The following is the problem put to me:
A 5-card poker hand is dealt from a well shuffled regular 52-card playing card deck. Find the probability that the hand is a Flush (5 nonconsecutive cards each of the same suit).
I am completely unfamiliar with poker, and just learning the principles of probability. I would appreciate some help understanding the problem, and figuring out how to proceed from there.
For example: when it says "5 nonconsecutive cards," does it mean no single card can be consecutive to any other, or does it only mean that they cannot all 5 be consecutive to one another? (Same goes for other combinations- are 4 in a row disallowed? 3 in a row? 2? Are 2 and Ace considered consecutive?)
For the sake of beginning the problem, I'm going to assume none of them can be consecutive to any other, and that the 2 and the Ace are not considered to be consecutive. If I'm wrong I would appreciate someone correcting me.
Here's how I think I might begin $\ldots$
First, we're choosing cards of the same suit, so we have 13 of any one suit.
None of these can be next to each other, so we'd have to consider the different ways they could be alternating.
They could either be in the alternating slots: 2,4,6,8,10,Q,A , of which there are 7 possibilities.
Or in the alternating slots: 3,5,7,9,J,K , of which there are 6 possibilities.
So I figure, the answer to this would be:
$(7C5 * 6C5) / 52C5$
Can anyone either verify or correct my assumptions about the nature of the question, and point me in the right direction if my logic/solution is incorrect? Thank you very much.
P.S. I assume this question is asking ONLY about the probability of getting a flush, rather than a straight flush or a royal flush (which I believe is a "thing"), since it's asking about nonconsecutive cards specifically.
Best Answer
First choose the suit: $\binom{4}{1}$ choices.
Next choose the ranks . . .
Start with any $5$ ranks: $\binom{13}{5}$ choices.
Subtract the straights: $10$ possible straights in that suit (assuming an Ace can rank as either low or high).
So the number of qualifying hands is $${\small{\binom{4}{1}\left(\binom{13}{5}-10\right)}}$$ Hence the desired probability is $${\large{\frac {\binom{4}{1}\left(\binom{13}{5}-10\right)} {\binom{52}{5}}}} $$