So Ive been toying a bit around today with this concept of card removal and how it is effecting opening ranges
My startion question basically was: How much more frequent are you gonna get the opportunity to open up AK on the button (suits don't matter for now) opposed to 32 for example so that's assuming everyone folded to you
Obv this is a very complex thing to calculate, and i'm not really interested in exact numbers, so i've been simplifying things quite heavily, and was wondering if this was (again overly simplified) a somewhat decent/valid first estimation
The first thing I did, was predefine some standard opening ranges for every position in front of the button, so for example UTG & UTG1 15%, MP 18% and so on
Next thing I did was calculating the chance of a player folding a certain card, so let's start with the Ace in a 15% UTG range (for example 55+, A8s+, A5s-A2s, K9s+, QTs+, JTs, T9s, 98s, ATo+, KJo+)
So that means the UTG player is folding 104 combos that consist an ace, namely A7s-A6s, A9o-A2o
So the chance that a UTG player is folding an ace is then 104/1326 combos = 0.078431 aka roughly 7.8%
So far I don't think I've made any mistakes, but now things get more complicated obviously. In reality once the UTG player folds, it should effect the probability of UTG1 folding a certain card as well (since we know what the UTG player is opening)
Anyway for the sake of simplicity, so far I haven't try to work this out, since it would take me too far for what I'm only trying to accomplish. So basically I did this for every position up till the button which gives this table
The next thing I did was to take the sum of each probability which gives us plus minus the chance that either UTG, UTG1, MP, LJ, HIJ or CO folds a certain card (I believe statistically you should also take into account the intersection but again simplifying things here)
So now as you can see, there is a +- 43% chance that an ace is folded in front of you and about a 57% chance that a king is folded in front of you
So now I've applied another very simplified logic to try to measure relative differences
Usually you have a 16/1326 = 1.2% chance of getting dealt AK
As I'm sure you are aware, if you fully remove one card, that chance is being reduced to 12/1326 or 0.9%
So given that information I computed the following calculation: 16 - (4*0.43) - (4*0.57) = +- 12 combos of AK left
If you want to calculate the chance of A2 you get 16 - (4*0.43) - (4*0.87) = +- 10.75 combos of A2 left
So relatively speaking, if you were to attach weights you could (very simplified!) say that the chance of getting to raise first in with A2 on the button compared to AK is about 10.75/12 = 89.5%
For pairs you can apply the same logic, only replace the 4 by 3 now, since the full removal of one card takes away 3 out of 6 combos so then you get AA = 6 - (0.43*3) = 4.71 combos and for deuces = 6 - (0.87*3) = 3.39 combos hence a relative difference of 3.39/4.71 = +- 72%
Any feedback or comments would be really appreciated, since it's a long time since I took my statistics class, and can def have made some thinking errors on the way
Edit: So let's say if you were to construct a BTN RFI range you could assign the plus minus following weights, making AA the benchmark 100% weight
AK = 95%
A2 = 85%
22 = 70%
Last edited by LittleGoliath; 05-12-2018 at 03:29 PM.