You are given the digits 0 to 9 in 4 poker suits.

Distribute them onto the 8 highest poker hands (make one of each) of 5 digits each. They are

  • Royal Flush (9 to 5 of one suit)
  • Straight Flush (consecutive digits of one suit)
  • Full House (3 identical + 2 identical)
  • Four of a Kind (4 identical)
  • Straight (5 consecutive)
  • Flush (5 of the same suit)
  • Three of a Kind (3 identical)
  • Two Pair (2 identical + 2 identical)

Can you do that so the sum of the digits in each poker hand is at least 17?

Bonus: Can you do the same with a 3 in the Four of a Kind hand?

  • $\begingroup$ Each hand is valued by the sum of the values you chose for the corresponding cards $\endgroup$
    – Nurator
    Mar 9 at 12:56
  • $\begingroup$ Because you have to distribute the deck to the different poker hands? But sorry if I used the wrong tag for this question $\endgroup$
    – Nurator
    Mar 9 at 13:06
  • $\begingroup$ Oh i see now what you meant. We have to make exactly one of each of the hands $\endgroup$
    – JLee
    Mar 9 at 15:01
  • $\begingroup$ Nah I'm just slow sometimes $\endgroup$
    – JLee
    Mar 9 at 17:56

1 Answer 1


Here is a partition where all hands are at least 17:

9♤8♤7♤6♤5♤ royal flush
7♧6♧5♧4♧3♧ straight flush
2♤2♡2◇2♧9♧ four of a kind
0♤0◇0♧9♡9◇ full house
8♡5♡3♡1♡0♡ flush
8♧7◇6◇5◇4◇ straight
1♤1◇1♧8◇6♡ three of a kind
4♤4♡3♤3◇7♡ two pair

Here is a partition with a 3 in the four of a kind hand:

9♤8♤7♤6♤5♤ royal flush
5♧4♧3♧2♧1♧ straight flush
0♤0♡0◇0♧3◇ four of a kind
6♡6◇6♧7◇7♧ full house
7♡5♡3♡2♡1♡ flush
5◇4◇3♤2◇1◇ straight
8♡8◇8♧9♧1♤ three of a kind
9♡9◇4♤4♡2♤ two pair

There are no partitions where all hands sum to at least 18 or that satisfy both the restrictions above. This is an exact cover problem; this is how the solutions above were found.

All hands at least 16 and 3 in the four of a kind:

9♤8♤7♤6♤5♤ royal flush
8♧7♧6♧5♧4♧ straight flush
3♤3♡3◇3♧5◇ four of a kind
0♤0◇0♧9◇9♧ full house
7♡6♡2♡1♡0♡ flush
8◇7◇6◇5♡4◇ straight
2♤2◇2♧9♡1♧ three of a kind
4♤4♡1♤1◇8♡ two pair

  • $\begingroup$ Nice! Can you get to at least 16 on all hands for the bonus question? $\endgroup$
    – Nurator
    Mar 9 at 14:01
  • $\begingroup$ @Nurator added. $\endgroup$ Mar 9 at 14:10

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