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....

RustyBrooks · #7 10-05-2007, 02:01 PM

You don't add the percentages like that. You have to do what I have above and figure out how often they each individually do NOT have a pair, then multiply to find out how often NEITHER has a pair. If the odds are 14% for one of them to pair, then it's 86% chance for each of them that they don't have a pair.

For 2: 26% one has a pair
For 3: 36% one has a pair
For 4: 45% one has a pair

Note that if you could add the percentages of independent events, then you would have to conclude that if two people flipped a coin, one of them would land on heads 100% of the time. And if 3 people flipped, 150% of the time.

If I had QQ and there was a K and A ahead of me I might still be tempted to raise. 26% is not so bad.

TheEye · #8 10-05-2007, 02:07 PM

Why not raise QQ with 3 or even 4 over behind you?

You get extra info, get rid of the overcards, or bad calls.

If the OP is talking about medium or small pairs, depends on kicker ofcourse, then 3 over behind you is a usual fold not even a call imo.

Red_Diamond · #9 10-05-2007, 02:09 PM

That's why I said 28% ROUGHLY.

In any case, I'll commit the 26% value to my stud memory of statistics. Thanks.

chucky · #10 10-05-2007, 02:10 PM

I tried to do the multiple high cards behind hero scenario but was unable to do the card removal aspect of the calculation. Namely if hero has split queens with K and A showing behind, you not only need to calculate the odds of K not having another K but also him not having an Ace under because that would affect the Ace's split pair odds. Is there a formula for this?