Probability that 5/9 players flop a heart flush in Texas Holdem need some help with this one.
Want to know the probability of 5 players in a 9 player game, all get dealt 2 hearts, and then the flop comes 3 more hearts.
Cheers for all the answers. Makes a lot of sense now.
I understand that there are 9c5 ways of arranging the 2card heart players out of 9. I am curious as to why it is this simple, and doesn't take into account the probability that the remaining hearts don't get drawn by the other players, or in the burn pile before the flop.
If there is any confusion. 9 Players in the game, 18 cards get dealt left to right 1 at a time to all players. 1 card is then burnt. 3 cards are then dealt (the flop) this is where the question ends.
 A: You use the fact that the order in which cards are dealt does not matter (and also that the burn card doesn't matter). 
You can find the probability that $5$ select players get the hearts, by simply assuming that you turn the flop and deal these players first. There's $13!$ ways the first $13$ cards will give only hearts out of a total of $52!/39!$ ways of doing that. That is probability of $13!39!/52!$
Now there are $9!/5!4!$ ways of choosing the players to get the hearts and all these are complementary events of propability of $13!39!/52!$. Consequently the probability is:
$${13!39!9!\over5!4!52!} = {9\over 45358111400}$$
One way to realize the ordering the dealing is done doesn't matter is to see that we're basically counting the number of arrangements (shufflings) of the deck that has hearts in prescribed positions. In my reasoning we're interested in those that has hearts at the first $13$ positions, but in fact these are exactly as many as those that has hearts in any select $13$ positions. 
We see this since we can craft a way $\Pi$ to rearrange the deck so that the cards at the select $13$ position are placed on the top (one should note that after constructing that rearrangement we can also create a scheme that will restore the order to the oringinal). 
The important thing here is that if we have an enumeration $S_j$ of all possible shufflings of the deck then $\Pi S_j$ will also enumerate all shufflings. But if $Sj$ has hearts on the select positions $\Pi S_j$ will have it on the first $13$ positions. 
This means that if we list all shufflings $S_j$ along the shufflings $\Pi S_j$ and count the number of shufflings in the $S_j$ column with hearts on select positions and $\Pi S_j$ with hearts at the first positions we see that we will count one on both sides simultaneously and therefore reach the exact same result.
A: There are ${9 \choose 5}=126$ ways to pick the players who get the hearts.  The chance of each of those coupled with the flop all being hearts is the chance you draw the $13$ hearts first out of a deck, which is the inverse of ${52 \choose 13}$ so the overall probability is $$\frac{126}{52 \choose 13}=\frac 9{45358111400}\approx 2 \cdot 10^{-10}$$
