Hand vs Hand Odds:

[rayrns]

I am holding Ks-Qs. My opponent has Jd-8o. Flop is 8s-4c-3h.
I am a 2.8-1 underdog to the pair of 8's. How do you arrive at hand to hand odds? I understand how to figure outs to odds but not hand to hand. I was going to try a search but don't know what I would enter.

[Louie Landale]

I assume when you say "hand-to-hand" odds you are refering simply to which hands will win. Its basically an adjustment to the "will I outdraw the opponent" odds ("how many outs do I have"), where you must consider the possibility that the opponent "redraws" you after you "outdraw" him.

It gets real sticky. In this case, you need not only calculate your chances of snagging a K or Q on the next two cards, but the other card must also not be a J or 8. To that you add your chances of backdooring a flush, but cannot count the Jh-8h as one of your out combinations.

With 7 seen cards there are (52-7)"choose"2 combinations of turn-and-river, or 45*44/2= 990 different ways the rest of the hand can play out. You have 3 different ways to win: [1] There are 6choose2 ways to snag a K or Q on both cards = 6*5/2=15. [2] You snag a single K or Q and the other card is not a K/Q/J/8: of the 45 unseen cards, 34 are none of those ranks. There are 6choose1 ways to snag the K/Q AND 34choose1 ways of snagging one of the other cards, or 6*34= 204 ways. [3] There are 10 of your suit out and you need two, or 10choose2 or 10*9/2=45 backdoor combinations, of which one is a loser (Jh8h), leaving 44 ways to make a winning flush. [Total] 15+204+44=263 winning combinations, and therefore 990-263=727 losing combinations, or 727:263 or 2.7:1 against.

I wonder if I made a mistake.

It would be easier with TTH, where you set up the heads up match and play a million hands, and figure out how often each won.

- Louie

[rayrns]

Thank you for your reply. The example was from Lee Jones book and I was wondering how he came up with the odds. Until I get much better with out odds, I think I will forgo worrying to much about the hand vs hand odds. I will copy your explanation for further study. I don't have TTH and before I purchase that, I think I would like to purchase pokertraker for my play in the microlimit games. Again thank you for your time.

[BruceZ]

You don't need TTH, use the free program at www.twodimes.net/poker (http://www.twodimes.net/poker) For this simple case it doesn't simulate a million hands, it checks every possible outcome.

The only thing wrong with Louie's calculation is that you don't have to worry about being outdrawn if you make a backdoor flush since you have spades not hearts. Here is an easy way to do it using just fractions rather than combinations:

(6/45)(34/44)*2 + (6/45)(5/44) + (10/45)(9/44) = 26.67% = 2.75-1.

The first term is for getting a single K or Q. There are 6 out of 45 cards you can catch on the turn, 34 out of 44 on the river, and then multiply by 2 since you can get the cards in the other order. The second term is for catching KK,QQ,KQ, or QK. Notice in this case we don't multiply by 2 since we are already covering every possible order of the two cards (6*5). The last term is for catching two running spades. Notice that the numbers are the same as what you get by Louie's method, except that since he is using combinations, he divides by 2 to get rid of order. Here we are effectively counting each order twice, and we are doing that consistently in both the numerator and the denominator. In other words, we are effectively using permutations instead of combinations.

Here is the output of the twodimes program which gives you exactly the same thing:

<pre><font class="small">code:</font><hr>
pokenum -h ks qs - jd 8h -- 8s 4c 3h
Holdem Hi: 990 enumerated boards containing 8s 4c 3h
cards win %win lose %lose tie %tie EV
Ks Qs 264 26.67 726 73.33 0 0.00 0.267
Jd 8h 726 73.33 264 26.67 0 0.00 0.733
</pre><hr>

[Louie Landale]

Calculating it exactly is rather tedius as you can tell. Never-the-less, you must not "forgo worrying" it. Otherwise, you will routinely confuse "making your hand" with "winning the pot", and that will prove disasterous as it will lead to a BUNCH of bad calls. You need to develope a crude feel for how often you may lose even when you improve.

My personal benchmark (when not drawing to the stone nuts) is to add 50% to the required size of the pot; then adjust (subjectively) from there.

- Louie

[BruceZ]

Louie,

Any spade flush IS the stone nuts here. The 8s is already on the flop.

-Bruce

[rayrns]

Thank you both. I have copied the information to work on at a later date.

[Don Quixote]

This is probably a dumb question or observation, but would someone set me straight? HOW do you know what your opponent holds? Wouldn't it be better to know how your hole cards stacks up against all other holdings in general?

I read an article some months ago on the easiest way to figure your odds for pairing up your overcards by the river, whether from preflop, flop, or turn. For instance, say you hold KQ after the flop and the flop did not pair you up. Any one of six cards will help you, so you multiply six times the number of cards yet to be put on the board and then multiply by two. So, 6x2 = 12x2 = 24. The percentage that you will hit either a K or a Q is 24% which is about 3:1 against your hitting either card. I have found this to be the simplist and quickest way to figure these type of odds in the heat of battle. Of course, this does not take into consideration backdoor flushes/straights. Number of outs times cards that make your hand times two will always give you the approximate odds. As someone else mentioned, making your hand and being second best is something else. If you have the ignorant end of a straight draw, you might want to forget about drawing. It depends :-)

[BruceZ]

For instance, say you hold KQ after the flop and the flop did not pair you up. Any one of six cards will help you, so you multiply six times the number of cards yet to be put on the board and then multiply by two. So, 6x2 = 12x2 = 24. The percentage that you will hit either a K or a Q is 24% which is about 3:1 against your hitting either card.

This is an approximation, and it makes the largest errors for large numbers of outs. For 15 outs, it would produce 60%, while the correct answer is 1 - (32/47)(31/46) = 54%. Of course with 15 outs you normally have a clear call anyway. You still need to convert percentage to odds. The quickest way is to simply memorize the odds needed for different numbers of outs, there aren't that many. The odds for 1 card to come are the most useful. With two cards to come the probability does not quite double.

[Louie]

I'm surprised that this simple calculation is as accurate as it really is, at least for small # outs (as Bruce suggested). Imagine a hand with 25 outs, your calculation would show you had a 100% chance to hit it, which obviously is not the case.

The detailed problem with your calculation is that it presumes there are 50 unseen cards (instead of 47 or less), and it doesn't take into consideration the fact that you may catch good on BOTH cards.

- Louie