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Poker nerd question here... I'll start out with some definitions of standard poker terms:

A flop is 3 cards chosen from a 52 card deck. For example: 789 (suits are irrelevant to this question). If you draw a fourth card from the deck you get a turn card.

A straight is 5 cards in succession, for example 56789. When making a straight, an Ace can count both as 1 and 14. For example, TJQKA is a straight, and so is A2345. However, you cannot "wrap around the ace": For example, KQA23 is not a straight.

Finally, a holdem hand is just 2 cards chosen from a 52 card deck.

Now, here's a definition I've made up myself. Let's say we are given a flop F. I define a straight making card for F as a turn card T for which there is a holdem hand H such that H didn't make a straight on F, but H makes a straight on F and T combined. A straight making rank is the rank of a straight making card.

For example, 5 is a straight making rank for the flop 789 because the holdem hand 46 makes a straight on 7895, but not on 789. (The straight being made is 45678). All in all, 789 has four straight making ranks: 5,6, T and J. (4 and Q don't count because 56 (and TJ) were already making straights.)

EDIT: In the context of this question, the holdem hand must use both of its two cards to make a straight. So, for example, the hand K6 making a straight on a 7895 board doesn't count.

Now I have this conjecture:

(*) No flop has more than 6 straight making ranks.

Having examined a number of different flop types, I'm pretty sure that (*) is true. I'm just wondering if someone can come up with an elegant, succinct proof? (I admit that "elegant" and "succinct" are subjective terms).

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"$5$ is a straight making rank for $789$ because the holdem hand $46$ makes a straight on $7895$ but not on $789$" But it makes the straight $56789$, which doesn't use both of the holdem cards. So I'm confused. – Gerry Myerson May 8 '11 at 12:43
@Gerry Myerson: Well, if the holdem hand is 46 and the board is 7895, there are two possible straights: 5 6 789 (as you mention), but also 4 5 6 78. The latter uses both holdem cards. It does not matter that 5 6 789 is a higher straight. I thought of this question while playing Omaha poker - in this context the restrictions make sense. Maybe I should have framed it as an Omaha question, but that would've required even more poker term definitions. – user10632 May 8 '11 at 13:32
OK, so, instead of "the holdem hand must use both of its two cards" you meant "it must be possible to use both of the holdem cards to make a straight." Anyway, I'm sure (*) is true, but maybe the only proof is by an exhaustive list of cases. I don't see any clever way. – Gerry Myerson May 9 '11 at 0:29
up vote 1 down vote accepted

Not a solution, but some potentially helpful observations.

Since you're looking for a maximum numbers of SMRs, you should assume your flop is in the "middle" of the ranks. Anything at an "edge" (near A or K) will only have fewer.

Let the flop ranks be $f_1, f_2, f_3$ in non-decreasing order, with differences $d_1 = f_2 - f_1$ and $d_2 = f_3-f_2$. In order to make a straight, you need $\min(d_1,d_2) \leq 4$. (The flop 2, 7, Q has no straight making ranks; the flop 2, 6, 10 has 6 SMRs.)

Looking at isolated cases (where one card is too far away to matter), a $d_i = 4$ has 3 SMRs (the ranks in between), $d_i = 3$ has 4 SMRs (the ranks in between, one below, and one above), $d_i = 2$ has 5 SMRs (the rank in between, two above and two below), $d_i = 1$ has 6 SMRs (ranks 3 below and 3 above), $d_i = 0$ has no SMRs.

To work through the combined cases, you can organize by $r = f_3 - f_1$. If $r \geq 9$, you just have one of the isolated cases above. For $r \leq 8$, you have to think through how the previous cases combine. At least, using symmetry, you can assume $d_1 \geq d_2$. There are 10 cases to work through, listed below as $r = d_1 + d_2$, two-part partitions of $r$ with largest part 4. Note that any $d_i = 0$ reduces to one of the isolated cases above.

8 = 4+4

7 = 4+3

6 = 4+2 = 3+3

5 = 4+1 = 3+2

4 = 3+1 = 2+2

3 = 2+1

2 = 1+1

So not elegant, but it's not really that many cases to work through.

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