View Full Version : Math for probability of flopping 2-pair or better

10-24-2004, 03:59 AM
Could someone explain how the calculations would work? I can't even begin to do the calculations because they seem to be pretty complex. For example Ad 7d?

10-24-2004, 05:43 AM
I don't know how to do the calculations, but I made a computer simulation. You're dealt Ad 7d and you look at a flop. The results are based on the ranking of your final hand (hole + flop) and also the state of the flop (so you know if "One Pair" means the board paired OR if you hit your card.)

My PC's RNG probably isn't that good, so these results aren't perfect... but they should be a reasonable ballpark.

<font class="small">Code:</font><hr /><pre>
200000 Trials
52.514% High Card
27.296% One Pair
13.472% One Pair (Pair on board)
2.103% Two Pair (Pair on board)
2.054% Two Pair
1.409% Three-of-a-Kind (Pair on board)
0.804% Flush
0.233% Three-of-a-Kind (Trips on board)
0.105% Full House (Pair on board)
0.011% Four-of-a-Kind (Trips on board)
</pre><hr />

10-24-2004, 05:56 AM
See the reason I want to know is because many times I'll have a hand I'm not comfortable with playing on the flop without a great draw or two-pair+ and I'd like to know what pot odds I'd need to call.

Example: I'm on the button with some trash. UTG raises and there's a few callers. What kind of pot odds would I need to call? If he raises big, my implied odds are a lot bigger too.

10-24-2004, 06:08 AM
I wouldn't count on flopping two-pair as to determining pre-flop play. Everyone else in the hand (except for players with pocket pairs) has basically the same chance to flop two-pair as you do. I see what you're getting at though - as the pot becomes more multiway it pays to loosen up and play more speculative hands - can we quantify this mathmatically? What sort of odds do I need to start playing mid range suited one gappers? and similar questions.

I don't know the answer, but I think finding the exact standard of loosening up to include marginal hands with lots of limpers also has a marginal EV.

Also, for the specific above question.. I recommend reading the pre-flop section of SSH /images/graemlins/smile.gif

10-24-2004, 11:45 AM
for the calculation of getting two pair or better, you could start by getting the probability of two pair or better without counting in straights or flushes, and then doing those seperately and adding them back in. for two pair or better,

p = (6/50)(5/49)+(6/50)(5/48)+(6/49)(5/48)+P(straight)+p(flush)

p(straight) = 0
p(flush) = (11/50)(10/49)(9/48)

therefore, p = .0122+.0125+.0128+0+.0084 = .0459

that number feels wrong. does anyone out there recognise a missing combinatorial term or something?

[edit]: i suppose a combinatorial term would make the number larger. it feels too big, perhaps the first 3 terms should have a (1/3)?

Lost Wages
10-25-2004, 09:26 AM
A7s isn't too hard since it can't flop a straight-flush/straight.

First, there are 50*49*48/(3*2) = 19,600 possible flops.

Total flops = 2 (AAA &amp; 777)

Full House
Flops of AA7 = ((3*2)/2)*3 = 9
Flops of 77A = ((3*2)/2)*3 = 9
Total flops = 18

There are 10 remaining cards in your suit
Total flops = 10*9*8/(3*2) = 165

Trips (using one hole card)
Flops of AAX (X not A or 7) = ((3*2)/2)*44 = 132
Flops of 77X (X not A or 7) = ((3*2)/2)*44 = 132
Total flops = 264

2 Pair (using both hole cards)
Flops of A7X (X not A or 7) = 3*3*44 = 396

Total flops of 2 pair or better = 2+18+165+264+396 = 845
Probability of flopping 2 pair or better = 845/19,600 = 4.31%

Lost Wages