Maniac Math

I've seen some threads in the BC lately that have me concerned. Someone will post a HH with villain's stats something like 50/30 over 30 hands but will discount that - could be just a TAGG on heater.

Not very likely. Using the binomial theorem, we can calculate the probability that a TAGG with 25/15 stats raises 30% of the time (or more) over 30 hands: about 2.75%. Over 50 hands, there's about 0.5% chance an actual 25/15 guy will raise 15 times.

My point is that when you see a couple dozen hands from a guy who's seein' the flop half the time, and raising a good bit, you don't need 1k HH's to assume he's pretty loose. It's time to mix up with him a bit before he donates his stack to someone else.

If anyone's interested in the math, I'll post something more detailed tomorrow.

[x] interested

I smell some standard deviations a-comin'.

Very interested in this. Run into this type of situation quite a bit.

http://en.wikipedia.org/wiki/Binomial_theorem

???

you mean bayes theorem?

Originally Posted by JKDS

http://en.wikipedia.org/wiki/Binomial_theorem

???

you mean bayes theorem?

Yeah, a lot of folks use Bayes Theorem here and treat this as conditional probability, but we can make a simple assumption and just compute the probability directly. The Binomial Theorem works for repeatedly "coin flip" situations, whether the coin is fair or not.

Suppose we flip a coin with a 20% probability of heads. Let x = .2 (probability of success) and y = 1 - .2 = .8. Now, this may seem trivial at first, but note that x + y = 1. So (x+y)^5 = 1 for instance. This happy fact allows us to simply expand the binomial power and calculate the probabilities we want. If you remember Pascal's Triangle in high school algebra, it's the simplified version of the Binomial Theorem. And the process works for any probability simulation we can create with fixed chance of success.

So, imagine a TAGG opening in the CO with a known chance of success (open-raising) of x = .2. And imagine that we see him in this situation 5 times (after learning the iron-clad fact he open-raises 20% in the CO). What is the probability he open-raises 4 times or more out of 5 trials?

The probability of 5 successes is a straightforward (.2)^5.

The probability of 4 successes is a bit more interesting:

(5 choose 4)(0.2^4)(0.8)

The 4 successes each have probability 0.2, the failure has probability 0.8. But there are 5 ways to choose one slot out of 5 for the failure to occur (or 5 choose 4 ways to choose the 4 slots for the successes to occur):

F S S S S
S F S S S
S S F S S
S S S F S
S S S S F

Recalling Pascal's Triangle:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

The bottom row allows us to expand the binomial (x+y)^5 by telling us the coefficients in the pattern where powers of x ascend from 0 to 5 and powers of y descend from 5 to 0:

(x + y)^5 = y^5 + 5 x y^4 + 10 x^2 y^3 + 10 x^3 y^2 + 5 x^4 y + x^5.

Again, recall that this value is just = 1, since x + y = 1. In the expansion, the pattern relates to the probabilities:

y^5 term - probability of 0 successes, 5 failures
x y^4 term - probability of 1 success, 4 failures
x^2 y^3 term - probability of 2 successes, 3 failures, and so on


Example: For a fixed chance of success x = .2, what is the probability of 5 successes in 8 trials?

P(X = 5) = (8 choose 5)(0.2^5)(0.8^3)

The "8 choose 5" simply calculates all the possible ways to permute the 5 successes into 8 slots.

Awright, given all that introduction, the setup for my post is simple.

Hypothesis: let's assume we have a TAGG reg who plays <25/15 over some reasonable sample of hands. We can think of his PFR = .15 as a chance of success, so x = .15. Then the probability of this event occurring 15 times OR MORE out of 50 trials is the sum:

P(X >= 15) = (50 choose 15) (.15^15)(.85^35) + (50 choose 16) (.15^16)(.85^34) + (50 choose 17) (.15^17)(.85^33) + ...

I did the sum in an Excel spreadsheet. There's an extension of the binomial theorem called the multinomial theorem which would allow us to take VPiP into account simultaneously with PFR, which I may get into later if you're still interested after all that.

Just to check the viability of this math, you can see the same calculations for other probabilities.

For x = .2, there is about a 6% of him open-raising 15 times (or more) out 50.

For x = .25, there is about a 25% of him open-raising 15 times (or more) out 50.

For x = .3, there is about a 55% of him open-raising 15 times (or more) out 50.

For x = .4, there is about a 94% of him open-raising 15 times (or more) out 50.

For x = .5, there is more than a 99% of him open-raising 15 times (or more) out 50.

Let me know what you think. If you have more questions, I can try to answer them.

After studying this theorem closely 1-2 months ago I can say my win rate has improved and my awareness has increased significantly.

I for one am prepared to accept your conclusion.

