<h1>TLDR:</h1>

<p>The probability is about 8%.</p>

<h1>Distribution of diamonds for any deal</h1>

<p>With no restrictions at all then dealing a pack between four hands results in the following probabilities for distributions of diamonds:</p>

<pre><code>(0, 0, 0, 13): 0.00% or ~1/158753389900
(0, 0, 1, 12): 0.00% or ~1/313123057
(0, 0, 2, 11): 0.00% or ~1/8697863
(0, 0, 3, 10): 0.00% or ~1/646948
(0, 0, 4, 9): 0.00% or ~1/103512
(0, 0, 5, 8): 0.00% or ~1/31948
(0, 0, 6, 7): 0.01% or ~1/17971
(0, 1, 1, 11): 0.00% or ~1/4014398
(0, 1, 2, 10): 0.00% or ~1/91236
(0, 1, 3, 9): 0.01% or ~1/9953
(0, 1, 4, 8): 0.05% or ~1/2212
(0, 1, 5, 7): 0.11% or ~1/922
(0, 1, 6, 6): 0.07% or ~1/1382
(0, 2, 2, 9): 0.01% or ~1/12165
(0, 2, 3, 8): 0.11% or ~1/922
(0, 2, 4, 7): 0.36% or ~1/276
(0, 2, 5, 6): 0.65% or ~1/154
(0, 3, 3, 7): 0.27% or ~1/377
(0, 3, 4, 6): 1.33% or ~1/75
(0, 3, 5, 5): 0.90% or ~1/112
(0, 4, 4, 5): 1.24% or ~1/80
(1, 1, 1, 10): 0.00% or ~1/252654
(1, 1, 2, 9): 0.02% or ~1/5615
(1, 1, 3, 8): 0.12% or ~1/851
(1, 1, 4, 7): 0.39% or ~1/255
(1, 1, 5, 6): 0.71% or ~1/142
(1, 2, 2, 8): 0.19% or ~1/520
(1, 2, 3, 7): 1.88% or ~1/53
(1, 2, 4, 6): 4.70% or ~1/21
(1, 2, 5, 5): 3.17% or ~1/32
(1, 3, 3, 6): 3.45% or ~1/29
(1, 3, 4, 5): 12.93% or ~1/8
(1, 4, 4, 4): 2.99% or ~1/33
(2, 2, 2, 7): 0.51% or ~1/195
(2, 2, 3, 6): 5.64% or ~1/18
(2, 2, 4, 5): 10.58% or ~1/9
(2, 3, 3, 5): 15.52% or ~1/6
(2, 3, 4, 4): 21.55% or ~1/5
(3, 3, 3, 4): 10.54% or ~1/9
</code></pre>

<h1>Distribution of diamonds given I have four of them</h1>

<p>Assuming instead that I can see four of the diamonds in my hand then the distribution for the other three hands becomes:</p>

<pre><code>(0, 0, 9): 0.00% or ~1/98795
(0, 1, 8): 0.05% or ~1/2111
(0, 2, 7): 0.38% or ~1/264
(0, 3, 6): 1.39% or ~1/72
(0, 4, 5): 2.61% or ~1/38
(1, 1, 7): 0.41% or ~1/244
(1, 2, 6): 4.93% or ~1/20
(1, 3, 5): 13.55% or ~1/7
(1, 4, 4): 9.41% or ~1/11
(2, 2, 5): 11.08% or ~1/9
(2, 3, 4): 45.16% or ~1/2
(3, 3, 3): 11.04% or ~1/9
</code></pre>

<h1>Distribution of diamonds given I have four and treating my partner differently</h1>

<p>Treating my partner's hand as distinct from my two opponents results in the following probabilities:</p>

