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VivaLaViking
08-04-2005, 03:19 PM
I am trying to evaluate the "outs" when holding cards that may produce a straight on the flop. I will evaluate the other possibilities later (an Integration by Parts lol) but I wanted to verify that my thought process is correct at the start or if anyone can suggest an easier way I would be grateful. LetyouDown my hat is off to the person who did this by bit shifting.

[ QUOTE ]

Assume you hold a connector not bound on the low side or high side. Example: holding 4 5. There are 24 value combinations yielding a straight.

6 7 8 or 6 8 7 or 7 6 8 or 7 8 6 or 8 6 7 or 8 7 6 or
1 2 3 or 1 3 2 or 2 1 3 or 2 3 1 or 3 1 2 or 3 2 1 or
3 6 7 or 3 7 6 or 6 7 3 or 6 3 7 or 7 3 6 or 7 6 3 or
3 2 6 or 3 6 2 or 2 3 6 or 2 6 3 or 6 3 2 or 6 2 3

(24)&lt;- Combinations * [(4 C 1)(4 C 1)(4 C 1)]&lt;- suits

[/ QUOTE ]

Could anyone reply with an affirmation or correction.

TY

LetYouDown
08-04-2005, 03:32 PM
It's bizarre but it looks right to me if you're counting EXACT ways a board can fall. As in...6/images/graemlins/club.gif 7/images/graemlins/diamond.gif 8/images/graemlins/heart.gif is different from 6/images/graemlins/club.gif 8/images/graemlins/heart.gif 7/images/graemlins/diamond.gif. It might be a lot easier for you to handle if you reordered the cards before you analyzed.

VivaLaViking
08-04-2005, 03:40 PM
You have obviously have had some exposure to programming, Regarding ordering, I was hoping to just be able to multiply based on the difference of the cards and the bounding limitation but "The ememy lines are fluid".

TY

LetYouDown
08-04-2005, 03:44 PM
Oh, so you're just hoping to get the correct # odds of hitting a straight on the flop? I don't see why you're using permutations here, is my point...is it necessary?

VivaLaViking
08-04-2005, 03:49 PM
I don't know if it is necessary but it is how I'm thinking of the problem now unless I can gleam a better way.

TY

LetYouDown
08-04-2005, 04:14 PM
Well, if you hold 4-5...the board can come the following ways:

A-2-3
2-3-6
3-6-7
6-7-8

For each of these combinations, there are 4 * 4 * 4 ways they can fall in terms of suits. So you have:

[4 * C(4,1) * C(4,1) * C(4,1)] out of a C(50,3) possible flops. If you're doing it via permutations, you end up with 6X that number...which you have.

VivaLaViking
08-04-2005, 05:39 PM
Yes, as it turns out algebraicly, if two cards have a separation D, and they are not bounded, it appears that (5 - D) * 384 is the number of combinations. The 5 - D value is decreased by each bound but I have not proven this to myself yet that is my present effort.

TY