If you draw three cards, each from its own standard 52-card deck how
many combinations are there for each of the following? Exclude
combinations which can also be part of a higher ranked hand (the
rankings are given in parenthesis next to each hand-type). For
example, if you draw three queen of diamonds, it will not count as a
pair, three of a kind or a flush because queen of diamonds has the
highest ranking.

* Three queen of diamonds (rank 8)- three queens, all diamonds.
* Royal Flush (rank 7)- Ace, King, Queen of same suit
* Straight Flush (rank 6)- 3 Sequential cards of the same suit
* Three of a kind (rank 5)- 3 cards of the same value
* Straight (rank 4)- 3 sequential cards but not of the same suit
* flush (rank 3)- 3 cards of the same suit
* pair (rank 2) - 2 cards of the same value
* None of the above (rank 1) - none of the above combinations.

Naturally, the total number of combinations is 52*52*52 so the sum of
all of the above should equal this.

8-1
7-24
6-254
5-831
4-4330
3-8458
2-28080
1-98630

I get slightly different nswers.

* Three queen of diamonds (rank 8)- three queens, all diamonds.
Evidently there is exactly one combination.
* Royal Flush (rank 7)- Ace, King, Queen of same suit
Pick a suit (4 choices). Each Royal Flush of a given suit can then be
drawn in six different ways (orders). So there are 24 combinations.
* Straight Flush (rank 6)- 3 Sequential cards of the same suit
I assume that Ace, 2, 3 of the same suit is allowed. Pick any suit (4
choices) and any card between a 3 and a K as the top card of the
straight flush (11 choices).There are 6 ways to draw any such straight
flush, for a total of 264 combinations. If Ace, 2, 3 is not allowed,
there are 240 combinations.
* Three of a kind (rank 5)- 3 cards of the same value
Pick the kind (13 choices). There are 4*4*4 = 64 ways to produce such
a hand, for a total of 832 combinations. One of these is three queens
of diamonds, leaving 831 combinations.
* Straight (rank 4)- 3 sequential cards but not of the same suit
Assuming again that Ace, 2, 3 counts as a straight, there are 12
choices for the top card of the straight. Each such hand can occur in
4*4*4 = 64 combinations of suits, of which 4 are actually straight
flushes or royal flushes - leaving 60 combinations of suits. Each can
be drawn in 6 different ways from the three decks. The total number of
combinations is 12*60*6 = 4320. If Ace, 2, 3 does not count as a
straight, the number is 11*60*6 = 3960.
* flush (rank 3)- 3 cards of the same suit
There are 4 suits, and 13*13*13 ways to draw a flush in any given
suit, for 4*13*13*13= 8788 combinations. Of these, 288 have higher
rank (royal or straight flush) if Ace, 2, 3 is allowed, and 264
otherwise, leaving 8500 combinations or 8524 combinations.
* pair (rank 2) - 2 cards of the same value
The card NOT of the same value can be in three different positions.
Next there are 13 possible values for the pair and 12 for the third
card. There are now 4*4*4 ways to produce such a hand (4 suits for
each card), of which 4 will produce a flush, for a total of 3*13*12*60
= 28080 combinations.
* None of the above (rank 1) - none of the above combinations.
The total number of hands is 52*52*52 = 140608, of which
1+24+264+831+4320+8500+28080=42020 have higher rank if Ace, 2, 3 is a
straight, and 1+24+240+831+4960+8524+28080 = 41660 have higher rank if
this does not count as a straight. This leaves 98588 hands of rank 1
in the former case and 98948 hands in the latter case.

hpe-ga,

is there an error in your math on flush computation? 8788 makes sense,
but don't you have to subtract straight flush, royal flush *and* three
of kind of the same suit? This would be 8788-264-24- (52*1*1), or 8448
or am I missing something?

londonplayer

You're right, londonplayer.