6 Card Poker hand with given probabilities Each player draws a hand of six cards from the standard pack.
two triples: three cards of one value, and three of another, for instance,
3♠ 3♥ 3♣ J♠ J♥ J♦;
three pairs: two cards of each of three different values, for instance,
Q♥ Q♣ 8♠ 8♣ 4♦ 4♣;
full house: four cards of one value, and two cards of another value, for instance,
5♠ 5♥ 5♦ 5♣ 7♥ 7♣.
(i)
What is the probability that you draw two triples?
I think the answer here is
$$\frac{\pmatrix{13\\2}\cdot\pmatrix{4\\3}^2}{\pmatrix{52\\6}}$$
(ii)
Then someone else draws from the 46 remaining cards. What is the conditional probability that they draw three pairs conditional on you having drawn two triples?
I did 
$$\frac{\binom{11}{3} \cdot \binom{4}{2}^3}{\binom{46}{6}}$$
but I don't think it is correct as you aren't just choosing from 46 random remaining cards, they must be 46 remaining cards given that two triples have already been drawn. 
(iii)
Another person then draws, from the 40 remaining cards. What is the conditional probability that they draw a full house, conditional on both you having drawn two triples and the other person having drawn three pairs?
This one was really confusing and I didn't know where to start
 A: It is good to question whether or not your sample space must change, however in these questions it will not change considerably.  Regardless what the hand for the first player is, there will be $46$ cards remaining to choose from for player two.  Given a specific hand from player one, there will be $\binom{46}{6}$ available hands for player two.  Although the specific cards themselves might change depending on which specific hand player one received, the number of cards (and hence the number of possible hands) does not change.
That being said, the answers you received for both questions (i) and (ii) are correct.
For question (iii), let $E$ denote the event that the first player received a two-triple and player two received a three-pair.
Break it into cases on whether the number used as a pair for player three has been seen before in someone else's hand or whether all four of the number are available in the deck.  Note, this can only occur if the number used in the pair for player three is one of the numbers used in one of the pairs for player two (there are not enough cards with that number in the deck otherwise).
$Pr(\text{player 3 gets a full house} | E) = Pr(\text{player 3 gets a full house and the pair is a previously unseen number}|E)+Pr(\text{player 3 gets a full house and the pair is a one of the numbers used by player 2}|E)$
Hopefully you can take it from here.

$\frac{\binom{8}{1}\binom{7}{1}\binom{4}{2}+\binom{8}{1}\binom{3}{1}}{\binom{40}{6}}$

