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What are the odds for a Gin on the first turn? That is, what are the odds for arranging 10 out of 11 random cards in melds.

In Gin-Rummy, at the start of a round, players draw 10 random cards. On the first turn the player decide to either pickup the face-up card or pass the turn. Thus, for the first player to make a Gin on the first turn, she must take the face-up card and arrange at least 10 out of the 11 cards in melds. In an earlier question we found out that the probability of 10 random cards to form a Gin is about 1 in 115,733. Great explanation for the counting process here.

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I modified that program by Andrew Inwood to count them (including all the three-of-a-kinds, still), and it finds 5,379,372 11-card hands that contain a 10-card gin hand.

Out of C(52,11) hands, that makes the odds about 1 in 11,228 or 0.00891%

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