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?


1 in ~118,000, according to 2+2 forum post. Using a brute force, that checked for Gin each of the 15.8B hands.

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

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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 this analysis, which includes source code so you can verify.

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  • 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 at 16:55
  • 2
    Because Aces are low-only, in the standard rules of Gin (en.wikipedia.org/wiki/Gin_rummy#Objective) – L. Scott Johnson Sep 22 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. – L. Scott Johnson Sep 23 at 13:29

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