# 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?

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

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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.

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Thanks for the explanation Andre. I wonder if you could clarify when you say "the patterns are disjoint, so in a strong sense not independant". I thought that the patterns being disjoint would mean that they are independant. (Assuming these are all dealt at once). Because the 64 combinations of the first pattern are totally distinct from the 64 combinations of the second pattern etc. –  Joe Oct 25 '11 at 20:20
@Joe: "Independent" events means that knowing whether or not one has appeared will not help you guess whether or not the other has (also) appeared. However, if you know that you have 78, then you also know with certainty that you don't have QK. That's more information about the possibility of QK than you had before you knew 78. Therefore the events are not independent. –  Henning Makholm Oct 25 '11 at 21:44
Confusing disjoint and independent is common, and understandable. The words both have the connotation of separate. However, the event "we get a Jack and an $8$" and "we get a Jack and a Queen" are disjoint, they cannot both happen on the same deal. They are not independent: if the first happened, for sure the second didn't. Two events are independent if knowledge about the first tells us nothing about the second. Independence of events $A$ and $B$ means that to find the probability they both happen, we multiply. Disjoint means probability that both happen is $0$. –  André Nicolas Oct 25 '11 at 21:51
There is no overlap. The calculations are correct; the figure need only be multiplied by $3$. –  Brian M. Scott Oct 25 '11 at 20:21