Guys, could you tell me if this thinking is correct?

NL Hold'em

We've got a 15/10 guy (VPIP/PFR)

Let's call his PFR% = Opening% + 3-bet% [Is this how PT calculates this stat?]

Let's assume because this player is so tight, his 4-betting% is negligible and we'll ignore it.

Let's say we know his 3-bet range is QQ+, AK. Using Pokerstove we find that's the top 2.6% of hands.

Using the above assumptions, can we say 10% total PFR - 2.6% he 3-bets should equal 7.4% which is his opening range (???)

If that's the case, we can use Pokerstove to determine a rough opening range by finding the top 7.4% of hands.

Is that right?

I think it's getting a little hairy because we're trying to apply a frequency (preflop raise %) to a range % (top X% of a total # of possible hands). I'm afraid I might be double-counting a hand like QQ or the calculation isn't this simple since we can never 3-bet when we're first to act.

Please point me in the right direction. Thanks!

The problem is also that the range varies by position with this kind of player, normally from very tight (UTG) 3-4 % pfr to quite loose (BUTTON) 15% pfr.

That means that you have to find a way to fit your average range into positional ranges to make any good use of it.

what if for the sake of simplicity we simply ignore position and consider 10% pfr being the overall average and therefore finding the average range? (basically that's true, right?)

Or said another way, what if we assume that from UTG he's very tight, to the button being very loose, so at exactly MP he's 15/10, therefore we try to find his range given the 15/10 stat from MP?

If we look at his 15/10 stat being exactly from MP, would the logic in the OP then hold true?

$50 on stars, full tilt, or UB to anyone can give me a definitive answer that seems to make logical sense to me and my friend

You're the one w/ the Mathematics of NL presentation, champ.

Yea that's why i want to make sure it's correct

I don't think you can figure this out without knowing how often he 3-bets. Without a raise ahead, a 15/10 is going to open 10% of his hands. With a raise ahead, he's 3-betting 2.6% of his hands, but he's only opening that 2.6% {3-bet percentage} of the time.

So he's opening 10*(1 - 3-bet%)% of hands when he's not 3-betting (I think).

PT says it calculates PFR using (pre-flop raises / total hands), FWIW.

This is closer, but not correct eighter.

The problem with the logic in OP is that opening opportunities and 3-betting opportunities occur in totally different situations, i.e. either it's been raised before hero, and then he can 3-bet, or it has not, and then he can open raise.

This means that PFR is not the added probabilities of him doing one or another, since they don't apply to the same situation. This is easy to see with an example: Say hero openraises 100% and 3-bets 30%. Using this method, we would believe his PFR to be 130%...

Since PFR is the frequency of raising in all situations, it has to be a weighted mean of his frequency in each specific situation. The distinguishable situations are (1) no one has raised, (2) it has been raised, (3) it has been 3-bet, and so on.

In order to make the connection between PFR and, for example, 3-betting ratio, we need the likelihood of each scenario. The connection looks like this:

PFR = P(no raise)*R(1) + P(raise)*R(2) + P(3-bet)*R(3) + [...] + P(n-bet)*R(n)

where

P(no raise) = probability that none of the villains has made any raise
P(raise) = probability that it has been raised once
P(3-bet).. and so on, you get it..
R(n) = The frequency by which hero puts in the n:th raise when he has the chance to do so. For example R(2) is his 3-betting frequency (note that the second raise is what we call a 3-bet).

So, to use your assumption that raises beyond 3-bets are negligible, our formula looks like this:

PFR = P(no raise)*R(1) + P(raise)*R(2)

So this is unfortunately bad news. In order to find anything out, you have to do some modeling on how his opponents have been playing. Opponents' playing style affects PFR.

Couldn't you also naively assume that P(raise) is 3-bet% and P(no raise) is (1 - 3-bet%). Using the logic: if someone is raising, they are either open raising or 3-betting, so if they're not 3-betting, they must be open raising.

I don't see why we would assume P(raise) to equal R(2). Just to check whether you got the variables right:

P(no raise) = Probability that it has not been raised when it's hero's turn to act
P(raise) = Probability that it has been raised (only one raise) when it's hero's turn to act. In this case, when we don't account for 3-bets before it's heros turn to act, P(raise) = 1-P(no raise).
R(1) = The frequency by which hero raises when it has not been raised before him
R(2) = The frequency by which hero 3-bets when it has been raised once before him

Also, in this case, P(no raise) = 1-P(raise). This means that we can simplify the formula to:

PFR = P(raise) * R(2) + (1-P(raise)) * R(1)

Inserting the known values, we get:

10% = P(raise) * 2.6% + (1-P(raise)) * R(1)
R(1) = (10% - P(raise) * 2.6%) / (1 - P(raise))

And here are the outcomes for some values of P(raise):

P(raise) = 10% -> 10.82%
P(raise) = 25% -> 12.47%
P(raise) = 50% -> 17.40%

yeah, in other words, the PFR statisctic is meaningless. The real numbers you'd like to see is the proba that the guy raises knowing that it has been folded (or limped) to him. I fear no one will be able to claim for the 50$

To keep in simple, i'd just assume yeah he'd 3bet with the top 2.6% of his range if theres a raise in front. Typical good players open like 10-13% in ep, so he's got a pfr of X where .13*.026 + .87*X = .1

so his opening range would be about 11.1% of his hands. I think this effect is pretty marginal/ unimportant to be honest in early position.

stealthcow-

No, this is not right, even given your assumption. The problem is, you're using the assumption, that people generally raise 10%, to say that the probability that someone raises is 10%. That's wrong, since there are more players before our hero, of which each have a 10% probability of raising.

In fact, given the assumption that every player on all the different positions raise 10% of their hands, the probability that it will be raised before it's hero's turn to act is

1 - 0.9^n

where n is the number of players to act before hero. This is because 0.9^n is the probability that there will be no raise.

So plugging that into my formula

PFR = P(raise) * R(2) + (1-P(raise)) * R(1)

we get

PFR = (1 - 0.9^n) * R(2) + 0.9^n * R(1)

And assuming there are 3 players before our hero, and plugging in the known values again we get:

10% = (1 - 0.9^3) * 2.6% + 0.9^3 * R(1)
10% - (1 - 0.9^3) * 2.6% = 0.9^3 * R(1)
R(1) = (10% - (1 - 0.9^3) * 2.6%) / 0.9^3
R(1) ~ 12.8%

So, what does this mean? If our hero has PFR 10%, and 3-bets the top 2.6% hands, and the 3 opponents that act before him each raise 10% of their hands, then hero open-raises 12.8% of his hands (i.e. raises 12.8% of the times when there is no raise before him).

The formula you used was actually the same as the one I've been using, but you plugged the wrong numbers into it.

Edit: I see now that you used 13% as your assumption of the villains' raising frequencies. Well, that does not matter, since the overall error in your post is still the same.

sorry, i just assumed this was 6max, cause thats all i play.

you'd probably have to change my assumption that most players open 10-13% in ep as well then.

stealthcow-

That was not the point. I'm sure the 10-13% assumption is good. What I meant is that if one player raises with probability P%, then any of n similar players raise with 1 - (1 - P)^n % probability. You just used the P% figure, as if that applied to several players just as well as to just one.