In Gin-Rummy, in the start of a round players draw 10 cards.

What is the probability to draw 10 cards such that all the cards are part of some meld?


Three Independent sources counted about 136,694 Gin hands, out of 15.8B possible 10-cards hands.

  1. 2+2 forum post: 1 in ~118,000. Using brute force which checked for Gin each of the 15.8B hands.

  2. Rulemonger's analysis: 136,694 in 15,820,024,220 or 1 in ~115733

  3. How to Win at Gin Rummy: 1 in ~117,000, according to the book How to Win at Gin Rummy: Playing for Fun and Profit which state that there are 136,694 Gin hands out of 15,820,024,220 hands. See page 65: gin hand - starting amount of deadwood


EDIT: this count is flawed, it doesn't recognize sets of three that include a spade. The correct number is given in the accepted answer. When the code is corrected, it gives the 136,694 unique gin hands, matching the accepted answer.


1 in 308,984, according to Andrew Inwood's analysis, which includes source code.

  • 1
    Does that include being able to pick up a visible discard to make Gin on your first turn? I'm inclined to think it doesn't, because I've done that twice, and I only play gin for an hour, once a week, for the past 4-5 years.
    – Zeiss Ikon
    Jul 15 '19 at 12:07
  • it does not. Just the first 10 cards. Getting a gin in the first turn (out of the first 11 cards) is a better (much harder question
    – Cohensius
    Jul 15 '19 at 12:14
  • This appears to count Aces as low-only, I assume this is why it disagrees with the other counts. See GinManager::SetUsableCards in bitbucket.org/ainwood87/gin/src/master/GinManager.cpp, which has nine types of length-five runs, when it should be 10.
    – DanTilkin
    Sep 22 '20 at 16:55
  • 2
    Because Aces are low-only, in the standard rules of Gin (en.wikipedia.org/wiki/Gin_rummy#Objective) Sep 22 '20 at 17:12
  • 2
    Found it: it doesn't recognize three-of-a-kinds that include a spade, so undercounts the number of gin hands accordingly. Sep 23 '20 at 13:29

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