Originally Posted by bjsaust

I for one am prepared to accept your conclusion.

lol same

I didn't quite understand the point of all the gibberish at first, but I just went and looked it over again, and I think I understand its usefulness now. A guy starting out 30/20 over 25 hands probably isnt going to be a 10/8 on a heater all that often, etc.

I don't know if thats the whole scope of the post, and maybe I'm missing the actual point of the thread, but thats what I got out of it. :/

Originally Posted by dranger7070

Originally Posted by bjsaust

I for one am prepared to accept your conclusion.

lol same

I didn't quite understand the point of all the gibberish at first, but I just went and looked it over again, and I think I understand its usefulness now. A guy starting out 30/20 over 25 hands probably isnt going to be a 10/8 on a heater all that often, etc.

I don't know if thats the whole scope of the post, and maybe I'm missing the actual point of the thread, but thats what I got out of it. :/

That was it - some folks wanted to be convinced, I guess.

25 hands of 60/25 is enough to be pretty doggone certain he's no TAGG. If you've got that, you're gtg.

Sweet, good post Robb. I've always known that to be the case, but makes you feel a lil better seeing the maffs that make it true.

moral of the story: low sample size (30hands) stats can usually (but not always) be interpreted to be roughly the same as bigger sample size stats. for extra points use showdown reads

Unless you're playing somebody who's randomizing their play. Then again, if you do, you're probably playing at a level where you don't need this forum's help.

Originally Posted by StillDeadMoney

Unless you're playing somebody who's randomizing their play. Then again, if you do, you're probably playing at a level where you don't need this forum's help.

A lot TAGG-res randomize lots of things about their play but would rarely have 60/35 stats over 50 HH's. I guess I don't really get what you're sayin', here.

Originally Posted by Robb

25 hands of 60/25 is enough to be pretty doggone certain he's no TAGG. If you've got that, you're gtg.

Thanks for this valuable info Robb. I don't know if a similar math can be applied for a TAGG. How much of a sample do we need in order to accept someone TAGG?

Originally Posted by Extremophile

Originally Posted by Robb

25 hands of 60/25 is enough to be pretty doggone certain he's no TAGG. If you've got that, you're gtg.

Thanks for this valuable info Robb. I don't know if a similar math can be applied for a TAGG. How much of a sample do we need in order to accept someone TAGG?

The problem w/ TAGG is that we're looking to narrow down the vpip to, say, 15 - 25. With LAGG's/maniacs, we have this huge range of vpip's > 35 or so.

But, usually ~60-75 HH's is enough to guess someone's got TAGG tendencies. You'll be surprised every now and then if a nit gets on heater or LAGG goes card dead.

But only the preflop stats are accurate early. Flop stats and things like AF take a lot longer to be reliable.

Robb, what function in excel did you use to sum

P(X >= 15) = (50 choose 15) (.15^15)(.85^35) + (50 choose 16) (.15^16)(.85^34) + (50 choose 17) (.15^17)(.85^33) + ...

wow thanks for bumping

yeah no prob still trying to figure out that excel shit tho

Originally Posted by JoeHaw

Robb, what function in excel did you use to sum

OK, the binomial coefficient "4 choose 2" gives the total number of combinations possible when selecting, say, 2 Aces from a total of four.

Excel has the formula:

= combin(1st,2nd)

where 1st is the number of things being chosen from, and 2nd is the number of things being chosen. I used the binomial theorem to determine the probability of a unfair coin flip situation. Here's how it works with another, simpler example. Suppose I flip a coin 10 times, but it has a 60% chance of Heads.

Then, the probability of 10 Heads in 10 trials:

P(X=10) = .6^10

The probability of 9 Heads in 10 trials:

P(X=9) = (10 choose 9) * .6^9 * .4 ^1

This can be thought of as lining up 10 slots. The "10 choose 9" selects a space for the Tails (and the other nine for the Heads). The ".6^9" is the probability of Heads occuring raised the 9th power since it happened 9 times, and same for Tails.

You'll see the pattern if I list the next few:

P(X=8) = (10 choose 8) * .6^8 * .4 ^2
P(X=7) = (10 choose 7) * .6^7 * .4 ^3
P(X=6) = (10 choose 6) * .6^6 * .4 ^4 ...

Now, the cool thing here is that these probabilities are EXACT, not estimates. So we can ask the question: "What is the probability that this coin is flipped 10 times, and we see AT LEAST 6 Heads?"

To answer, we simply create a simple spreadsheet that indexes through the 11 outcomes, calculating the probability for each one, then sum up the five in question. Knowing that everywhere I wrote (n choose k), I just used the Excel formula "=combin(n,k)" makes it straightforward, if you're used to Excel.

If not, I created a spreadsheet in Google Docs which demonstrates the above example, and I will post the screenshot later today, with the formula visible. Google docs works in this instance exactly like Excel.

Here's the spreadsheet:



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