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Some kind person on here said that when dealing out hands from a standard 52-card deck in the game of Rummy, there are 136,694 possible 10-card hands that can be declared "Gin." I am just wondering how that number was found - can it be done using pencil and paper and combinatorics or does a computer program have to be written to do it?

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    The number seems to come from this question, for reference. – ConMan Jun 2 at 1:46
  • And in a comment on that post I've found the solution, which I'll post as an answer. – ConMan Jun 2 at 1:47
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According to the accepted answer on the post containing that number, it originates in the book How to Win at Gin Rummy: Playing for Fun and Profit by Pramod Shankar. Unfortunately, the book doesn't seem to have a source or proof for it.

Thankfully, though, a comment on that post does give an answer. A blog called Entropy Clay steps through the calculations, breaking down all the possible ways the hand could contain Gin and a full calculation of the number of such hands in terms of fairly standard combinatorics. I won't repeat the whole thing, but the top level results are:

Hand composition Number of hands
Two runs of 5 526
Three sets (4, 3, 3) 13,728
Three runs (4, 3, 3) 25,452
Set of 4, two runs of 3 6,636
Set of 3, two runs (4, 3) 47,272
Two sets (4, 3), run of 3 17,120
Two sets (3, 3), run of 4 25,960
Total 136,694
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  • That seems to be missing some combinations, such as a run of 6, 4 and a run of 7, 3. – Joe W Jun 2 at 14:18
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    In that scenario, the 6 & 7 runs would be split into separate runs of 3s and a 4. – Joe Kerr Jun 2 at 14:29
  • I was initially skeptical, but after looking over the calculations in the linked blog post they do appear correct. The common mistakes that students make with overcounting seem to all be handled correctly. – JMoravitz Jun 2 at 15:42
  • Wow, I could have worked on this until Doom's Day and never solved it ! – Ken Bannister Jun 2 at 23:40

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