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I'm hoping to generalize this solution, but for the purposes of simplicity let's solve it first for just three cards.

For my purposes I'm not interested in removing "higher order" solutions such as is common for 5-card poker deck probabilities and others (e.g. Royal or Straight Flush) I want to collectively determine the probability of getting any flush.

I'm fairly confident the frequency of getting a full flush is fairly straightforward:

(13 choose n)*(4 choose 1) 

where n is the number of cards being dealt, in the case of three cards, n = 3.

However, what about the frequency of getting a partial flush. That is, 2 of the three cards are a flush.

Would this be:

(13 choose 2)*(4 choose 1)*(39 choose 1)

Since the last card can be anything I want? So... let's say I'm dealt k cards with n of them having a matching suit (and therefore partial flush). Is the probability going to be

(13 choose n)*(4 choose 1)*(39 choose [k-n])

By the way, deriving the probability from these is assumed to be

frequency/(52 choose k)

where, again, frequency is the number of ways to get the hand, and k is the total number of cards dealt.

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You are correct for full flushes and for the flush missing one card. You can get into trouble in your case of $k$ cards, $n$ of which match in suit. As long as $2n \gt k$ you are fine, but if $k \ge 2n$ you can have have two suits with each one having $n$ cards and you will double count these.

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  • $\begingroup$ Ah, right. Since the additional cards can also match suits. $\endgroup$ – Meshach Jun 6 '14 at 0:14

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