Are these events independent? I have the following question:
If I hold a Jack and a nine, what is the chance of making a straight (five cards in a row) when the next three cards have been dealt?
My attempt at an answer:
There are three possible 5 card hands that make a straight:
7 8 9 T J
8 9 T J Q
9 T J Q K
There are 50 cards unseen, and there are 4 each of the 7's, 8's and 10's, so the possibility of making the first straight is the joint probability:
12/50 x 8/49 x 4/48 = 0.00327
What I can't work out is, are the probabilities for each of those 3 straights independent or not?
I was also trying to work this out using combinations, by the following method:
(4C1 x 4C1 x 4C1) / 50C3 = 4/1225 = 0.00327
and doing it that way, it's easier to see that there 64 possible combinations for the first straight and 64 combinations each for the other 2. So I'm pretty sure that this means that make one type of straight or another are independent events, but I'd be grateful if someone could maybe shed a bit more light on exactly why.
Many thanks
 A: We need a $10$.  We also need either (i)Q and $8$ or (ii) Q and K or (iii) $8$ and $7$.
Each of the patterns (i), (ii), and (iii) is equally likely.  So we find the probability of getting a $10$, Q, and $8$, then multiply by $3$. Note that the event of getting pattern (i), for example, is not independent of the pattern (ii). In fact the patterns are disjoint, so in a strong sense not independent. It is because these $3$ events are disjoint that we can add the probabilities of (i), (ii), and (iii) to get the required probability.
For the probability calculation, what you did is correct. In each case, you were finding correctly the probability of pattern (i) (or of (ii), or of (iii)). The probability of pattern (i) is
$$\frac{\binom{4}{1}\binom{4}{1}\binom{4}{1}}{\binom{50}{3}}.$$
Then the required probability is 
$$3\frac{\binom{4}{1}\binom{4}{1}\binom{4}{1}}{\binom{50}{3}}.$$
The calculation can also be done using your first expression. Again, we need to multiply by $3$.
Comment: In poker terms, the calculation makes sense only if the Jack and the $9$ are of different suits. If they are of the same suit, then the calculation has included the straight flushes, which are a much better hand.
A: There are 64 possible combinations for each straight, but the straight events can't all be independent because there is some overlap between them. For instance, the first two straights can't be independent because they both involve drawing an 8 and a T. Hopefully this illustrates how the second and third straight and first and third straights are not independent of each other either. 
