Open ended straight draw question

[kingstalker]

If you J-10 or 6-5 what are the odds that you will flop a open ended straight draw?

[M.B.E.]

Take JT as an example. To flop an openender, the flop will have:

(1) K and Q but no A and no 9 (592 combinations); or
(2) Q and 9 but no K and no 8 (592 combinations); or
(3) 9 and 8 but no Q and no 7 (592 combinations).

So the probability is 1776/19600 = 9%. The odds against are 10:1.

[Lost Wages]

If you hold JTo, KQ can make 16 combinations, which can combine with 52-(2 in your hand)-(2 for the KQ)-(4 aces)-(4 nines)=40 other cards. 16*40=640 flops of KQX. Note that "X" can be a king or queen. Q9X and 98X also make 640 combinations for 1920 total or a probability of 9.8% for a open end straight draw.

In addition, flops of AQ8 or K97 give you an 8 out double gutshot straight draw. there are 4*4*4=64 combinations each. So the probability of flopping an eight out straight draw is (1920+128)/19,600 = 10.5%.

Also, flops of AKQ, KQ9, Q98 and 987 (64 combinations each) will give you a straight. So the probability of flopping a straight or 8 out straight draw is 11.8%.

These probabilities are the same for all of the offsuit max-stretch connectors, JTo thru 54o. If your hand is suited, you would have to subtract the times that you flop a straight-flush or flush.

Lost Wages

[M.B.E.]

[ QUOTE ]
If you hold JTo, KQ can make 16 combinations, which can combine with 52-(2 in your hand)-(2 for the KQ)-(4 aces)-(4 nines)=40 other cards. 16*40=640 flops of KQX. Note that "X" can be a king or queen.

[/ QUOTE ]
No, this method double-counts some flops. You say that "X" can be a king or queen. Well, in reality Ks-Qh-Kc (where X is the Kc) and Kc-Qh-Ks (where X is the Ks) are the same flop, but your method counts it as two.

There are only 592 combinations, not 640.

[ QUOTE ]
In addition, flops of AQ8 or K97 give you an 8 out double gutshot straight draw.

[/ QUOTE ]
The question specifically asked about open-ended straight draws. You're right that it is more useful to include double-gutters, however.

[Lost Wages]

Please show your calculation for arriving at 592 combinations.

Lost Wages

[M.B.E.]

There are 24 combinations of KKQ (six ways to choose a pair of kings, multiplied by four ways to choose a queen). Likewise there are 24 combinations of KQQ. Your method arrived at 640 combinations, but it counted each combination of KKQ and KQQ exactly twice. So subtract 48 from 640 to get 592.

Or, you could note that there are 544 combinations of KQx where x is not an A, K, Q, or 9 (4x4x34=544). 24+24+544=592.

[Bozeman]

Do people ever use this method:

y=544+x=640-x ==> x=48 ==> y=592 ?

That is, count the possibilities with and without double counted hands, and solve for the difference? It seems that the 544 and 640 type calculations are easier in general than enumerating the special cases.

Craig

[bigpooch]

Right! Or, more simply, ignoring the factor of 16, you
could even say:

y=34+x=40-x ==> x=3 ==> y=37.

One can also just take 40 and subtract 1/2(double counted
Kings and Queens).