[BonJoviJones]

Just for kicks and mental exercise, last night I tried to calculate the odds of flopping an open-ender in hold'em, given the starting hand of 67. To simplify my calculations, I didn't worry about "accidentally" getting a real straight.

I got stuck after awhile, and couldn't figure it out. Help on my thinking process would be appreciated.

There are three sets of two cards that give me an open-ender (45, 58, 89). Each of the three is equally likely. Thus I figured that the answer was:

Let P(OE) = Probability of flopping one of the three success hands.

( 3 * P(OE) ) / (50c3)

This seems fairly obvious to me. However I'm having trouble finding P(OE). I don't particularly want an answer to what P(OE) is, but just a hint or two on where to go next.

Thanks.