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BonJoviJones
09-11-2002, 12:58 PM
Just for kicks and mental exercise, last night I tried to calculate the odds of flopping an open-ender in hold'em, given the starting hand of 67. To simplify my calculations, I didn't worry about "accidentally" getting a real straight.

I got stuck after awhile, and couldn't figure it out. Help on my thinking process would be appreciated.

There are three sets of two cards that give me an open-ender (45, 58, 89). Each of the three is equally likely. Thus I figured that the answer was:

Let P(OE) = Probability of flopping one of the three success hands.

( 3 * P(OE) ) / (50c3)

This seems fairly obvious to me. However I'm having trouble finding P(OE). I don't particularly want an answer to what P(OE) is, but just a hint or two on where to go next.

Thanks.

09-11-2002, 01:14 PM
I believe I wrote a solution to this a few weeks ago. I think it shoudl still be up.

But if you like if you wanted P(of an openended say 9 high straight| 67))=

P(998)+ P(889) +P(89X | X is not 8, 9 T or 5).

BruceZ
09-11-2002, 06:03 PM
In your equation P(OE) should actually be the number of ways to flop and open ender with two of the cards, not the probability. For 45x, the number of ways is 4*4*48 since there are 4 ways to get a 4, 4 ways to get a 5, and 48 ways to get the other card. Now this includes 458, so when we do the same thing for 58x, we shouldn't count this again, so we can't just multiply by 3. So for 58x the number of ways is 4*4*44 since we don't want 584. Now one of these ways includes 589, so when we do 89x we don't want 895, so we have again 4*4*44. So all together the probability is:

(4*4*48 + 4*4*44 + 4*4*44)/C(50,3) = 11.1%.

Buzz
09-11-2002, 07:33 PM
BonJovi - I think you get the probability by dividing the number of possibilities of getting one of the boards by 50C3.

In other words, rather than dividing the probability by 50C3 to get the probability, divide the number of flops which have the card combinations in which you're interested by 50C3 to get the probability.

(the # of flops you like)/(50C3) = (the probability of flopping a board you like).

Thus this is an exercise in "counting." You're going to count the number of flops that satisfy your requirement (a flopped open ender). Counting is a bit of a misnomer. What you're doing is really tabulating, but it's called "counting."

How many ways can you put 45 together with another card? Then how many ways can you put 58 together with another card? Finally, how many ways can you put 89 together with another card. You obviously don't want to count any of the hands twice. That's where the counting gets sticky.

For example, 458 includes an eight and thus would be included both in
your count for 45 + any other card and
your count for 58 + any other card. Right?.

There are two ways to proceed. You can count hands twice (or more) and then go back and subtract duplications, or you can avoid duplications as you proceed.

In any event you should probably would make a chart. Start with any of the combinations you are seeking, say 45. Below is the start of the chart. (X can be any card but 4, 5, 8, or 9)

454 6*4 24 ways to do this
455 4*6 24 ways to do this
458 4*4*4 64 ways to do this
459 4*4*4 64 ways to do this
45X 4*4*34 544 ways to do this

I think the total of the above is the number of possible three card combinations (all of them) which include a 4 and a 5. Note that the order in which the three cards are flopped doesn't matter. In other words, 445 is the same as 454, which is the same as 544. In each case there is a five and two fours on the flop. The 6*4 comes about because there are 6 different ways to combine two fours to make a pair, and 4 different fives.

Now work in all the possible three card combinations which include 5 and 8. I'll leave figuring out how many times each listing occurs to you this time.
585
588
589
58X
(I left out 584 because it's already listed above as 458).

Finally work in all the possible three card combinations which include 8 and 9.
899
898
894
89X
(I left out 895 because it's already listed above as 589).

Add it all together to get the number of three card combinations which make an open-ender + a flopped straight. If you want to exclude flopped straights from your total, go back and subtract them.

Then you divide your total by 50C3 to get the probability.

That's how I'd do the problem.

As an aside, I didn't know anything about these problems before I retired a couple of years ago and started playing casino poker. I read David Sklanski's "The Theory of Poker" (which I highly recommend to you), and also read some of Barbara Yoon's posts over on the r.g.p. forum (and some other posters posts on r.g.p. and here on 2+2 and some other books) - and I learned. It's a little tricky knowing whose posts are credible, who is a truth seeker, who to believe.

I think learning is a matter of interest. If you're interested in learning about something, then you learn about it. If not, then you don't. At least that's how it has always been for me.

Good luck to you.

Buzz

BruceZ
09-11-2002, 10:04 PM
I avoided double counting cases like 458 and 584, but I am still double counting cases like 454. This is because when we say there are 4*4*48 ways to get 45x, some of those 48 cards will be a 4 or a 5, and those cards were already counted in the 4*4. So we have to go a step further and separate out the cases where there is a pair on the flop. Sooooo...proceding as before:

45x where x is anything:
4*4*42 + 6*4 + 6*4 = 720.
The 6*4 correspond to a pair of 4s or a pair of 5s.

58x where x is anything but 4:
4*4*38 + 6*4 + 6*4 = 656

89x where x is anything but 5:
4*4*38 + 6*4 + 6*4 = 656

(720+656+656)/C(50,3) = 10.4%.