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.

  • $\begingroup$ Completely wrong. A flush only requires that all the cards be of the same suit (e.g. 2347Q♤). $\endgroup$ – Parcly Taxel May 6 '17 at 1:52
  • $\begingroup$ Yes, in this context, non-consecutive means a flush but not a straight (i.e. not a straight flush). So as long as all cards are of the same suit, and not a straight (in some order), then the hand ranks as a flush. $\endgroup$ – quasi May 6 '17 at 1:52
  • $\begingroup$ @Parcly Taxel -- Yes, but the question specifically asks for the probability a flush which is not also a straight. $\endgroup$ – quasi May 6 '17 at 1:54
  • $\begingroup$ @quasi I know right? $\endgroup$ – Parcly Taxel May 6 '17 at 1:54
  • 1
    $\begingroup$ @quasi Completely wrong reasoning of "non-consecutive". $\endgroup$ – Parcly Taxel May 6 '17 at 1:56

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 $${\large{\frac {\binom{4}{1}\left(\binom{13}{5}-10\right)} {\binom{52}{5}}}} $$

  • $\begingroup$ Thank you @quasi , that absolutely makes sense $\endgroup$ – BabaSvoloch May 6 '17 at 2:04
  • $\begingroup$ @BabaSvoloch -- See my latest edit. In poker, $A,2,3,4,5$ is a $5$-high straight (with Ace regarded as $1$), but $10,J,Q,K,A$ is an Ace-high straight with Ace regarded as the next rank up from King. Hence there are $10$ straights in a given suit, not $9$. $\endgroup$ – quasi May 6 '17 at 2:29
  • $\begingroup$ Thank you, @quasi! It was marked correct by the online system $\endgroup$ – BabaSvoloch May 6 '17 at 2:40

I believe the answer should actually be $${{\frac {\binom{4}{1}\binom{13}{5}} {\binom{52}{5}}}}$$

I understand quasi is subtracting all the straight flushes but as the name implies straight flushes are flushes. Therefore, should not be subtracted.

  • $\begingroup$ The question itself says "5 nonconsecutive cards each of the same suit", so straight flushes should not be included. $\endgroup$ – Kevin Long Nov 3 '17 at 23:36
  • $\begingroup$ My bad. You are correct. $\endgroup$ – EFiore Nov 5 '17 at 15:21

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