so i was sweating my little brother play, and i told him to fold J7s, which he did. the flop came 8 9 T giving him the straight and he was pretty mad. We tried to figure out the probability of flopping a straight with a 3 gapper like j7 or k9, are our methods sound:
there are 52x51x50 flops ~133k, assuming both hole cards are used, there are 64 combos that give us straights (four 8's * four 9's * four Ts)
64/133,000 = the answer?
somewhere in the back of my mind i think theres like, the probability of each card which is 1/13 aka .076, so you can Pvalue(.076,.076,.076) ?

also, in the total calculation of flop possibilities, despite you and your opponents holding cards it's still 52x51x50, right? since you're not calculating for card removal/where the cards are, since its out of the total possibilities (other than the negligible removal effects your 2 known cards would make)

Given hero hand, there are C(50,3) flop combinations. To flop a 3 gapper straight, there are 4^3 successful combos. So the probability is 4^3/C(50,3) = 64/19600 = 0.33% or about once in every 300 hands.

4 / 50 * 4 / 49 * 4 / 48 * 6
= 0.33%

you know your own two cards
and there are 6 orderings 89T, and 8T9, 98T, 9T8, T89, T98

With total nerdiness, I should point out that the above result is based on assuming that hero's 3 gapper is off-suit.

If it is suited, there is one flop combo that is a straight flush, so the probability of a straight decreases from 0.327% to 0.321%.

HA HA @ straight flush :'D

so the 50x49x48 to find total flops is incorrect because it counts multiple rearrangements of the same cards?
im not sure i understand why the total flop combos isnt 117k

No, it is not incorrect assuming it is used correctly. In your use above, you used combinations in the numeerator and permutations in the denominator in your probability calculation. See below:

There are two basic ways you can count an arrangement of things – permutations and combinations. Permutations consider different orderings, first, second, etc. and combinations do not.

So, assume we are considering a flop consisting of Kh, Qh and Jh. There are 6 permutations, KQJ, KJQ. QKJ. QJK. JKQ. JQK. But, there is only one combination as order is ignored.

You can use either in calculating probability as long as you are consistent with all the counts in the equation.

For 50 cards and a flop 50x49x48 is the numberof permutations. Since there are 6 possible orders, divide that product by 6 and you get the number of combinations, 19,600.

OOOOOOOOOOOOHHH, TYVM !!!!!!

out of curiosity wouldn't it be 52x51x50, since were looking at all total possible flops, and whats in my hand is random

No, you specified a 3 gapper held by hero so there are only 50 remaining cards in the deck. Though no specific 3 gapper is specified that will have no effect on the calculation.

Both Statmanhal and Longspring receive an A+ on their math exam.