Omaha Hi: Playing unpaired starting hands

[tvdad]

If it's correct that a pair will hit the board about 50% of the time (and if it's not, then ignore the rest of this question and tell me I'm wrong), then is it worth it to play starting hands containing no pairs? In other words, should I even bother with hands like AKQJ and AQJT, suited or not, when the best I can do is make a straight or flush and have a 50% chance of a pair hitting the board?

I ask because I'm obviously doing something very wrong with my flopped (or turned) straights and flushes. I usually bet them hard, but in the low limit games that I play it's nearly impossible to get the 2-pairs and sets to fold, so I lose a lot of straights and flushes when the board pairs up.

A lot of people in these low limit games love to play any pair, so any pair hitting the board with 5-6 people in the hand is scary when I've got a straight or flush.

T

[joeg]

A pair will make a set 11.97% of the time

[tvdad]

[ QUOTE ]
A pair will make a set 11.97% of the time

[/ QUOTE ]

Hmmm, rereading my post makes me think perhaps I wasn't quite clear enough. I was talking about two cards of the same rank appearing on the board, not a pair in your hand hitting a match on the board for a set.

T

[chaos]

A pair will flop about 17% of the time, not taking into account the cards in your hand.

Here is how I got it:
There are C(52, 3) = 2210 ways to choose the three cards for the flop from a 52 card deck.

For a flop with a pair:
there are 13 ways to choose the rank of the pair
there are 6 ways to choose the suits for the pair
there are 48 ways to choose the odd card
13*6*48 = 3744

3744 / 2210 = 0.1694

No guarantee that my logic or math is correct.

[tvdad]

Sorry for the confusion everyone. Maybe it would help if I stated that part my question another way. What are the odds that two cards of the same rank (a pair) will appear on the full board, not just on the flop? I thought this was around 50%.

T

[Buzz]

TV Dad -

When you are dealt a hand that doesn't have a pair, the five cards on the board fall into six separate categories, as follows:

a. The board has no cards of the same rank as your hand.
b. The board has exactly one card of the same rank as your hand.
c. The board has exactly two cards of the same rank as your hand.
d. The board has exactly three cards of the same rank as your hand.
e. The board has exactly four cards of the same rank as your hand.
f. The board has exactly five cards of the same ranks as your hand.

You decide what interests you to answer your question.

I’m interested in some aspects of c., d., e., and f as follows:

c. 4*3*36*35*34/6 ways to make a set
d. 4*3*9*36*35/2 ways to make a full house
d. 4*1*36*35/2 ways to make quads
e. 4*3*3*3*36/2 + 4*3*9*6*36/2 ways to make a full house
e. 4*1*9*36 ways to make quads
f. 4*3*9*6*3/6 ways to make a full house
f. 4*1*3*3 + 4*1*9*6/2 ways to make quads

c. 85680 ways to make a set
d. 68040 ways to make a full house
d. 2520 ways to make quads
e. 1944 + 11664 ways to make a full house
e. 1296 ways to make quads
f. 324 ways to make a full house
f. 36 + 108 ways to make quads

Combining, I see
85680 ways to make a set (the board is paired),
81972 ways to make a full house (the board is paired), and
3960 ways to make quads (the board has trips).
That’s all out of C(48,5) = 1712304 possible boards.

It’s noteworthy that there are a number of other winning hands you can make without the board pairing when you hold cards that do not contain a pair, as for example with AcKdQdJc.

Never a guarantee about my math. You can see there are a lot of places to go wrong.

Buzz

[chaos]

"No guarantee that my logic or math is correct."
This time it wasn't completely! Buzz pointed out in a PM that I had a typo in my original when copyng numbers from the spreadsheet I used to do the calculations. "2210" should read "22100." Final answer was correct. I revised the post below.


A pair will flop about 17% of the time, not taking into account the cards in your hand.

Here is how I got it:
There are C(52, 3) = 22100 ways to choose the three cards for the flop from a 52 card deck.

For a flop with a pair:
there are 13 ways to choose the rank of the pair
there are 6 ways to choose the suits for the pair
there are 48 ways to choose the odd card
13*6*48 = 3744

3744 / 22100 = 0.1694

[ThinkQuick]

Following chaos' logic:

There are C(52,5) = 2598960 ways to choose the five cards for the board from a 52 card deck.

Pairs:
there are 13 ways to choose the rank of the pair
there are 6 ways to choose the suits for the pair
with 52-2=49 cards left, there are C(49,3)= 18424 choices for the other cards, I think.

13*6*16215 = 1437072

1437072/ 2598960 = 0.5529 = 55.29%.

Close to 50, if the math is right. Someone check it please.
Shoot I must have made a mistake somewhere. Can you even follow 'logic' from 'chaos'?

[chaos]

I do not follow your logic about paired boards.
You do not need a pair in your hand to make better than a straight or a flush. You need the board to pair and to match one of your hole cards. Then having another hole card match a board card gives you a full house.

For example you hold TJQK and the board is KKJ83. You have Kings full of Jacks for the nuts.

[chaos]

52-2 does not equal 49.

I'm not sure what exactly you are trying to calculate. The board contains exactly one pair or the board contains one pair or better.