4 of us are playing a game with a traditional deck of 52 cards.  9
cards are dealt to player 1 (p1) and 13 cards are dealt to each of the
remaining players (p2, p3, p4), leaving 4 cards that are discarded. 
Assume that 6 of the 9 cards dealt to p1 are of the same suit (ie
hearts) and the remaining 3 cards are non-hearts.  What is probability
that one of the other players (p2, p3 or p4) are dealt the jack of
hearts along with 3 other hearts?  It doesn't matter if that other
player has more than 3 hearts in addition to the jack, just that they
have the jack and at least 3 other hearts.

First of all, note that there's a 6/13 chance that the jack of hearts
is in p1's hand.

On the assumption that p1 does not have the jack of hearts, there is a
13/43 chance for each of the other players to be dealt the jack, and a
4/43 chance that it winds up in the discard pile.

If we assume that a specific player other than p1 has the jack - say
p2 - then we have 6 remaining hearts and 42 cards left in total (p1's
nine cards are all accounted for). What's the probability that of p2's
12 remaining cards, at least three are hearts?

Well, he can have -
 three hearts in 6C3.36C9 = 20.94143280 = 1882865600 ways
  four hearts in 6C4.36C8 = 15.30260340 =  453905100 ways
  five hearts in 6C5.36C7 =  6.8347680  =   50086080 ways
   six hearts in 6C6.36C6 =  1.1947792  =    1947792 ways
for a total of 2388804572 ways.
The probability is therefore 2388804572 / 42C12 = 161889 / 749406 = 21.60%.

But remember, this only applies on the assumption that p2 has the jack
of hearts. If we still assume that p1 doesn't have it, the chances of
p2 having it are 13/43, so we get 13/43 * 21.60% = 6.53%. But p3 and
p4 have the same chance, so the probability that one of them has it is
3*6.53% = 19.59%.

We are still assuming that p1 doesn't have the jack of hearts. The
probability of this is 7/13, so the final answer to your question is
7/13 * 19.59% = 10.55%.

Wow!  I tried a couple of ways trying to come up with an answer but
none were close and none were as thoughtfully explained as yours. 
Thanks, manuka.