Probability Question about Full Houses So I've figured out the probability of getting a full house.  I want to show that P(getting a full house | my first card is the 9h) is the same.
Essentially, I want to show that getting a full house and having your first card be the 9h are independent.  Then I can take it from there.
Intuitively, I'm not getting any information about whether I'm going to get a full house after the first card since there's no card I can get that makes it more / less likely that i end up with a full house.  Put another way, each card is equally likely (when the first card) to result in a full house.
I feel like this isn't a particularly rigorous way of going about showing independence.  Anyone have a better argument?
Thanks,
Mariogs
 A: We can just do a brute force approach - the only more elegant approach I can think of is your symmetry-based one.  Let's consider "ordered hands", in which it matters which order the five cards are picked in.  Of those, the number where the first card is the 9H is:


*

*those in which the 9H is part of a three of a kind: there are 3 ways to pick the other two nines, 12 ways to pick the rank of a pair, 6 ways to pick their suits, and 4! ways to order the whole thing.

*those in which the 9H is part of a pair: 3 ways to pick the other nine, 12 ways to pick the rank of the three of the kind, 4 ways to pick their suits, and 4! ways to order the whole thing.
So the number of ordered hands which are full houses and which start with a 9H is
$$ (3 \times 12 \times 6 + 3 \times 12 \times 4) \times 4! = 360 \times 4! $$.
The total number of ordered hands which start with 9H is of course 51 \times 50 \times 49 \times 48 = 51!/47!$.  So the probability that an ordered hand is a full house, given that it begins with a 9H, is
$$ {360 \times 4! \times 47! \over 51!}. $$
You can work out that the number of ordered hands that are full houses, overall, is $13 \times 4 \times 12 \times 6 \times 5! = 3744 \times 5!$ (pick the rank of the three of a kind, the suits, the rank of the pair, the suit, and the order).  The total number of ordered hands is $52!/47!$, so the probability that an ordered hand is a full house is therefore 
$$ {3744 \times 5! \times 47! \over 52!}. $$
Finally, you just need to check that these two probabilities are the same.  They are - that reduces to the fact that $3744/360 = 52/5$. 
A: If the first card of a possible full house is $9$ of hearts, there are $3$ more of that same rank in a standard deck.  The same is true for any rank in the deck.  Since a full house is $3$ of one rank and $2$ of another rank, it should be clear that getting a $9$ of hearts as the first card doesn't change the probability of a full house for that hand (assuming a fair deck and random card draws without replacement during the hand).
The probability of a full house in a fair $52$ card deck is:
$13\choose 1$ $4 \choose 3$ $12 \choose 1$ $4 \choose 2$ / $52 \choose 5$ = $3,744$ / $2,598,960$ = $1$ / $694.1666666$...
