same 30% chance to piss while they're up there? Here is a professional-grade diagram to help us:



As you can easily see, the chance of both puppies pissing on the bed is 9%, the chance of puppy 1 pissing on the bed and puppy 2 not pissing on the bed is 21%, the chance of puppy 2 pissing on the bed and puppy 1 not pissing on the bed is also 21%, and the chance that neither puppy pisses on the bed is 49%. Now what confuses the shit out of people is that they think the chance will double since there are two puppies, but they're stupid so they don't know why this makes no sense at all. What if there were 4 puppies, then the chance is supposed to be 120%?

The easy way to solve this problem is to realize that having 1 puppy piss on your bed is the same as having 2 puppies piss on your bed since you're going to have to wash your shit either way. For this type of situation it's going to be easier to find the chance that your bed does not get pissed on, then subtract that from 100%. So the chance of puppy #1 not pissing on the bed is 70%, and the chance of puppy #2 not pissing on the bed is 70%, then the chance of both not pissing on the bed is 70% times 70% which is 0.7 * 0.7 = 0.49, so 49%. Then obviously the chance that at least one of them pisses on the bed is 51%.

Okay so poker example. Say we have pocket 8s AKA SNOWMEN or whatever stupid nickname Mike Sexton gave them preflop and we want to know the chance that at least one 8 comes on the flop. It's easier to find the chance that no 8s come on the flop and subtract that from 100%. There are 50 cards left in the deck and 48 aren't 8s, so the chance the first card isn't an 8 is 48/50. Similarly, the chance the second and third cards aren't an 8 are 47/49 and 46/48, so we multiply them together to get 48/50 * 47/49 * 46/48 = 0.8824 or an 88.24% chance that no 8s come on the flop. We subtract that from 100% to get 11.76%, which is the chance of hitting a set or quads.

Alright another poker example. Say we have the powerful hand pocket Yellow Draw 2s in early position with 7 people left to act behind us and we put in a raise to try to take down the blinds. There are 4 tags who will play 12% of hands, 1 supernit who will play 4% of hands, and 2 other people who will play 18% of hands. What is the chance that someone else enters the pot?

Again it's easier to tell what the chance is that everyone folds instead. So the 4 tags fold 88% of the time each, the supernit folds 96% of the time, and the 2 other dumbasses fold 82% of the time each. So the chance that they all fold is 0.88 * 0.88 * 0.88 * 0.88 * 0.96 * 0.82 * 0.82 = 0.387 or 38.7%. Then the chance that someone else enters the pot is 100% - 38.7% = 61.3%.

[drmcboy]

wait, don't be hating on mike sexton, not his fault poker players are lame.

On a scale of 1-10, 1 being Zen of poker or any book by TJ Cloutier ever wrote and 10 being Theory of poker, where does the Mathematics of poker rank? It is the only poker book I have ever failed to read after buying.

[spoonitnow]

wait, don't be hating on mike sexton, not his fault poker players are lame.

On a scale of 1-10, 1 being Zen of poker or any book by TJ Cloutier ever wrote and 10 being Theory of poker, where does the Mathematics of poker rank? It is the only poker book I have ever failed to read after buying.

It's not really a poker book, per se. It's really a math book (think along the lines of a textbook without groups of example problems for each concept presented) that took me a lot of laboring over just to make it through, and I've taken all of the math except 1 class required for a B.S. in Applied Mathematics. I can't say I gained anything from it that has helped my game that stands out.

The main part of the book goes through a number of poker heads-up toy games and finds the unexploitable strategy for each player, which sounds all cool and stuff but it's really just a glorified algebra problem. For example, it shows the process of solving for unexploitable push/fold no-limit hold'em strategies for given stacks. Since I'm bored as shit and can't fall asleep even though it's 3 am, I'll talk about one of the most basic toy games it solves for and some of the consequences of the findings.

The Rules of the Half-Street AKQ Game: Okay imagine a heads-up game with fixed-limit betting and both players post an ante of P/2. The deck has three cards: one Ace, one King, and one Queen. Each player is dealt a card from the deck. Hero is in position, and Villain is forced to check in the dark. Hero has the choice to bet 1 unit or check to showdown. If Hero bets, Villain has the choice to fold or call (he cannot raise). At showdown, the highest card wins.

Side note: A dominated strategy is a strategy that always performs better-than or equal-to all other options. For example, if you're last to act on the river with the nuts, betting dominates checking.

