What's the probability of a straight in $5$ card poker given $5$ and $7$ of hearts?

Question

Using a standard $52$ card deck, if you are given the $5$ and $7$ of hearts from it, what is the probability that you end up with a straight if $3$ additional cards from that same deck are given to you?

I was trying to set this up and got a little confused as to how to do this problem because if the next card is a $3$ or $9$ of any suit, then there are $2$ possible ranks for the following card and if the next card is a $4$ or $8$ of any suit, then there are $3$ possible ranks for the following card. Finally, if the next card is a $6$ of any suit, then there are $4$ possible ranks for the following card.

I'm not sure how to compute this because the probability changes for different cards. Do I add them all up?

Edit:
My attempt at a solution:

$3\cdot\frac{({4\choose 1}(3\cdot2\cdot 1) - 1)}{{50 \choose 3 }} \approx 0.35\%$

Answer 1

We definitely need the $6$ to make a straight. There are $4$ of those cards. For the last $2$ cards, we need either a $8$, $9$ combo, a $4$, $8$ combo, or a $3$, $4$ combo.

There are $16$ choices for each needed combo.

We have to subtract out the $3$ cases where we get straight flushes.

So the answer is: $$\frac {4*(16+16+16)-3} {50 \choose 3} = \frac {189} {19,600} \approx 0.00964$$

Answer 2

I think a simpler way to proceed would be to observe that if you have a straight, it must be one of

• 5–6–7–8–9
• 4–5–6–7–8
• 3–4–5–6–7

You can calculate the probabilities for each of these three separately (each is straightforward since you know just what three cards you need) and then, since they are disjoint, add them up.

Answer 3

This should be self explanatory considering the comments made above.

$$P(straight) = 3\frac{ (12\cdot 8\cdot 4-6)}{50\cdot 49\cdot 48} = \frac{27}{2800}$$