Okay suppose you have the following in 27 OFC

A
23457
QQT

On the third draw you receive A62 and are deciding whether or not to put the A up top. What are the odds of making a Q, T, or a pocket pair in the final three card pull? Your two discards were 99.

For those of you unfamiliar with 27 OFC the question is essentially this:

You have seen 223456799TQQAA (fourteen cards)

You are then dealt a three card hand, what is prob that hand contains at least a Q, a ten, or a pocket pair?

Combinatorics seemed like easiest way but maybe I have quarantine brain and I'm missing something but I'm getting an unreasonable answer.

The denominator is C(38,3) = 8,436

Just starting out with the Qs and Ts I have:

Pair Queens, QQX where X could be anything = 2*1*36 =72

Pair Tens, TTX where X can be anything but Q = 3*1*34 = 204

QTX where X can't be a Q or T = 2*3*33 =198

QXX where X can't be Q or T = 2*33*32 = 2112

TXX where X can't be Q or T = 3*33*32 = 3168

I'm massively overstating this so I'm doing something wrong, can someone point me in right direction?

You basically did permutations in the numerator for the XX cases and several minor errors.

The easiest way IMO is to see how many ways you will not select a Q or T (33) and subtract that from 1.0

Pr = 1 - C(33,3)/C(38,3) = 35.3%

Thanks man, yeah I can match that 35% from doing it from a probability standpoint using 1-Probability you miss all three.

But the pocket pairs seem tougher so I tried to do the entire thing from combinatorics and testing to see if I could match the 35%.

Could you assist with the pocket pairs?

For example what is the probability of JJX where I imagine X is not a Q or a T to avoid double counting.

I get 2.1%. It will be different for other ranks if some have removed cards.

Awesome, what would it be for a pair of deuces then since there are two out?

Can you show me some work so I can learn?

Thank you for all of this

I agree that it's easier here to find the chance of it not happening.

C(33,3)/C(38,3) = chance of no T's or Q's, from which we need to subtract P(pocket pair | no T or Q)

Of the other 11 ranks:
3 of the them have 2 cards remaining, so 1 two-card combo
7 of them have 3 cards remaining, so 3 combos
1 has all 4 cards remaining, so 6 combos

So P(pocket pair | no T or Q) = 3(3 + 7*3 + 6)/C(33,2) - 2(7+4)/C(33,3) = 16.64%
(If having a pocket set doesn't count, the bold 2 should be a 3.)

I multiplied the 2-card probability by C(3,2) since we get 3 pocket cards. However, the sets are then triple-counted, so I subtracted them twice. Seven of the ranks have 1 set combo and one rank has 4 combos.

If you're not comfortable with that approach, you can also have a single denominator of C(33,3) and the following numerator: 3*31 + 7(3*30 + 1) + 6*29 + 4

And now we're done: 1 - C(33,3)/C(38,3) + .1664 = 51.97%

Oh wait, I think it's probably 46.09%

I said P(not) = P(no Q or T) - P(pocket pair | no Q or T)

But I suspect that's a fallacy, because it disagrees with: P(not) = P(no Q or T)*P(no pocket pair | no Q or T)

We can either multiply C(33,3)/C(38,3) by (1-.1664), or from the start we can do:

[C(33,3) - (3*31 + 7(3*30 + 1) + 6*29 + 4)] / C(38,3)

Subtracting that from 1, we get 46.09%

Yeah definitely 46.09%

You can say N(not) = N(no Q or T) - N(pair | no Q or T)

but you can't do that with probabilities because the given condition isn't a given unless you're multiplying. It would be valid to do:

P(not) = P(no Q or T) - P(non-QQ/TT pair)
or
P(not) = P(no Q or T)*P(no pocket pair | no Q or T)
or
P(not) = [N(no Q or T) - N(pair | no Q or T)] / N(total)

Thanks heehaw, but I’m sorry I had a little trouble following some of this. So your final conclusion is that the probability associated with getting a pocket (or trips) eg. JJX (where X not Q or T) is around 11%?

Yup, 10.76% is the chance of a pocket pair/trips (other than QQ/TT) and no Q or T.
35.33% is the chance of any Q or T and possibly any pocket pair.

Thank you very much

I know you didn't ask about best play, but what's the opponent's hand? Is he in fantasy? If he is not and the 6 is live, a 3-out kicker on the last draw always has a better chance of pairing than making a pocket pair. The extra royalties and hand equity from FH/Quads is not worth the extra foul chance.

However I don't think you should play the A on top here. You have 8 royalties, fantasy and you can never be scooped so risking a 50%-ish foul seems bad. What are the FL rules if you get both AA and a nut wheel?

Thanks brown, good post man. I wanted to get this probability to do some calcs and figure out the best play.

I assumed the opponent was in FL. And I calculated that playing it safe does better by around 2 points than choosing to gamble here. However, if we have 23467 we would gamble.

AA and nut wheel would give you a 15 card FL. Do you have any opinions on the value of a 14 card and 15 FL? I assumed 6 points for 14 card FL and 8 for 15 card FL, is that reasonable iyo?

And you are indeed correct that it is correct to play a live kicker on the bottom as that increases the odds of not fouling.

I haven't played much of the low variant. In standard OFCP 14 card FL is considered by most sources to be about 7.5-8 points. I really can't guess how much the extra card is worth or how much FL is worth in 2-7 in the middle.

Ok thanks, according to people who have studied it FL is worth less in 27 and many think its around 6. Couldn’t find anything on 15 card so I just bumped it up a few points

heehaw, can you confirm the following?

Prob totally live pair w/ no Q/T is: C(4,2)*(29,1)/8436 = 2.1%

Prob pair with one card dead w/ no Q/T is C(3,2)*C(30,1)/8436 = 1.1%

Prob pair with two card dead w/ no Q/T is C(2,2)*C(31,1)/8436 = .4%

For example, prob 99 with no Q/T is .4% but prob 88 with no Q/T is 2.1%, is this correct?

Exactly, assuming trips aren't allowed.