Analysis: Immediately we see some dominated strategies. If Hero is dealt an Ace, he must bet. If Hero is dealt a King, he must check since better never folds (since that means Villain has the nuts) and worse never calls (since that means Villain has the worst hand possible). If Villain has an Ace he always calls, and if he has a Queen he always folds.

All of these dominated options lead us to realize that each player only has one element to their strategy. Hero has to decide how often he bets with a Queen, and Villain has to decide how often he calls with a King. It's a match-up of a bluffing hand vs. bluff catcher.

The starting pot size is P (since we had two antes that were P/2 each) and our bet size is 1. Suppose Hero bets. Now Villain is facing a bet size of 1 with a pot size of P+1, so he should call with a King if he is winning more than 1/((P+1)+1) = 1/(P+2) of the time for it to be profitable in a vacuum. Similarly, if Hero is dealt a Queen, he should bet if Villain is folding more than 1/(P+1) of the time for it to be profitable in a vacuum. Conceptually, it's easy to think of the unexploitable strategies as being then are when Villain folds a King exactly 1/(P+1) of the time and Hero bluffs with a Queen exactly 1/(P+2) of the time, since that is which the opponent will have a break even decision and can gain no edge. But this is a math thread so we must use algebra instead.

To find the unexploitable strategy, we must make our opponent's options have an EV such that he has no single correct play. That is, the best two (or more) of his options all have the same EV. So if Hero bets and Villain is dealt a King, Villain has two options. The EV of Villain folding (from Villain's perspective) is 0. The EV of Villain calling is (% of time Hero has a Queen * (P+1)) + (% of time Hero has an Ace * -1). So we have to figure out when the EV of Villain calling is the same as the EV of Villain folding to know what the unexploitable strategy is for Hero. So we want to know when 0 = (% of time Hero has a Queen * (P+1)) + (% of time Hero has an Ace * -1).

0 = (% of time Hero has a Queen * (P+1)) + (% of time Hero has an Ace * -1)
0 = (% of time Hero has a Queen * (P+1)) - (% of time Hero has an Ace)
(% of time Hero has an Ace) = (% of time Hero has a Queen * (P+1))
(% of time Hero has an Ace)/(% of time Hero has a Queen) = P+1
(% of time Hero has a Queen)/(% of time Hero has an Ace) = 1/(P+1)

Ta da! This gives us the ratio of Queens to Aces that Hero will be betting when he is playing unexploitably. So for example, suppose the pot size is 3 units. This gives us 1/4, which means 20% of our betting range is a Queen and 80% of our betting range is an Ace. Since we're betting Aces 100% of the time we're dealt them, that means that we're only betting a Queen 25% of the time we're dealt them, which is 1/4, etc. We can solve for Villain's calling range in a similar fashion.

Okay so this isn't a very complicated game but we get some important ideas out of it, like how you should generally bet your strongest hands, check your middle hands, and bluff sometimes with your worst hands. If you play it out with someone, you also get to see a very raw example of playing exploitably vs playing unexploitably, which is a fun balance when you're playing against good players. You also see a good poker example of dominated strategies. So now we'd take this game and expand it to a full-street game (which means Villain isn't forced to check in the dark), raise the cap on the fixed-limit betting, or change the betting structure entirely to pot-limit or no-limit or whatever we feel like playing with. We can even add another street of betting where the worst hand will suck out some set % of the time. A few expanded versions of the AKQ game are looked at later in the book.

[siltstrider]

How do I set up the EV equation to find out how often my opponent has to give up (whether c/f, bet/fold, c/c flop c/f turn, etc.) postflop to make it at least EV neutral to flat his 3bet if he's 3betting a given percentage of his hands preflop?

We'll be in position, as I'd like to know this for heads up. I just don't really know how to set it up.

I mean, I know postflop, we have:

EV = p(fold)(pot) - (1 - p(fold))(stack)

But how do we take into account our preflop call? This equation shows us how much we need to profitable when we make the decision on the flop, but I want to know the possible EV of the whole strategy.

I realize there will be several variables, like his 3bet %, he c/f %, b/f%, etc. but assuming we knew those, how would we figure it out?

edit: Just saw that you did something similar to this above. In that case, can you just check this for me to see if I've done it right?

Assuming he is 3betting 10.7%: 88+, ATs+, AJo+, JTs - 43s.