<pre><code>(0, (0, 9)): 0.00% or ~1/148192
(0, (1, 8)): 0.02% or ~1/6333
(0, (2, 7)): 0.13% or ~1/792
(0, (3, 6)): 0.46% or ~1/216
(0, (4, 5)): 0.87% or ~1/115
(1, (0, 8)): 0.02% or ~1/6333
(1, (1, 7)): 0.27% or ~1/365
(1, (2, 6)): 1.64% or ~1/61
(1, (3, 5)): 4.52% or ~1/22
(1, (4, 4)): 3.14% or ~1/32
(2, (0, 7)): 0.13% or ~1/792
(2, (1, 6)): 1.64% or ~1/61
(2, (2, 5)): 7.39% or ~1/14
(2, (3, 4)): 15.05% or ~1/7
(3, (0, 6)): 0.46% or ~1/216
(3, (1, 5)): 4.52% or ~1/22
(3, (2, 4)): 15.05% or ~1/7
(3, (3, 3)): 11.04% or ~1/9
(4, (0, 5)): 0.87% or ~1/115
(4, (1, 4)): 6.27% or ~1/16
(4, (2, 3)): 15.05% or ~1/7
(5, (0, 4)): 0.87% or ~1/115
(5, (1, 3)): 4.52% or ~1/22
(5, (2, 2)): 3.69% or ~1/27
(6, (0, 3)): 0.46% or ~1/216
(6, (1, 2)): 1.64% or ~1/61
(7, (0, 2)): 0.13% or ~1/792
(7, (1, 1)): 0.14% or ~1/731
(8, (0, 1)): 0.02% or ~1/6333
(9, (0, 0)): 0.00% or ~1/296385
</code></pre>

<p>From this we can see that the probability of my partner and one opponent having at least four diamonds each is given by these rows:</p>

<pre><code>(4, (0, 5)): 0.87% or ~1/115
(4, (1, 4)): 6.27% or ~1/16
(5, (0, 4)): 0.87% or ~1/115
</code></pre>

<p>Adding these together gives a probability of about 8%.</p>

CC BY-SA 4.0
2019-09-10T16:48:09.933Z
This looks a bit high to me. My calculation for (4, (1, 4)) is about 4.5%
Forget I was ever here
CC BY-SA 4.0
2019-09-10T17:15:02.170Z
I dealt with the partner then the two opponents. I get P(Partner gets exactly 4 of 9 diamonds) = 0.22 and P(Opponent gets 1 or 4 diamonds from 5 remaining) = 0.14 + 0.14 = 0.28.  Working this out gives P((4, (1, 4)) = 0.22 * 0.28 = 0.06.
tttppp
CC BY-SA 4.0
2019-09-10T17:33:58.370Z
Okay, my calc is wrong. [Casual Enthusiast](http://www.occasionalenthusiast.com/bridge-partnership-two-hands-probability-analysis/) agrees with you on the odds of partner having exactly 4, and the 28% odds for a 4-1 break of the opponents' five cards is a well known value.
Forget I was ever here
CC BY-SA 4.0
2019-09-10T16:12:41.530Z
2019-09-10T16:12:41.530Z
tttppp
<p>In Spades, (or any other trick taking game), when I hold <strong>n</strong> cards of a certain suit, what is the distribution of the other <strong>13-n</strong> cards in that suit?</p>

<p><strong>In particular</strong>, what is the probability that ♦'s are distributed either 4-4-4-1 or 5-4-4-0, given I hold 4 ♦ cards? (my partner and one of the opponents both hold at least four ♦'s, and one opponent is holding either 0 or 1 ♦'s)?</p>

<p><strong>The situation that led to this question:</strong>  I bid nil with 4 ♦'s cards, after 3 ♦'s tricks were played, the opponent led a fourth ♦, I had to play my A♦ hoping that by the fourth trick my partner will be able to cut, however my partner had also 4♦'s, thus the nil was set :( .</p>

