5card draw high odds question

[DOMIT]

Hi,

I've read Caro's section on draw in Super System and looked at his tables of odds, but the one that I don't see is drawing two to a pair w/ a kicker.

I see that he states drawing 2 to trips will fill up 6.11% of the time; meaning that you'll draw a pair on a two card draw 6.11%. Then drawing 3 to your kicker and 2 to your pair: 5/47 * 2 draws is 21.28%. This gives a combined total of 27.39; which makes you 2.65-to-1 to improve.

Do I have the math correct on this?

Then of course, I might ask... do I every really want to draw to a kicker? Perhaps in late position w/ A55xx after a couple limpers? Say I know that the limpers would have raised if they had aces or better.

Just curious if I'm playing too tight by throwing away these hands in late position or in the blind when it's been raised. I understand that the 6% of the time that I do draw another pair, it may be no good (say I end up w/ 5522A), so do I really want to add in that 6% at all or reduce it down to 3% for half the time that I'll get higher pairs?

Thanks for any advice.

~DOMIT

[Buzz]

Domit - Five card draw is not my game but maybe I can answer your questions.

"I see that he states drawing 2 to trips will fill up 6.11% of the time; meaning that you'll draw a pair on a two card draw 6.11%."

Yes. Here you have been dealt, for example, AsAcAh2s3d. Note that you should be more interested in the probability of improving to either a full house or quads than simply the probability of making a full house - but that isn't what your question is about.

"Then drawing 3 to your kicker and 2 to your pair: 5/47 * 2 draws is 21.28%."

No. Now you don't have trips any more. Without the trips, the odds of catching a pair are a bit different.

5/47 is the probability of drawing one card (and ending up with four cards) that is the same rank as your pair or kicker. However, you're drawing two cards (and ending up with five cards). You can't just multiply the probability of catching a card in a one card draw by two when you draw two cards. It may seem to you that it should work that way, but it simply doesn’t.

"Then drawing 3 to your kicker and 2 to your pair: 5/47 * 2 draws is 21.28%. This gives a combined total of 27.39; which makes you 2.65-to-1 to improve.

Do I have the math correct on this?"

No. Your math doesn't look correct to me.

(5*4/2+5*42+2*3+9*6)/1081 = 280/1081 = 0.259 is the true probability of improving when you draw two cards to a pair plus a kicker. The chances of improving when you draw two cards to a pair plus a kicker are lower than when you draw three cards to a pair. This should make intuitive sense to you.

"Then of course, I might ask... do I every really want to draw to a kicker? Perhaps in late position w/ A55xx after a couple limpers? Say I know that the limpers would have raised if they had aces or better."

The negative side of keeping the ace kicker is you decrease your chances of making quad jacks, jacks full, or trip jacks - but you didn’t have much of a chance of making any of these hands anyway. To be more specific, the probability of making trip jacks or better drops from 3270/16215 to 92/1081, or from 0.202 to 0.085 (94% of that being a decrease in the probability of making trip jacks).

Why, then, would anyone draw two cards to a pair plus a kicker?

Couple of reasons, as I see it: (1) deception, and (2) if you keep an ace kicker, for example when you have a pair of jacks, you have a greater chance of making two pairs, aces over jacks or better. The probability of making aces over jacks or better goes up from 3396/16215 = 0.209 to 344/1081 = 0.318.

So do you keep an ace kicker or not? The answer, in my humble opinion, is it depends on how your opponents play. In a looney tunes game where everyone takes a draw, some drawing four cards to an ace, others drawing three cards to a pair of sevens, and still others drawing one card to an inside straight - in a game like that, I don’t think you want to decrease your chances of making trips or better. Translation: don’t keep a kicker.

However in a tight game where most individuals with less than a pair of queens fold rather than taking a draw - in a game like that, you probably don’t need trip jacks or better to win. In other words, two pairs, aces over, will probably be a winner. Therefore try to increase your chances of making such a hand. Translation: keep the ace as a kicker.

"Just curious if I'm playing too tight by throwing away these hands in late position or in the blind when it's been raised. I understand that the 6% of the time that I do draw another pair, it may be no good (say I end up w/ 5522A), so do I really want to add in that 6% at all or reduce it down to 3% for half the time that I'll get higher pairs?"