So his range looks like:

Group A: 29.91% pocket pairs
Group B: 44.85% premium broadway
Group C: 22.43% suited connectors

We would go through through all three cases, A, B, C, and for each one go through the list:

1. Probability we both miss the flop and he cannot continue to aggression.
2. Probably he can continue, but I cannot.
3. Probably I outflop him (depends on my cards).

And further, to determine the probability that he misses with, for example, pocket pairs, we go through each pocket pair and find the probability that two overcards come on the flop.

Is this the process I should go through? And then the total EV is just the sum of the EV for each case?

[spoonitnow]

How do I set up the EV equation to find out how often my opponent has to give up (whether c/f, bet/fold, c/c flop c/f turn, etc.) postflop to make it at least EV neutral to flat his 3bet if he's 3betting a given percentage of his hands preflop?

We'll be in position, as I'd like to know this for heads up. I just don't really know how to set it up.

I mean, I know postflop, we have:

EV = p(fold)(pot) - (1 - p(fold))(stack)

But how do we take into account our preflop call? This equation shows us how much we need to profitable when we make the decision on the flop, but I want to know the possible EV of the whole strategy.

I realize there will be several variables, like his 3bet %, he c/f %, b/f%, etc. but assuming we knew those, how would we figure it out?

edit: Just saw that you did something similar to this above. In that case, can you just check this for me to see if I've done it right?

Assuming he is 3betting 10.7%: 88+, ATs+, AJo+, JTs - 43s.

So his range looks like:

Group A: 29.91% pocket pairs
Group B: 44.85% premium broadway
Group C: 22.43% suited connectors

We would go through through all three cases, A, B, C, and for each one go through the list:

1. Probability we both miss the flop and he cannot continue to aggression.
2. Probably he can continue, but I cannot.
3. Probably I outflop him (depends on my cards).

And further, to determine the probability that he misses with, for example, pocket pairs, we go through each pocket pair and find the probability that two overcards come on the flop.

Is this the process I should go through? And then the total EV is just the sum of the EV for each case?

@bold, yes. The total EV is just the sum of the EVs for each case. I'm a bit pressed for time but I'll give you a simple example that should illustrate the concept well enough for you to figure out what you want to on your own.

Suppose we're playing heads-up and open to 3bb in the SB, Villain 3-bets to 10bb, and we have the option to call 7bb. We'll pretend for this example that we don't even know what our cards are, and we're just going to either bet when he checks the flop or raise his flop c-bet.

Now let's say we think (because of how we have broken down his range, etc) that he's going to check about 25% of the time on the flop, and c-bet 14bb the other 75% of the time. Of the times he checks, he will fold to a 14bb bet 50% of the time, and raise all-in the other 50% of the time, to which we will fold. If he c-bets, he will fold to a raise to 35bb 40% of the time and 3-bet all-in 60% of the time, to which we will fold. So now we break down the cases:

Case 1, He check/folds the flop: He checks 25% of the time and folds to our bet 40% of the time, and we profit 13bb from our call (the 10 he put in plus the 3 we put in preflop) so our EV for this case is 25% * 50% * 13 = 1.625bb.
Case 2, He check/raises the flop: He checks 25% of the time and raises the flop 60% of the time, and we profit -7bb preflop plus -14bb on the flop for a total of -21bb. Then our EV for this case is 25% * 50% * -21 = -2.625bb.
Case 3, He bet/folds the flop: He c-bets 75% of the time and folds to a raise 40% of the time. Our profit from this line is the 13bb preflop plus the 14bb he put in on the flop for a total profit of 27bb. So our EV for this outcome is 75% * 40% * 27 = 8.1bb.
Case 4, He bet/raises the flop: He c-bets 75% of the time, reraises 60% of the time, and our profit is -7bb preflop plus -35bb on the flop for a total of -42bb. The EV for this outcome is 75% * 60% * -42 = -18.9bb.

Now for the total EV we add the EV of each possible case/outcome to get 1.625 - 2.625 + 8.1 - 18.9 = -11.8 which will be our average profit for the call if we play according to this game tree postflop. (As an aside, you can note that we're playing pretty shitty postflop as well since we're losing more with this strategy than the 7bb we were calling preflop, which implies we're losing money postflop as well, etc.)

This is a pretty stripped-down example, but I just wanted to show you how to do the EV calculation once you have broken down his range and how he plays it and what your strategy is postflop, etc.

Now if you wanted to set variables for how often he has to take various lines, you can do that also, but it's not that helpful once the game tree gets to have more than 4-5 outcomes or so since you can't break it down into anything that's easy to work with.

