Raise your hand if you are ready for micropoker, which minimalistically resembles 5-card poker.

  • The deck has just 8 cards with 2 suits of 4 cards each.
  • A hand is dealt as 3 cards that are final, with no further play.
  • The two suits —“b” (bitties) and “i” (itties) — have different sets of consecutive integers to be determined. The bitties are  ‘ (b0)b  (b0+1)b  (b0+2)b  (b0+3)b ’  while the itties are  ‘ (i0)i  (i0+1)i  (i0+2)i  (i0+3)i ’  with  -∞ < b0 < i0 < ∞ .

Six types of hands are possible.

  • Dud – Three cards that do not qualify as any other type of hand.
  • Plain flush – Three non-consecutively numbered cards from one suit.
  • Straight flush – Three consecutively numbered cards from one suit.
  • Mixed straight – Three consecutively numbered cards with at least one from each suit.
  • Mixed pair – Two cards have the same number from different suits.
  • Flush pair – Two cards have the same number from the same suit.

A flush pair seems impossible as each card is unique in its suit. Here is a break from tradition.
   Two cards may be combined to form one new card in their place. The new card’s number is the sum of the original two cards. The resulting suit is inherited from the original cards if their suits match. The resulting suit is “o” (other) if the original suits differ.
   A flush pair can derive from a straight flush or a plain flush, suppose ‘1b 3b 4b’, which may become the flush pair ‘4b 4b’ by combining 1b+3b = 4b.
   A mixed pair can derive from a mixed straight or a dud, suppose ‘1b 3i 4b’, which may become the mixed pair ‘4o 4b’ by combining 1b+3i = 4o.

Q.  How to number the suits so that a dud hand is the third-likeliest type of hand to be dealt?
(Two types of hands would be more likely than a dud hand while three types would be less likely. In a game with such a deck, duds would outrank two more-glamorous-sounding types of hands.)

  • 1
    $\begingroup$ If you can combine to a pair, does that automatically count (for purposes of "likeliest") as the pair? $\endgroup$
    – msh210
    Apr 10, 2023 at 13:20
  • $\begingroup$ @msh210 , that is part of the hidden hint. They do. $\endgroup$
    – humn
    Apr 10, 2023 at 13:40
  • 1
    $\begingroup$ So when you write "may be combined", you mean "are combined"? $\endgroup$
    – msh210
    Apr 10, 2023 at 14:00
  • 1
    $\begingroup$ Does the hand from your first example count for both plain flush and flush pair? Or only flush pair? $\endgroup$
    – justhalf
    Apr 10, 2023 at 15:25
  • 4
    $\begingroup$ These things that you call "hint" or "hidden hint" are not hints, I think. They're part of the question. There can equally likely be another question which does not count both, but only the final one, for example. So it's not a hint, since it's not solvable without that information (it would lead to a different answer) $\endgroup$
    – justhalf
    Apr 11, 2023 at 6:32

1 Answer 1


Hope I have this right...

There are 8 x 7 x 6 / 6 = 56 different three-card combinations in total. The proportion of some hand types does not depend on what number we choose. I'll use 1-4b and 5-8i as examples for these hands.

Regardless of the numbers chosen, there are
- 4 different straight flushes (123b, 234b, 567i, 678i), and
- 4 different plain flushes (124b, 134b, 568i, 578i).

There are eight possible selections for a suit that allow for flush pairs. They are
- −5 to −2 (−5=−3+(−2))
- −4 to −1 (−3=−2+(−1) and −4=−3+(−1))
- −3 to 0 (−3=−2+(−1))
- −2 to 1 (0=−1+1)
- −1 to 2 (0=−1+1)
- 0 to 3 (3=2+1)
- 1 to 4 (3=2+1 and 4=3+1)
- 2 to 5 (5=3+2)

In each of these cases there is a maximum of two possible flush pairs per suit. The maximum number of flush pairs is therefore 4.

No mixed straights are possible if there are any unused numbers between bitties and itties. If itties follow directly after the bitties, there are two possible mixed straigts (say, 2i3i4b and 3i4b5b). The maximum number of mixed straights is achieved when i0 = b0+1 (eg. 0-3b and 1-4i). In that case, the number of possible mixed straights is 16 (eg. four different 012, eight different 123, four different 234).

The mixed pairs are a little trickier since there's an infinite number of possible ways of making them (for any b0 >2, just pick an i0 so that the itties include 2b0+1). So far we've covered at most 4 + 4 + 4 + 16 = 28 different combinations out of the 56, so we're left with at least 28 more. We need duds to account for fewer than 14 combos to make them the exactly third-most likely hand, so mixed pairs should take up at least 15 of the remaining combos.

Let's test our 0-3b and 1-4i version, which would allow for the maximum number of mixed straights and three flush pairs (leaving 29 combos for mixed pairs and duds). Counting how many combos of mixed pairs we get:
- 1 combo of 0 + 1 = 1
- 1 combo of 0 + 2 = 2
- 1 combo of 0 + 3 = 3
- 2 combos of 1 + 1 = 2
- 8 combos of 1 + 2 = 3
- 4 combos of 1 + 3 = 4
- 1 combo of 2 + 2 = 4

This is a total of 18 combinations for mixed pairs, which would leave them the second-most likely in the final tally, leaving 29-18 = 11 combos for duds. So our pick would be

0b 1b 2b 3b 1i 2i 3i 4i

Giving the final hand ranks:
- Mixed straight (18 combos)
- Mixed pair (16 combos)
- Dud (11 combos)
- Straight flush (4 combos)
- Plain flush (4 combos)
- Flush pair (3 combos)

Re: uniqueness

I don't think the solution is unique. It seems the same numbers but negative should work just as well (although I haven't checked all that thoroughly). Maybe there are other possible solutions as well.

  • $\begingroup$ I should've done more homework. Thank you. $\endgroup$
    – humn
    Apr 11, 2023 at 10:31
  • 1
    $\begingroup$ Why isn't 2+3=5 (from 2,3,4,5) a possibility for a flush pair? $\endgroup$
    – fljx
    Apr 11, 2023 at 14:49
  • 2
    $\begingroup$ @fljx Because I'm a donkey... Lemme see if I can fix it. $\endgroup$
    – Jafe
    Apr 11, 2023 at 22:44
  • $\begingroup$ Rakastan (beloved) @Jafe . You are right about negative duals. The solution in mind is its own while you bent the poles. $\endgroup$
    – humn
    Apr 26, 2023 at 5:30

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