I’m trying to figure out what the odds of getting a royal flush is in the following situation:
I’m holding Ah and Kh
The flop is Qh 2c 7d
Now I need the Jh and 10h to be dealt on the turn and river. There are 47 unseen cards. I’m thinking the formula is this:
2/47 x 1/46
That leaves me with a .09% chance of getting the royal flush. I think I’m doing something wrong as it seems only needing two cards, I should have a better chance than that.
Does anyone know if I’ve done something wrong?
I think what you want to do is to compute:
(the chance that the 4th card isn’t one that I want) * (the chance that the 5th card isn’t one that I want)
Then subtract that combined chance from 1.
So I get:
1 - (45/47)*(46/47)
which works out to about 6.3%
bstenger, your answer looks good to me. Another way to consider it–There are 1081 different possible pairs of cards that can turn up next. Only one of those pair will give you the royal flush.
ebb, the mistake you’re making is that you’re calculating the probability of getting either the ten of hearts or jack of hearts (but not necessarily both).
Okay, I need exactly two cards of which there are ony two of in that scenario.
Lets say the board is 1H 4D 8S
I’m holding KC and JH
If I use the formula I used in my OP to find out what the odds are of getting at least 3 kings are, it should give me a result that is greater than getting a royal flush in the other example because there are three kings left:
3/47 x 2/46
That leaves me with a .00277% chance of getting at least 3 Kings.
I must be doing something wrong. It has to be harder to get get the exact two cards in the first example as it is in getting any two kings in this one. 
It’s actually a .00277 chance, or .277% chance.
You forgot to convert the decimal to a percent. It is actually .2775% in the kings example.
Oh yeah, forgot about those zeros. :o
Thanks.