To show you what I mean by the equations getting extremely complex as you add to the game tree, I have an example from my blog that is the source of a lot of fun poked at me where I prove a short-cut for solving the fold % needed for a heads-up all-in semibluff to work. Note the complexity even though there are only 3 possible outcomes to the game tree (villain folds, villain calls and we win, villain calls and we lose). Now imagine doing something like this for something with 7-8 possible outcomes.

Suppose there is some pot P and we make some bet B all-in that our opponent has the option to call. Our opponent will fold F % of the time, and against his calling range we will have an equity of E. We would like to know under what conditions this play will be +EV based around how often our opponent folds and what our equity is against his calling range.

To start off, there are three cases. In the first case, our opponent folds and we profit P. In the second case, our opponent calls, we win at showdown, and we profit (P+B) since we pick up the pot as well as the bet he calls. In the third case, our opponent calls, we lose at showdown, and we profit (-B) since we lose our bet. The EV of the first case is F * P. The EV of the second case is (1-F) * E * (P+B). The EV of the third case is (1-F) * (1-E) * (-B). We add the EV of these three cases together to get the total EV of the semi-bluff. Now let’s find for what fold % it’s +EV:



When our opponent’s fold % is greater than that big glob of stuff at the end (which we’ll simplify immensely in a moment), our semi-bluff is +EV. So now we have the task of figuring out how to do something with this.

If we let B – E(P+2B) be an entity of its own (call it x), then we’d have F > x/(x+P). This is exactly like the equation we solved for earlier when looking at a pure bluff! Now let’s figure out what B – E(P+2B) means since it seems like it might be important. If you notice, (P+2B) is the total size of the pot after we bet and Villain calls. Then, E(P+2B) is our equity % of that total pot size. Finally, B – E(P+2B) is our bet size minus our equity % of the total pot size. So let’s see this in an example.

Say it’s 25nl and the pot is $30 and we have $20 behind heads-up. If we shove our $20 and we think we’ll have 20% equity when we’re called, how often does Villain have to fold for it to be +EV? First, we’ll find the total pot size after Villain calls, and that’s $70. Second, we’ll find our equity’s percent of that pot, and 20% of $70 is $14. Now we subtract $14 from our bet size of $20 to get $6, which is our x value. To finish, our Villain’s fold % has to be greater than x/(x+P) for our semi-bluff to be +EV, which is 6/(6+30) here, or 1/6 = 16.7%

[siltstrider]

Wow, that's incredible. Basically, your calculation there means that I can flat a threebet and shove any pair (5 outs to trips/2pair) if I'm getting 3:2 from the pot if my open is 3betting a range that consists of even 20% bullshit, right? Since I pair 1/3 of the time, it really isn't even that important to know what my cards are, I can just say "1/3 of the time, this will be profitable."

I mean, since the specific action of shoving our 2/3 psb is profitable whenever we have a pair against someone who 3bet/cbets too much, then the whole game tree is profitable 1/3 of the time, right?

I wonder if it would be better to balance this with monsters, like KK+, AKs? I think this might also imply that we want to fold easily dominated hands like K2o when 3bet at shortish stacks, as hitting our K will often have us drawing to 3 outs instead of 5 when called.

This should also increase our fold equity after villain sees we can flat AA, KK, AKs, but now I'm wondering if it may be more profitable to just start 4bet bluffing. It seems like it's wasting equity to do that with like KQs, though, and if we're only flatting big broadway, that should become kind of obvious.

[spoonitnow]

If you want to get more sophisticated with this type of thing you can download a piece of software (7 day trial) called flopzilla which breaks down exactly how often you'll flop different holdings based on individual hands or ranges, etc.

[Outlaw]

Hey Spoon, thanks for the thread. I have read it through even though most of what you write is way beyond me.

So for poker players that flunked their way through math in school while getting A's in English/History.. could you list what you feel are the minimum "must know" math concepts related to poker for people crazy bad at math? Maybe going from the simplest to the most complex.

[spoonitnow]

Implied odds and some level of understanding of the EV of bets.

Edit: And I'd say the least useful for microstakes is understanding how to find unexploitable strategies.

[Sasquach991]

Every time I flip a coin and get heads, I get $1 from M2M.
Every time it's tails, M2M gets $1 from me.

Since M2M is such a luckbox he gets tails 50 times in a row.

On the 51st flip, are my odds getting better to get heads or is it still 50/50?

