Stud/8: Quads Freeroll Nightmare!

Here's a problem I worked out recently. I'll post the answer later.


In an eight-or-better stud game, on sixth street, the Cincinnatti Kid's hand is (Kh Kd) Kc Ks Qd Jh, while Lancey Howards is (XY) 3h 4h 5h 6h. The table is full, and the other third-street upcards were Ts, Tc, Td, 9c, 9s, and 9d. Needless to say, only the Kid and Lancey saw fourth street. The Kid believes that Lancey would play any hand that Ray Zee identifies in HIGH-LOW SPLIT POKER FOR ADVANCED PLAYERS as playing well heads-up.


How small must the pot be for it to be correct for the Kid to check and fold to Lancey's sixth-street bet?

Let's see:


Known cards = 16, leaving 36


Opponent will have 3 unknown cards, though one would surmise that the two he already has are not utterly random, but if they were I think this makes the odds 2 in 36 3 times, so there would have to be less than 6 big bets in the pot? And since it is not quite random, you might up that to 8 or something. In either case, not very likely to be that small.

Oh wait, of course I was just calculating for high. Given that it is hi-lo that makes the calculation much more difficult (not that I have much confidence my original calculation was properly formulated.


Of the 36 unknown cards, a whopping 28 of them are low! So odds are huge that low is locked up (though strating trips or two pair and no made low are possibilities), so playing for half the pot...I'd say if the pot has something like 16 or fewer BB, fold.


It's not entirely clesar to me how to factor in the worth of playing for 1/2 the pot: does it exactly halve the necessary pot odds? Since there is a slight possibility that you are playing for the whole pot, the calculation is obscure to me.


Curious to see the math for the exact answer, but my seat of the pants answer is if it were say 15 BB or small I'd fold.

This is a good brain teaser.

6 dead cards plus 6 known cards plus 4 opponent cards = 16 known cards

52 - 16 = 36 unknown cards

Only 8 high cards left leaving 28 that give opponent half the pot.

Only 2 let him scoop the whole pot (2h,7h)

2 out of 36 cards, odds 17:1 against him scooping

What are the chances he has these two cards given that 7CS8FAP determines they are a qualifier?

XX/3 could be: AA/3, 33/3, A2/3, 54/3

If his hand were A2/3 the combos would be:

A*4 suits + 2*4 suits = 16 combinations possible of which 4 would include 2 of hearts.

Now I am confused. David, what is the next step?

Well considering whenever Lancey and the Kid play, it's No limit and you can bet everything you have in front of you plus whatever money you or anyone else you know has in their pockets/bank accounts, a discussion of pot odds is not really significant.

Hands:


Kid: (Kh Kd) Kc Ks Qd Jh


Lancey: (X Y) 3h 4h 5h 6h


Folded on 3rd St: Ts, Tc, Td, 9s, 9c, and 9d


The Kid believes that Lancey is playing a hand which plays well heads-up, according to Ray Zee's book.


Lancey's possible starters:


(3x 3y) 3h -- (3 hands)

(Ah 2h) 3h -- (1 hand)

(Ax Ay) 3h -- (6 hands)

(Ax 2y) 3h (*) -- (15 hands, including 3 with 2h)

(Ax 5y) 3h (*) -- (12 hands)


Total starting hands 37


(*) If Lancey were playing three small straight cards, he'd probably bail when he paired up, ruling out hands like (5x 4y) 3h or (4x 2y) 3h; however, he might well stick around if his wheel draw paired up on 5th street, making (Ax 5y) 3h a real possibility. I doubt (2x 5y) 3h in this spot, but I could be wrong.


Four of Lancey's 37 possible starters *already* scoop.


Lancey surely has a made low, locking up half the pot. Ignoring 6th and 7th street action, the Kid's equity is half the pot, reduced by the chance that he's already beat: E = 33/37 * 0.5 = 33/74.


If the Kid's hand is currently good for high, Lancey has two scoop outs: 2h and 7h. The kid has seen 16 cards (his 6, Lancey's 4, the other players' 6), leaving 36 unseen cards in the deck. (Aside: does this get modified by putting Lancey on a limited range of hands?) 34 cards leave the Kid good for high, and two scoop him. His OVERALL equity in the pot is thus 0.5 * 33/37 * 34/36.


Pot size is currently P. By checking and calling on 6th street and the river, the Kid is paying 2 bets. Pot size will then be P+4.


Let A = 33/37 * 34/36.


Then the Kid's expectation is 1/2 * A * (P+4) - 2.


The break-even point will be where expectation is exactly 0. We can solve this for P to find that

P = (4/A)*(1 - A).


A=0.84234, and thus P = 0.749.


The pot must contain at least 3/4 of one big bet for the Kid to gain the most/lose the least by paying off to see the showdown. There is no conceivable circumstance with these cards where this is not the case.


The Kid should essentially always check and call.