4
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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

4
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.