[LawDude]

Every time I flip a coin and get heads, I get $1 from M2M.
Every time it's tails, M2M gets $1 from me.

Since M2M is such a luckbox he gets tails 50 times in a row.

On the 51st flip, are my odds getting better to get heads or is it still 50/50?

50/50, except at some point you have the right to start wondering whether M2M loaded the coin.

[spoonitnow]

Every time I flip a coin and get heads, I get $1 from M2M.
Every time it's tails, M2M gets $1 from me.

Since M2M is such a luckbox he gets tails 50 times in a row.

On the 51st flip, are my odds getting better to get heads or is it still 50/50?

Actually flipping a coin isn't 50/50. The more rotations in the air, the closer it gets to 50/50, but it never quite gets there. Whichever side starts facing up is at a disadvantage, and here's why.

Suppose it starts with heads facing upward. Here is the side facing up once the flip begins:

Tails - Heads - Tails - Heads - Tails - Heads.....

Now after each half-rotation, let's list the % of the "facing up sides" that have been heads:

Tails (0%) Heads (50%) Tails (33%) Heads (50%) Tails (40%) Heads (50%) Tails (43%) Heads (50%) Tails (44%).....

And so on. If you're spinning a coin instead (like if you hold it on it's edge on a table top and thump it and it spins) there's a much larger bias. You can search Google for more information.

[Sasquach991]

Ok. I agree (esp the part about the loaded coin)

Now, what are the odds that M2M will win 50 more times in a row?

The same as the odds that he would win the first 50 times in a row. And what are the odds of that?

[LawDude]

By the way, an easy way to correct for the bias spoon identifies is to place the coin in the starting position opposite to the outcome of the last flip, i.e., if it was tails, start the next flip with heads. If you do this a few times, you are going to move the percentages very close to 50 percent even if you don't put enough rotations on the coin in the air.

[LawDude]

Ok. I agree (esp the part about the loaded coin)

Now, what are the odds that M2M will win 50 more times in a row?

The same as the odds that he would win the first 50 times in a row. And what are the odds of that?

1 in 2 to the 50th power.

[Sasquach991]

~9 x 10E-16

He is a luckbox.

So if past events have no affect on the probability of future events then what are the chances that my parachute will not open twice?

[Robb]

To find the unexploitable strategy, we must make our opponent's options have an EV such that he has no single correct play. That is, the best two (or more) of his options all have the same EV.

NH, spoon. I think I get what you mean: if your "best two (or more) strategies" refers to all active strategies in opponent's optimal solution set. =)

Can't add much except a couple examples on unexploitable strategies. Suppose a pitcher in baseball has a good fastball and curve. But his change-up sucks. He's almost always better off to throw the change-up some small percentage of the time. The possibility of throwing the bad pitch makes it harder for the batter to "sit on" his best two pitches.

This is roughly the mathematical argument for playing a hand like 87s UTG along with an otherwise mega-strong range. A few combos of the "change-up" can really cross up our opponents who expect premium hands.

There's also a really cool example of Evolutionary Stable Strategies, where a species develop "fight-or-flight" patterns. I don't have time right now to get into it, but I'll try to post later. It shows how a certain strategy is "good enough to play" when only 2 strategies are considered, exits when a third strategy is introduced that dominates it, and becomes active again when a fourth strategy joins the mix.

I guess my point is that game theory definitely gives lots of credibility to shania theory. And explains why a lot of styles from TAG to LAG can be profitable.

Here's two poker math questions for you. Suppose Hero plays Heads Up SnG's and has a probability of winning of 55%.

1. What is the probability he wins at least 6 matches out of ten?
2. What is the probability he wins least 6 before losing 4?

[spoonitnow]

To find the unexploitable strategy, we must make our opponent's options have an EV such that he has no single correct play. That is, the best two (or more) of his options all have the same EV.

NH, spoon. I think I get what you mean: if your "best two (or more) strategies" refers to all active strategies in opponent's optimal solution set. =)

Yeah, I was trying to explain it in a way that would make sense to someone who has no idea what I'm talking about. If he has two options then we make them both equal in EV, if he has three options then we make his best two (possibly all three?) performing options have the same EV, etc.

Here's two poker math questions for you. Suppose Hero plays Heads Up SnG's and has a probability of winning of 55%.

1. What is the probability he wins at least 6 matches out of ten?
2. What is the probability he wins least 6 before losing 4?

I hate this type of thing since I can never remember the nice compact simple ways to do it.