You're not using the figures you should be using, in my humble opinion. I think that whether you throw away these hands or not depends on how your opponents are playing.

Hope this helps. You might get better advice for these types of questions on the probability forum. Interest in probability here is mostly as it relates to a game somebody is playing (Omaha-8 in my own case). I probably won't answer another question about probability as related to five card draw.

Buzz

[Al Mirpuri]

Dear Domit,

Forget the math.

It is very simple.

Keeping a kicker harms your chances of making quads, a boat, trips and any other two pair than the one you are going for.

The improvement to the two pair that you are going for is enough to make it worth your while to draw to it if that two pair will win post draw.

Keeping a kicker is pointless if your pair is higher than your opponents.

You should not be playing pairs smaller than your opponent's pairs. At Jacks or better, you should not be playing less than Aces up front but loosening to Jacks at the back. You should never be playing short pairs (shorts - pairs lower than Jacks against an opener who must by definition have at least Jacks.). You never play shorts in Jacks or better whether you are heads up or multiway.

If you do not raise and keep a kicker then everyone knows that you have a pair with a kicker. You are bluffable.

If you raise and keep a kicker then everyone knows that you have a pair with a kicker. Why? Well, when a player has trips he draws two multiway but disguises them by drawing one when heads up or short-handed.

Raising and keeping a kicker with a short pair may be effective as an occasional bluff but do it too often and it will cost you money. However, it will allow you to draw two cards to your trips when heads up or short-handed.

Do not play short pairs (any pair lower than Aces-Jacks -accounting for position) in any form of draw poker.

Purchase a copy of Winning Concepts in Draw and Lowball by Mason Malmuth. I own one. Incredible book. It might be my favorite book by Mason Malmuth.

[DOMIT]

Thank you both for your replies.

First, I'd like to say that the draw game that I'm playing is double blind with a 52 card deck; i.e., no joker, and NOT jacks-or-better.

Second, Buzz, can you please break down the numbers:
"(5*4/2+5*42+2*3+9*6)/1081 = 280/1081 = 0.259 is the true probability ". What do the numbers represent; where do you get them from? I like math and would like to understand what you're trying to show me above.

Finally, the game I'm playing. There are some good players and some fish. When I started reading Mason's book, I tried to always play nothing less then jacks-or-better, even though, using the double blind, there's no openning qualifier. Reading Caro's section of Super System on draw, he has limited detail, but states you can play down to 8's on the button (if no one else has openned). After reading this, I was wondering if I was playing too tighly for my game. I do see people keeping kickers at times, and I do understand the need for deception. If I raise as first in w/ QQKxx and get a caller or two, I'll go ahead and keep my king kicker and still bet out after the draw (if my competition draws 1 or 3) to represent trips. Sometimes it works, sometimes not.

Once again, thanks for the replies.

~DOMIT

[Buzz]

“can you please break down the numbers:
"(5*4/2+5*42+2*3+9*6)/1081 = 280/1081 = 0.259 is the true probability ". What do the numbers represent; where do you get them from? I like math and would like to understand what you're trying to show me above.”

Domit - Okay, I’ll break down the numbers for you. But first let me make it clear that I agree with Al Mirpuri. Get yourself a copy of Winning Concepts in Draw and Lowball by Mason Malmuth.

Second, I have to warn you that I’m not a mathematician and advise you that you might get a better explanation from one of the mathematicians who post over on the 2+2 probability forum.

Third, you should understand there are a variety of ways to solve probability problems. I might (likely would) set up the problem differently the next time I did it. But here goes.

Think of some particular holding in a setting where you might want to keep an ace kicker. Note that I’m not advocating keeping an ace kicker. But suppose, for example, you were dealt JhJdAc2s3h in a very tight game and decided to keep JhJdAc and draw two cards.

You know the whereabouts of five cards, the three you are holding and the two you just mucked. In a 52 card deck there are 47 cards you haven’t seen. When you draw two cards, you could get any two of these forty seven cards.

The order in which you are dealt the two cards for your draw is immaterial. This allows calculation of two-card “combinations” rather than two-card “permutations.”

The total possible number of two-card combinations of forty seven cards, the total number of possible two card draws is given by the expression 47 choose 2. (There are several ways mathematicians write this, but 47 choose 2 will do). 47 choose 2 turns out to equal 1081. Let me try to explain.

