My friend and I were arguing and trying to find out the odds of hitting a runner-runner royal flush after the flop. We came up with the following formula, that I think can be applied to any runner-runner situation.

x= # of outs at turn.

[(x^2)-x]/2162

we arrived at that from the runner-runner royal question

at turn:
2/47
at river:
1/46
since the probabilities of both occurring are independent of each other, we multiply the probabilities, right?
so we have:
(2/47)(1/46)= 2/2162= 1/1081
or using the above formula:
[(2^2)-2]/2162=[4-2]/2161

Does everything look good? I'm super rusty from probability classes that I took ages ago.

There are 1081 unordered pairs of 47 possible turn cards and 46 possible river cards. Exactly one of these makes your royal flush, so the answer is 1/1081.

Your formula doesn't always work.

I assume you derived (x^2-x) by saying, I have x outs on the turn, and then after I hit one, I have x-1 outs on the river, so I can hit both in x(x-1) ways?

This is not always true for runner runner draws. Suppose you have a runner - runner straight draw. On the turn you have eight 'outs', on the river you have four outs, not seven.

It does work for runner runner flush draws though, since after you hit a flush card on the turn, there is one less flush card in the deck for the river.

Thank you.

OH, good call. Thanks!

Keep in mind that the situation is different if e.g. we are heads-up and all-in for a showdown at a tournament. Now we know two more cards, which reduces the number of outstanding combinations to 990 (45 choose 2), but may also reduce our outs.