raising with draws - how much FE needed?

[Galwegian]

I'm sure that many of the more mathematically inclined posters might have
worked this out for themselves, but I thought that some people might
find the following useful. If you don't like math, you probably
don't want to read this. If you are a math geek (like me), you might
check my calculations to see if I am right. I worked this out while
I was analyzing a play that I had made in a live game last night and
I came up with a nice little formula that people might find helpful.
The basic situation is the following.

In NL holdem you have a 8/9 out draw on the flop and your single
opponent bets into you. How much fold equity do you need to make it
correct to reraise all in?

Say that

P = pot on the flop

S = effective stack

B = opponents bet


Now E(fold) = S. That is, the expected value of your stack given
that you fold is S. We want to compare this with E(all in) - the
expected value of your stack if you raise all in. Say that your
opponent will fold x of the time and call 1-x of the time (0<= x <=
1). We want to know how big x must be in order to make raising all
in at least as good as folding.

I will assume that if he calls, you have a 1 in 3 chance of winning
the hand. This is approximately true for many common draws (e.g
flush draw, oesd with a backdoor flush draw)

Then

E(all in) = x(P+B+S)+1/3(1-x)(2S+P)

since if opponent folds, your stack will be P+B+S and if he calls,
your stack will on average be 1/3(2S+P). Therefore raising all in is
better than folding if

x(P+B+S)+1/3(1-x)(2S+P) > S

Rearranging this yields

x>(S-P)/(S+2P+3B)

The nice thing about this is that this formula is just about useable at the table (there is only one division involved - my real time arithmetic skills struggle to cope with most formulae that involve lots of divisions).

As an example, here is the hand that prompted me to work this out.

I was at a final table of the local cardroom tourney and I picked up
Th9h in the big blind. There was 12K in the pot and the flop was
QhJs5c. I had 29K behind and I checked to my opponent who bet 7K
into the pot (he had me covered). So S=29, P = 12 and B=7. Therefore
(S-P)/(S+2P+3B) = 17/74 which is about 0.23.

Thus I needed him to fold at least 1/4 of the time to make the all
in raise better than folding (at least in terms of tournament
chips). Given my knowledge of this particular opponent, I think that
he won't fold 1/4 of his range here so I should have folded. In the
actual event, I reraised all in and busted from the tourney when he
called with QT.

This formula only works if you have close to a 1 in 3 chance of
winning when called, but it would not be hard to modify to other
drawing situations as well. Any comments?