Pairs behind you (mathematics)

Red_Diamond · #1 10-04-2007, 10:53 PM

One rule of thumb I remember from Chip was when starting with a pair, always look at how many up cards are BEHIND you. The magic number was 2 or more = DON'T RAISE.

I also read in other literature that this is a solid rule to play by.

However, what I have never seen, is even a simple mathematical model to explain why the cut-off number is 2. I had assumed back in the day that it was a simple rule of thumb, but by now I'm positive there must be some sort of mathematics behind it.

I just have never seen it.

RustyBrooks · #2 10-04-2007, 11:15 PM

I'm not sure what you mean... Is this seven card stud? Razz? hi/lo? What do you mean by "behind you", between you and the bringin? Up cards smaller than your pair?

Red_Diamond · #3 10-04-2007, 11:20 PM

Oops,I could have explained better.

On third street, look at how many up cards behind you are HIGHER than the pair you have. If there is TWO or MORE, then the rule is not to trap yourself.

RustyBrooks · #4 10-05-2007, 12:41 AM

Does anyone know the odds of someone having paired their door card in 7cs? I tried briefly to look it up but didn't find it right away. If you know that, it's easy to know, given N people with a doorcard above your pair, what percentage of the time at least one of them will have a higher pair. (You can also calculate how often TWO of them or THREE of them will have a higher pair if you want)

chucky · #5 10-05-2007, 01:39 AM

Lets say there is a full table where bring-in has 2, then 7, J, Hero has K3K, Q, 4, 9, A. Hero wants to know what are the odds that A has an ace under. Well the odds that the Ace doesn't have another Ace is:
(39/42)(38/41)= 86.1%

Therefore Ace has split aces 14% of the time. Andy B will point out that in practice you can find out a truer percentage by completing and seeing if Ace raises your complete.

Red_Diamond · #6 10-05-2007, 01:46 PM

Alright, so if you have a pair of queens, and behind you is a single King, with a single A behind him as well... then you are on average up against a higher pair 14% + 14% = 28% of the time roughly.

And I suppose 28% is asking for TOO much trouble? I wonder, where the THEORETICAL cut-off percentage would be here, and why. Naturally this theoretical value cant happen in real life, because you can only be on 14% or the other (no such thing as fractional pairs), but for the sake of arguement, let's suppose....