<p>According to the <a href="https://boardgames.stackexchange.com/a/47959/5356">accepted answer</a> on the post containing that number, it originates in the book <a href="https://books.google.com/books?id=eS4IswEACAAJ" rel="noreferrer">How to Win at Gin Rummy: Playing for Fun and Profit</a> by Pramod Shankar. Unfortunately, the book doesn't seem to have a source or proof for it.</p>
<p>Thankfully, though, a comment on that post <em>does</em> give an answer. A blog called <a href="https://entropyclay.wordpress.com/2020/09/28/gin/" rel="noreferrer">Entropy Clay</a> 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:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Hand composition</th>
<th style="text-align: right;">Number of hands</th>
</tr>
</thead>
<tbody>
<tr>
<td>Two runs of 5</td>
<td style="text-align: right;">526</td>
</tr>
<tr>
<td>Three sets (4, 3, 3)</td>
<td style="text-align: right;">13,728</td>
</tr>
<tr>
<td>Three runs (4, 3, 3)</td>
<td style="text-align: right;">25,452</td>
</tr>
<tr>
<td>Set of 4, two runs of 3</td>
<td style="text-align: right;">6,636</td>
</tr>
<tr>
<td>Set of 3, two runs (4, 3)</td>
<td style="text-align: right;">47,272</td>
</tr>
<tr>
<td>Two sets (4, 3), run of 3</td>
<td style="text-align: right;">17,120</td>
</tr>
<tr>
<td>Two sets (3, 3), run of 4</td>
<td style="text-align: right;">25,960</td>
</tr>
<tr>
<td><strong>Total</strong></td>
<td style="text-align: right;"><strong>136,694</strong></td>
</tr>
</tbody>
</table>
</div>
CC BY-SA 4.0
2021-06-02T14:18:12.857Z
That seems to be missing some combinations, such as a run of 6, 4 and a run of 7, 3.
Joe W
CC BY-SA 4.0
2021-06-02T14:29:55.683Z
In that scenario, the 6 & 7 runs would be split into separate runs of 3s and a 4.
Joe Kerr
CC BY-SA 4.0
2021-06-02T15:42:33.637Z
I was initially skeptical, but after looking over the calculations in the linked blog post they do appear correct.  The common mistakes that students make with overcounting seem to all be handled correctly.
JMoravitz
CC BY-SA 4.0
2021-06-02T23:40:50.703Z
Wow, I could have worked on this until Doom's Day and never solved it !
Ken Bannister
CC BY-SA 4.0
2021-06-02T01:56:41.400Z
2021-06-02T13:24:36.767Z
2021-06-02T13:24:36.767Z
ConMan
<p>Some kind person on here <a href="https://boardgames.stackexchange.com/questions/47900/in-gin-rummy-what-are-the-chances-to-start-with-a-gin">said</a> 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 &quot;Gin.&quot;
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?</p>

CC BY-SA 4.0
2021-06-02T01:46:35.373Z
The number seems to come from [this question](https://boardgames.stackexchange.com/questions/47900/in-gin-rummy-what-are-the-chances-to-start-with-a-gin), for reference.
ConMan
CC BY-SA 4.0
2021-06-02T01:47:49.690Z
And in a comment on that post I've found the solution, which I'll post as an answer.
ConMan
<statistics><gin-rummy>
I am wondering how in the world the number 136,694 of directly dealt out Gin hands was arrived at
Ken Bannister
# I am wondering how in the world the number 136,694 of directly dealt out Gin hands was arrived at

(Asked by: username_0 on 2021-06-02)

Some kind person on here [said](https://boardgames.stackexchange.com/questions/47900/in-gin-rummy-what-are-the-chances-to-start-with-a-gin) 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?

**Tags:** statistics, gin-rummy

> **Comments:**
> **username_1**: The number seems to come from [this question](https://boardgames.stackexchange.com/questions/47900/in-gin-rummy-what-are-the-chances-to-start-with-a-gin), for reference.  
> **username_1**: And in a comment on that post I've found the solution, which I'll post as an answer.  

---

## 1 answers

### Answer by username_1 on 2021-06-02. Score: 7

According to the [accepted answer](https://boardgames.stackexchange.com/a/47959/5356) on the post containing that number, it originates in the book [How to Win at Gin Rummy: Playing for Fun and Profit](https://books.google.com/books?id=eS4IswEACAAJ) 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](https://entropyclay.wordpress.com/2020/09/28/gin/) 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 |
| **Total** | **136,694** |

> **Comments:**
> **username_2**: That seems to be missing some combinations, such as a run of 6, 4 and a run of 7, 3.  
> **username_3**: In that scenario, the 6 & 7 runs would be split into separate runs of 3s and a 4.  
> **username_4**: I was initially skeptical, but after looking over the calculations in the linked blog post they do appear correct.  The common mistakes that students make with overcounting seem to all be handled correctly.  
> **username_0**: Wow, I could have worked on this until Doom's Day and never solved it !  

---

