2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $3744/360 = 52/5$.

$\endgroup$
0
$\begingroup$

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:

$13\choose 1$ $4 \choose 3$ $12 \choose 1$ $4 \choose 2$ / $52 \choose 5$ = $3,744$ / $2,598,960$ = $1$ / $694.1666666$...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.