#1: The chance of winning exactly 6 would be (0.55)^6 * (0.45)^4 * 10!/(6!4!) I think. So then the chance of winning n or more would be the sum of (0.55)^k * (0.45)^(10-k) * 10!/(k!(10-k)!) where k = n to 10. Since that doesn't mean a lot to 95%+ of the people reading this, I'll explain the process of finding the chance of winning exactly 6. Also you're an asshole because now I miss being in school.

The chance of winning once is 0.55 (55%) and the chance of losing once is 0.45 (45%). So the chance of there being a string of six wins and four losses is (0.55) * (0.55) * (0.55) * (0.55) * (0.55) * (0.55) * (0.45) * (0.45) * (0.45) * (0.45). But there are a lot of different ways you can win 6 and lose 4 inside a sample of 10 games. For example, it could look like WWWWWWLLLL or WLWWWWWLLL or LLWWWWLLWW and so on. So basically we're wanting to know how many different ways we can arrange six W's (wins) and four L's (losses).

For some insight on how this works, imagine instead what if we were trying to find different ways to arrange the letters ABCD. There are 4 ways to place the first letter, then 3 ways to place the second letter, 2 ways to place the third letter, and 1 way to place the fourth letter. Then there are 4 * 3 * 2 * 1 = 24 different orders we can write the four letters.

Now suppose we change these letters to ABCC. There are still 24 different orders we can write the four letters, but two of the letters are the same, so we have to adjust. Suppose in any way we right the letters there are 2 possible C's that can be written first and 1 possible C that can be written second, which means there are 2 * 1 = 2 total possible ways to write the C's in each arrangement. Then we divide 24 by 2 to get 12 possible ways to write the letters ABCC.

But what if it's ABBB? There are 24 ways to write four letters, but now three of them are the same. In any one arrangement of the letters, there are 3 ways to write the first B, 2 ways to write the second B and 1 way to write the third B, so there are 3 * 2 * 1 = 6 ways to write the B's within each arrangement. That means there will be six different times within those 24 arrangements that we see ABBB, BABB, BBAB, and BBBA each. So logically we think 24 / 6 = 4, which is rather obviously the real total number of possible arrangements for the letters ABBB.

So now we get back around to how to arrange the letters WWWWWWLLLL. The total number of ways to write ten letters is 10 * 9 * ... * 2 * 1, which we write in short-hand as 10! (meaning 10 factorial). There are six W's and four L's, so we'll divide 10! by 6! and 4! to eliminate the duplicate arrangements, so the total number of ways to arrange those 10 letters are 10!/(6!4!), whatever the hell number that comes out to be.

Since we know the chance of getting a specific line of 6 wins and 4 losses is (0.55)^6 * (0.45)^4 from earlier, and we know there are 10!/(6!4!) orders those 6 wins and 4 losses can go in, we have the final answer which is (0.55)^6 * (0.45)^4 * 10!/(6!4!).

Holy shit that's long.

2. What is the probability he wins least 6 before losing 4?

The chance of winning the first 6 + chance of winning 6 of the first 7 + chance of winning 6 of the first 8 + chance of winning 6 of the first 9 as per above. LDO. I need to get away from this shit before I type a novel.

[linaker]

So if past events have no affect on the probability of future events then what are the chances that my parachute will not open twice?

Past events have no effect on future events, if the events are independent. Parachutes failing to open are not independent events, because if that event occurs, then you are highly unlikely to get the chance to open a parachute again (either with success or failure) .

Funnily enough, your parachute opening successfully is an independent event. You are no more likely to plummet to your death on the 1000th jump than the first.

[celtic123]

[You are no more likely to plummet to your death on the 1000th jump than the first.

Thats intresting, I was told by a health and safety officer, that I was allowed to work off ladders, so Long as I am on the ladders for no more than 20 minutes per hour.

She said , statistically , I was less likely to fall off than someone who was on a ladder for a whole hour.


So isnt the 1,000 jumper more at risk than the 1 timer ?

[Robb]

[You are no more likely to plummet to your death on the 1000th jump than the first.

Thats intresting, I was told by a health and safety officer, that I was allowed to work off ladders, so Long as I am on the ladders for no more than 20 minutes per hour.

She said , statistically , I was less likely to fall off than someone who was on a ladder for a whole hour.


So isnt the 1,000 jumper more at risk than the 1 timer ?

