PDA

View Full Version : Odds of improving a set

Code Red
06-18-2003, 05:05 PM
4.47 to 1

Is this the correct odds against making a fullhouse or better when you have a set in hold'em with two cards to come?

Thanks

Copernicus
06-18-2003, 05:27 PM
It depends on whether there are draws to flushes and straights in addition to the quads and boat, but even without those I think its much better than that...bbiam with the answer.

Copernicus
06-18-2003, 05:53 PM
the computer answer is only 2/1 against, but manually it looks even a little better than that, but I must be duplicating some hands: I used 7,7,7,2,J rainbow to avoid straights and flushes.

There are 1081 combinations of 2 cards out of the remaining 47 after the flop C(47,2)

Of those 46 give you quads (a 7 plus any other of 46 cards).
6 cards match either the J or 2, and then any of the remaining non-7s (45 of those) work for the other card, so 6 times 45 is 270.
Then you can pull a pair from the 10 remaining ranks not yet seen C(4,2)x10 =60.

270+60=330 boat hands +46 quad hands = (1081-376)/376 or 1.875/1 against.

Somewhere I am duplicating 15 boat hands though, since there are apparantly only 315, total with quads of 361. (1081-361)/361 = 1.99/1

Cyrus
06-20-2003, 07:17 PM
"I used 7,7 | 7,2,J rainbow to avoid straights and flushes.

There are 1081 combinations of 2 cards out of the remaining 47 after the flop. Of those, 46 combinations give you quads (a 7 plus any other of 46 cards)."

All correct.

"6 cards match either the J or 2, and then any of the remaining non-7s (45 of those) work for the other card, so 6 times 45 is 270. Somewhere I am duplicating 15 boat hands though."

To make a boat, the two cards to come will have to belong to one of the following categories:

22
JJ
J2

2x
Jx

yy

In the first three categories, there are 3 Jacks and 3 deuces left, so C(6,2)=15.

In the next two categories, of Jack-something and deuce-something, each of the 3 Jacks and the 3 deuces combines with 1 of the remaining 40 cards to produce 6*40=240 combinations.

Note that (52 cards - 2 in your hand - 3 on the flop - 1 Seven - 3 Jacks - 3 Deuces) = 40 and this is where the mistake in your calculation lies ("45 remaining non-sevens").

In the last category, where the Turn and the River cards are a pair, a non-Seven, non-Jack, non-Deuce pair, you get as you say

"a pair from the 10 remaining ranks not yet seen C(4,2)*10=60"

So, 46+15+240+60=361 and the probability of getting at least a boat by the River is 361 / 1081 = 0.333950046253469010175763182238668 or 33.4% or 1.99:1 against.

--Cyrus

PS : Note that even with that little mistake above, the answer is still for all practical purposes 2:1. I have learned a long time ago, playing a completely different casino game, that mathematical accuracy can only take you so far. Knowing when the time is or isn't right to push the money across the felt is the premium characteristic of advantage play.