Prob. of flopping one ace and one jack

[mdb]

Sklansky says in his no limit hold'em book on page 36 that the probablility of flopping exactly one jack and one ace are 0.1% when one player has a pair of jacks and the other a pair of aces.

I think it should be [2/48 * 2/47 * 44/46]=0.177%.

[jay_shark]

It should be 2*2*44/48c3 =0.0101

That is , there are two aces and two jacks left and you must choose one of each and 44 non jack-ace cards .

or , 2/48*2/47*44/46*(3!)=0.0101 . You multiply by 3! because of the following reason .

Card 1 is an ace and card 2 is a jack and card three is x
card 1 is a jack and card 2 is an ace and card three is x.
card 1 is an ace and card 2 is x and card three is a jack
card 1 is a jack and card 2 is x and card three is an ace
card 1 is x and card 2 is an ace and card three is a jack
card 1 is an x and card 2 is a jack and card three is an ace

Also , your computations work out to 0.00169 which is 0.169%