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?
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|