This rule says it's OK to keep a 7 and discard a 6 (6 - 2 == 4) but that we shouldn't keep a 9 but dump an 8 if we have a 2 in the hand (9 - 2 > 4).
But, it is possible to construct situations where this heuristic excludes the best draw. Suppose you have 289TT, villain has a pat J, and there are just six cards left in the deck: three sixes and three eights. Then drawing one to 289T wins 50% of the time. But drawing two to 29T wins 60% of the time: out of the 15 possible combinations, you will get 88 three times, 66 three times, and 86 the remaining 9 times.
Unfortunately, this construction can be applied to smaller draws too. With 23588 (or 23577 or 23599), if the deck contains three fives and three sixes you are in the same situation and must break my proposed heuristic and ditch a 5 to have the best chance of winning.
However, I think there's some hope. Our intuition is that there ought to be some general rule about not throwing away good cards (and keeping worse ones.) With an infinite deck it should be possible to prove such a rule. The question then becomes how small we can make the deck without invalidating the rule.
Suppose EV(abce/d) > EV(abcd/e) where a < b < c < d < e in rank. Such a case is a counterexample of our desired property. (It might be better yet to draw to abc, of course.)
A draw to a particular set of cards results in a set of result tuples (p,H,v) == (probability of the result, Hand made, and EV against opponent's hand or hand range.) Denote this set by R(draw).
Thus EV(abce/d) = sum over (p,H,v) in R(abce) of p * e.
Compare R(abce) and R(abcd). The only hand they have in common is abcde. Otherwise for every result (p,abce[x],v) there is a corresponding hand (p,abcd[x],v'). The probabilities for these corresponding results must be identical by the assumption of an infinite deck. So for the second EV equation to hold, there must be some subset such that v > v'.
What values of [x] can be in this subset? [x] = a, b, c does not work since each of the resulting pair hands is stronger with the smaller kicker (d) so v <= v'. We also note that abce[e] is weaker than abcd[d], and that abce[d] and abcd[e] are identical. So if we modify our correspondence to match up these two pairs (instead of d<->d and e<->e; it is still 1:1 and the probabilities match up) we still have v <= v'. So [x]=d or e is out, too. If [x] does not form a straight or flush then abce[x] will always be weaker than abcd[x] because d < e. So the subset with v > v' must contain hands which are straights and/or flushes. Thus abcd must be four to a straight and/or flush.
We can apply the same reasoning to abe vs. abd, and ae vs ad The results are in one to one correspondence, with some fiddling in the 'd' and 'e' cases to compare pairs with pairs, trips with trips, etc. The only way drawing to 'e' can be better is if drawing to 'd' forms the basis of a straight or flush.
We showed that in the infinite-deck case, if drawing to 'e' is superior, it is because 'd' forms a straight or flush. Thus if 'd' does not form a straight or flush (Tim's rule), drawing to 'e' cannot be superior. But the inverse is not true--- drawing to 'd' might still be better even with the risk of a very high hand.
The above proof shows that the key step is the 1-1 correspondence between result sets. But this 1-1 correspondence is broken in a finite deck; the probability of drawing a 'd' is not the same as drawing an 'e'. However, all the other probabilities do match up! So it is really only the kernel consisting of draws with a 'd' or 'e' which matter.
This gives us one constraint: if the probability of drawing a 'd' and the probability of drawing an 'e' are identical, then the infinite-case reasoning holds and the heuristic is valid.
We can also bound the effect of the mismatched 'd'/'e' probability. Let's assume there is no straight or flush here so that the only cases which might differ are those containing d's and e's.
We care about abce[d] and abcd[e] which are of equal strength, and abcd[d] and abce[e], where the ee is the worse hand. The difference between P(e) and P(d) can be no more than 3/N, where N is the number of cards in the deck. (N=1 and N=2 are truly degenerate cases.) The best we can do is 'swing' that probability from a EV=0.0 hand to a EV=1.0 hand, the maximum increase in our expectation is also no more than 3/N.
So, that tells us how much we can possibly 'improve' by keeping the higher card. For example, with 44 cards in the deck, following the heuristic can be no more than an 7% error. This is still quite a bit.
We could tighten up that bound somewhat if we can arrive at an estimate of how much that the higher card hurts us on all the other corresponding hands. Are there degenerate cases where it doesn't hurt us at all? It seems unlikely unless standing pat is our best move.
For example: Villain has T9875 and is pat. You have T9732. Standing pat is obviously your best move. But if the other two tens are dead, drawing to T732 will win 40% of the time while 9732 will only win 35%. I haven't been able to translate this example into one that makes sense. But it suggests that there are cases in which the heuristic is invalid but in which its use won't affect the correct decision, and maybe those can be specified.
So... I identified one case where the heuristic can safely be used. Maybe a little further thought will solve this completely.
Also I didn't really solve the "draw two" vs. "keep a high card" case for the infinite deck.