<p>I modified that <a href="https://canadianandrew.wordpress.com/2016/09/13/odds-of-dealing-gin/" rel="nofollow noreferrer">program by Andrew Inwood</a> 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.</p>
<p>Out of C(52,11) hands, that makes the odds about 1 in 11,228 or 0.00891%</p>

CC BY-SA 4.0
2020-09-29T15:00:31.363Z
2020-09-29T17:49:15.427Z
2020-09-29T17:49:15.427Z
L. Scott Johnson
<p>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.</p>
<p>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 <a href="https://boardgames.stackexchange.com/questions/47900/in-gin-rummy-what-are-the-chances-to-start-with-a-gin">earlier question</a> we found out that the probability of 10 random cards to form a Gin is about <strong>1 in 115,733</strong>.  Great explanation for the counting process <a href="https://entropyclay.wordpress.com/2020/09/28/gin/comment-page-1/?unapproved=1&amp;moderation-hash=07e8cc6adbf1dd11332e3ce4e1e5e822#comment-1" rel="nofollow noreferrer">here</a>.</p>

<probability><gin-rummy>
In Gin-Rummy, what are the chances to make Gin on the first turn?
Cohensius
# In Gin-Rummy, what are the chances to make Gin on the first turn?

(Asked by: username_0 on 2020-09-28)

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](https://boardgames.stackexchange.com/questions/47900/in-gin-rummy-what-are-the-chances-to-start-with-a-gin) 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](https://entropyclay.wordpress.com/2020/09/28/gin/comment-page-1/?unapproved=1&moderation-hash=07e8cc6adbf1dd11332e3ce4e1e5e822#comment-1).

**Tags:** probability, gin-rummy


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## 1 answers

### Answer by username_1 on 2020-09-29. Score: 1 | Accepted answer

I modified that [program by Andrew Inwood](https://canadianandrew.wordpress.com/2016/09/13/odds-of-dealing-gin/) 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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