A thousand one-time jumpers would have the same risk-level as 1 thousand-time jumper, provided he lived long enough to risk all 1k jumps. Given the same level of training, expertise, equipment, weather conditions, etc.

[spoonitnow]

Thats intresting, I was told by a health and safety officer, that I was allowed to work off ladders, so Long as I am on the ladders for no more than 20 minutes per hour.

She said , statistically , I was less likely to fall off than someone who was on a ladder for a whole hour.

And statistically you're more likely to fall than someone who never gets on a ladder.

2. What is the probability he wins least 6 before losing 4?

The chance of winning the first 6 + chance of winning 6 of the first 7 + chance of winning 6 of the first 8 + chance of winning 6 of the first 9 as per above. LDO. I need to get away from this shit before I type a novel.

It hit me out of the blue in the car earlier that this isn't right. The chance of winning 6 of the first 7 includes the chance of winning the first 6, etc. So it should just be the chance of winning 6 of the first 9 + chance of winning 7 of the first 9 + chance of winning 8 of the first 9 + chance of winning 9 of the first 9 + (chance of winning 5 of the first 9 + winning the 10th). I think. There's probably a simpler way to do this, but a lot of things (especially in statistics) I've just learned how to figure out without necessarily knowing the most efficient way to do it. I've never taken a class on statistics, though I've basically taught a few low-level statistics classes, which I find hilarious.

[Robb]

Evolutionarily Stable Strategies (ESS's) are my favorite game theory example. Suppose we have a population of a certain species, part "hawks" and part "doves." Whenever a conflict over scarce resources occurs, the hawk will fight, the dove will flee. Say the resource is worth 50 "survival of the fittest points." When two hawks fight, someone is wounded while the other (less seriously wounded) wins the resources. So the population gains 50 SFP's but loses 100 SFP's from battle wounds. In hawk vs. dove, the hawk wins, and no one is injured, so the population gains +50 without loss. In dove vs. dove, there is lots of dancing until one gets bored and leaves so +50, and both are healthy but they wasted tons of time, so each loses -10. We make a matrix of the results for the population remembering that the resources are allocated to the population, so any one member's life or death or maiming is irrelevant. Sad and tragic, perhaps, but irrelevant to the process of evolution.

- -- - Hawk - Dove
Hawk: -25 - +50
Dove: - 0 - - +15

In the matrix above, the "scores" are given in terms of the ROW player, so the ROW hawk wins +50 vs. the COLUMN dove.

It's pretty obvious that you want to play a Hawk strategy against doves, but it's a horrible coin flip against another Hawk. Sort of like losing not only this MTT but the next as well when your AK < 88. Doves do better by not killing each other, just wasting time.

To analyze the game, we have to get back to some evolutionary theory for a second. We'll assume that a genetic mutation introduces a new behavior into the species, say, a "hawk" mutation invades a population of doves. Then we can think of the ROW player as the FOCAL player playing against the population. The focal player should randomly select a strategy (hawk or dove) based on a correct percentage, and the population should adjust its play accordingly.

We assume the FOCAL player faces x proportion of hawks and 1-x doves, and do a poker EV calculation:

FOCAL plays Hawk: -25x + 50 ( 1 - x ) = 50 - 75x
FOCAL plays Dove: 0x + 15 ( 1 - x ) = 15 - 15x

Like Spoon said above, the optimal mixed strategy is to choose x so that the EV's are equal. This prevents the FOCAL Player from being able to gain survival points by shifting his randomizing percentage. So we set the expression above equal to each other:

50 - 75x = 15 - 15x
35 = 60x
x = 7/12

This means the population should "play" hawk 7/12 and dove 5/12 of the time to optimize its chances of survival.

OK, so that's all fine and good. Some are lovers. Some are fighters. And they can coexist happily, it appears.

Suppose a "bully" mutation joins the population. He acts like he's going to fight, so he scares off doves, but he flees when actually hit, so he loses to Hawks. Against another bully, one of the bullies gets scared first and flees, so the other wins the +50 points without a fight. The new matrix looks like this:

- -- - Hawk - Dove - Bully
Hawk: -25 - +50- - +50
Dove: - 0 - - +15- - -- 0
Bully: - 0 - - +50- -- +25

Now, remember the payoffs are based on the ROW player, so the COLUMN player or population are actually minimizing their outcome while the ROW or FOCAL is maximizing. So looking at the Dove strategy, well, it's sucks.