Take a deck of cards and remove JhJdAc2s3h. You should have 47 cards left. Now take the first of these cards, whatever it is. How many cards can you put with it to make some two-card combination? The answer is 46. If the order in which the cards are combined (which comes first and which comes second) is unimportant, then there are 46 ways to combine the first card with some other card. Take a minute and make sure this concept makes complete sense to you. Okay? Now we’re done with that first card. There is no other way to make a two-card combination with that first card. Take it out of the pack and put it with the cards you have already mucked.

How many cards are left? The answer is 46. Next take the first of these 46 cards, whatever it is. How many cards can you put with it to make some two-card combination? This time the answer is 45.

Perhaps you can see ahead. There are going to be 46+45+44+43+42+41+40+
39+38+37+36+35+34+33+32+31+30+
29+28+27+26+25+24+23+22+21+20+
19+18+17+16+15+14+13+12+11+10+
9+8+7+6+5+4+3+2+1 two-card combinations of the 47 cards. All those numbers added together equal 1081.

Probably no one would calculate 47 choose 2 by actually adding all those numbers together. 47 choose 2 also equals (47*46)/(1*2).
47 choose 2 also equals 47 factorial divided by 2 factorial and divided by 45 factorial.
In other words, 47 choose 2 also equals (47!)/(2!*45!) or 47!/2!/45!.
And any of these equals 1081.

At any rate, when you know the whereabouts of five cards, and are drawing two of a possible 47 cards, there are 1081 possible outcomes for you.

That’s where the denominator, 1081, of the expression “5*4/2+5*42+2*3+9*6)/1081” comes from. 1081 is the number of possible two card combinations in 47 unknown cards. Now for the numerator.

If you have kept JhJdAc and discarded 2s3h, next we need to know how many of these 1081 two-card combinations would improve your hand. There are four ways a two card draw combination could improve JhJdAc to a hand better than a pair of jacks with kickers. You could make quad jacks, aces full, jacks full, trip jacks, two-pairs-aces-over-jacks, or two-pairs-but-not-aces. That’s the full extent of it.

(1) A jack or an ace plus a jack or an ace makes either quad jacks or a full house. (2) A jack or an ace plus any other card makes either trip jacks or two pairs, aces over jacks. (3) Any pair (other than aces or jacks) makes two pairs.

(1). 5*4/2 represents the number of possible two-card combinations that have a jack or an ace plus a jack or an ace.

(2). 5*42 represents a jack or an ace plus any other card.

(3). 2*3+9*6 represents any pair (other than aces or jacks).

Breaking category (3) down a bit further, there are three ways to be dealt a pair of deuces, since you have discarded the deuce of spades. They are 2h2d, 2h2c, and 2d2c. Similarly there are three ways to be dealt a pair of treys. Thus 2*3.

Continuing breaking category (3) down a bit further, there are six ways to be dealt a pair of fours. They are 4s4h, 4s4d, 4s4c, 4h4d, 4h4c, and 4d4c. Similarly there are six ways to be dealt a pair of fives, sixes, sevens, eights, nines, tens, jacks, queens, and kings. There are nine ranks of cards here. (Fives are one rank, sixes another, sevens another, etc.). Thus 9*6.

Thus 5*4/2 + 5*42 + 2*3 + 9*6 = 280 represents the total number of two-card draw combinations that will improve JhJdAc to a hand better than a pair of jacks with kickers.

Lastly, divide the 280 by 1081 to get the probability of improving JhJdAc to a hand better than a pair of jacks with kickers.

Hope this makes it clear.

Buzz

[DOMIT]

[ QUOTE ]
"Domit - Okay, I’ll break down the numbers for you. But first let me make it clear that I agree with Al Mirpuri. Get yourself a copy of Winning Concepts in Draw and Lowball by Mason Malmuth."

[/ QUOTE ]

OK, now let me make something clear; when I stated:
"When I started reading Mason's book...", I meant Winning Concepts in Draw and Lowball by Mason Malmuth.

That out of the way.. /images/graemlins/grin.gif

Let me say "thank you" again for explaining. You're breaking down all the permutations. I understand now.

Thank you /images/graemlins/smile.gif