Poker odds: Chances of a straight flush, given H4,H5

Question

I'm trying to learn Bayes's formula, and am coming up with some poker problems to learn this.

My problem is as following: given a $H4,H5$ ($4$ of hearts, $5$ of hearts) hand, what are the odds that I'll hit a straight flush?

My reasoning is like this:

$$\Pr(\text{straight flush}|H4H5) = (\Pr(H4H5|\text{straight flush}) \cdot \Pr(\text{straight flush})) / \Pr(H4H5)$$

Now, off of wikipedia, I learnt that:

$$P(\text{straight flush}) = 0.00139$$

Given that there are 36 ways to achieve a straight flush, and only 4 ways to have a straight flush with $H4,H5$ (namely $HA-H5, H2-6, H3-7, H4-8$), I calculated that:

$$\Pr(H4H5|\text{straight flush}) = 4/36 = 1/9$$

Now, how do we find $\Pr(H4H5)$? My reasoning was: There's a $2/52$ chance that we get dealt $H4$ or $H5$ as the first card, and then a $1/51$ chance that we get dealt $H4$ or $H5$ as the second card.

However, filling out those numbers says there is a 15% chance that this will happen. That numbers seems way to high to me. Surely, somewhere in my reasoning I'm making a mistake. Who can help?

Answer 1

Hint: You still have three cards to draw out of $50$. How many combinations of three cards result in a straight flush? How many total draws are there?

Answer 2

What are you asking is what is the probability of drawing two cards, without putting them back. Probability of drawing first card is 1/52. Now, there are 51 cards left. Therefore the probability of drawing second card is 1/51. Therefore PR(A & B) = PR(A)*PR(B).