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).