Straight By River Given Middle Connector

OOPS!!!

I ignored the case of duplicate ranks such as 4 5 5 6 7 8 T , which, of course, is a 4 5 6 7 8 straight. So, the 6.54% result is only for straights without any duplicate ranks.

After adjusting, I got 9.44% for a straight holding 67 with at least one of these two dealt cards included in the straight. I refuse to guarantee that this is correct.

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Since all straights require a T or 5; would the probability increase with 54,56,T9, or TJ?

@Bigstack no − while it's true that 54/JT etc have a head start in that regard, they don't have a head start overall compared to other middle connectors like 76. For either 76 or 54 to make a straight, they need the board to contain at least one of four three-card combos:

54 needs either A23, 236, 367, or 678
76 needs either 345, 458, 589, or 89T

So 54 needs just as much help.

In general, what matters is the number of five-card straights the connector can be part of, as well as how many hole cards are used by those straights.

Connectors 54 through JT can form 4 different five-cards straights using two hole cards;
43/QJ can form 3;
32/KQ can form 2;
A2/AK can only form one straight using both, plus one straight using just one (eg A2 with a board of TJQK), which is harder because using just one hole card requires a four-card combo on the board. For this reason, A2/AK are less likely to make a straight than 32/KQ.

THREE GAPPER STRAIGHT

I wanted to calculate the probability a NLHE 3 gapper would hit a straight by the river using both hole cards. If the 3 gapper is 5 9, then out of the five board cards you need at least one 6, 7 and 8.

I came up with a simple equation that gave a result that differed from a simulation result of 2.7%.

After much too much thought, the only -way I could analytically solve the question was to write out each possible good combination, such as 678xx, 6678x, 66678, etc., and add up the number of combos. I was then able to confirm the 2.7% result.

I woud think there is a more elegant way than the above.

Anybody?

Your 2.7% allows a 5/9 on the board; did you mean to?

To avoid casework we can use inclusion-exclusion, which is a little cleaner here. I'll show both, under the assumption we don't want a 5/9 on the board.

Casework: [4³(32C2) + 3(4C2)(4²)32 + 3*4³ + 3*4(4C2)²] / 50C5

PIE: (44C5 – 3(40C5) + 3(36C5) – 32C5) / 50C5

Both get the same result of ~1.96%

With PIE what we're doing is starting with all the boards lacking a 5/9, subtracting boards lacking a specific connecting rank, adding back the boards lacking two connecting ranks, then subtracting the boards lacking all three connecting ranks.

Yes

If I had to guess who would provide a neat analytical solution, heehaw would be he(haw).