# 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

$$(3 \times 12 \times 6 + 3 \times 12 \times 4) \times 4! = 360 \times 4!$$.
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$3744/360 = 52/5$. 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:$13\choose 14 \choose 312 \choose 14 \choose 2$/$52 \choose 5$=$3,744$/$2,598,960$=$1$/$694.1666666\$...