The dove strategy is an example of a dominated strategy. It is always better to be a bully than a dove, because the bully loses to the hawk without injury but wins against doves and half the times against other bullies. So the doves die out. You can redo the EV calculations and equalize them yourself. You do so with the new 2 x 2 payoff matrix with the dove row and dove column eliminated. The result is half hawks, half bullies.

Now here's the fun part. Retaliators. These cool little critters will act like doves until attacked, then fight back like a hawk, ready to kill or be killed rather than back down. The Retaliator earns a hawk's points vs. hawks and doves points vs. doves. Against bullies, however, they win 50 points. The bully postures long enough that the retaliator thinks he's in a fight, so he fights back. The bully runs away. The new matrix looks like this:

- -- - Hawk - Dove - Bully - Retal
Hawk: -25 - +50- - +50 - - -25
Dove: - 0 - - +15- - - 0 - -- +15
Bully: - 0 - - +50- -- +25 -- - 0
Retal: -25 - +15- - - +50 -- +15

Now, the Retaliators rule the day, beating up bullies, holding their own against hawks, and stealing from doves. They treat each other like dove vs. dove. So Retaliators are a Pure Strategy ESS. But they can live peacefully with doves. So a Dove-Retaliator population could live equally well, as long as the the Doves were 30% or less. Once the Doves are more than 30% of the population, Bullies can invade.

The interesting thing about the Retaliator vs. Dove population is that all confrontation is symbolic. Which explains a lot about why birds puff out their feathers, rams lock horns, and wolves have a sort of "tap out" procedure to keep their alpha male contests from killing off an otherwise stout member of their tribe.

If anyone's still reading, thanks. I just like this game and it's actually the type of analysis used in a lot of mathematical biology papers right now. It also is used in anthropology and sociology to help explain the evolution of superstitions, taboos, and other human culturalisms.

[celtic123]

I think I get from this, Aggression at the table should win.

[Robb]

I think I get from this, Aggression at the table should win.

Well, ESS's don't exactly apply to poker. However, it does explain something that confuses some beginners. Should I open pp's UTG in 6m games? The answer is yes for (probably) 25nl and certainly 10nl and below. However, somewhere around 50nl or 100nl, life changes. The 3betting and squeezing becomes problematic. Many players open-fold pp's UTG, or play only a small, randomly selected percentage of them. But somewhere around 200nl, it seems, a lot of the regs toss the pp's back into their UTG ranges, because the 3betting and 4betting dynamics are very different.

I guess my point is that ESS's demonstrate that no strategy (in poker or anywhere else) is optimal in a vacuum. Optimal strategies depend on your other alternatives and your opponents' likely reactions.

[spoonitnow]

Found this in an old blog entry of mine. It was my reply to someone who asked what a general outline would be of what poker math I think people should learn:

* Using PokerStove to develop the ability to estimate your equity against your opponent’s range. This is a very crucial skill that there is no way to get around needing to learn.
* Calling All-Ins: You absolutely must know how to quickly and effectively calculate if you can make a call that will close the action on the hand (whether it’s calling an all-in or just calling a river bet as last to act). This is so important that I wrote a script to help people practice it.
* The basic poker math behind value betting and bluffing. It’s not very hard to learn, but takes a little practice.
* Ideas in pot odds and implied odds behind playing draws when there will be money put in on later streets. Can be found in almost any 2+2 book.

So I guess I'll hit the highlights on a few of these. I've got a thread somewhere where I talk about calling all-ins and another where I talk about pot odds and implied odds when playing draws, but the value betting and bluffing thing eludes some people. Now this isn't the whole story, but it's a decent enough way to think about it for now while a new player develops their understanding of what's going on when they bet.

Okay so bluffing and semibluffing in a nutshell: If you have 0% equity in the pot (which rarely happens in holdem) and you make a bet into some pot, then if your opponent folds more than bet/(bet+pot) then your bluff is profitable. So if the pot is $10 and you bet $6 and they fold more than 6/16ths of the time, then your bluff is profitable. The more equity you have (semibluffing etc) the lower their fold % needs to be for a bet to be +EV.

Now here's value betting in a nutshell: if you beat most of what your opponent matches your bet with heads-up, you're probably +EV. This isn't strictly true if you can be raised off of your hand, but if you're going all-in or there's no chance you'll be raised off of your hand then it's a good guide. So if while heads-up I shove some amount into some pot on some street in some hand on some game and I have 60% equity against his calling range, my value bet is +EV.

Keep in mind this isn't all there is to it, but if you get these ideas you're on your way.