CC BY-SA 4.0
2019-09-10T13:42:45.917Z
Do you require that the 0 or 1 be your partner's holding?
L. Scott Johnson
CC BY-SA 4.0
2019-09-10T13:53:07.087Z
@L.ScottJohnson I require it to be one of the opponents
Cohensius
CC BY-SA 4.0
2019-09-10T13:57:04.927Z
Clarification: Are you specifically interested in **Partnership Spades** rather then **Cutthroat (ie Solo) Spades**?
Forget I was ever here
CC BY-SA 4.0
2019-09-10T14:00:10.123Z
In general I am interested in both, In this praticular question I think that the probabilities are the same.  Isn't it?
Cohensius
CC BY-SA 4.0
2019-09-10T14:00:34.330Z
The specific probability of a particular suit breaking 4441 is about 0.74% - a rather unusual occurrence. I would recommend ***not*** allowing this to affect your bidding strategy in future - as the opportunity cost of protecting against this unusual situation is too high.
Forget I was ever here
CC BY-SA 4.0
2019-09-10T14:02:01.790Z
@Cohensius: ***NO!*** Partner is a specific seat at the table, thus a privileged calculation. Read the Conditional Probability link in my answer to your previous question. It's the difference between any two of three equal seats, or of one specific seat AND one of the remaining two equal seats.
Forget I was ever here
CC BY-SA 4.0
2019-09-10T14:03:52.220Z
4441 is about 0.74%, however given I hold 4 cards from the suit, what is the probability?
Cohensius
CC BY-SA 4.0
2019-09-10T14:07:00.650Z
@Cohensius: As your edits have already invalidated two false starts by me on this question. Please confirm that the specific question you wish an answer for is this single scenario. Note that it will, alone, probably take over an hour to compute and write up with confidence.
Forget I was ever here
CC BY-SA 4.0
2019-09-10T15:05:31.130Z
@ForgetIwaseverhere confirmed
Cohensius
<probability><spades><trick-taking-games>
In Spades, what is the probability a suit is distributed in a way, given I hold n cards from that suit?
Cohensius
# In Spades, what is the probability a suit is distributed in a way, given I hold n cards from that suit?

(Asked by: username_0 on 2019-09-10)

In Spades, (or any other trick taking game), when I hold **n** cards of a certain suit, what is the distribution of the other **13-n** cards in that suit?

**In particular**, what is the probability that ♦'s are distributed either 4-4-4-1 or 5-4-4-0, given I hold 4 ♦ cards? (my partner and one of the opponents both hold at least four ♦'s, and one opponent is holding either 0 or 1 ♦'s)?

**The situation that led to this question:** I bid nil with 4 ♦'s cards, after 3 ♦'s tricks were played, the opponent led a fourth ♦, I had to play my A♦ hoping that by the fourth trick my partner will be able to cut, however my partner had also 4♦'s, thus the nil was set :( .

**Tags:** probability, spades, trick-taking-games

> **Comments:**
> **username_1**: Do you require that the 0 or 1 be your partner's holding?  
> **username_0**: @username_1 I require it to be one of the opponents  
> **username_2**: Clarification: Are you specifically interested in **Partnership Spades** rather then **Cutthroat (ie Solo) Spades**?  
> **username_0**: In general I am interested in both, In this praticular question I think that the probabilities are the same.  Isn't it?  
> **username_2**: The specific probability of a particular suit breaking 4441 is about 0.74% - a rather unusual occurrence. I would recommend ***not*** allowing this to affect your bidding strategy in future - as the opportunity cost of protecting against this unusual situation is too high.  
> **username_2**: @username_0: ***NO!*** Partner is a specific seat at the table, thus a privileged calculation. Read the Conditional Probability link in my answer to your previous question. It's the difference between any two of three equal seats, or of one specific seat AND one of the remaining two equal seats.  
> **username_0**: 4441 is about 0.74%, however given I hold 4 cards from the suit, what is the probability?  
> **username_2**: @username_0: As your edits have already invalidated two false starts by me on this question. Please confirm that the specific question you wish an answer for is this single scenario. Note that it will, alone, probably take over an hour to compute and write up with confidence.  
> **username_0**: @username_2 confirmed  

---

## 1 answers

### Answer by username_3 on 2019-09-10. Score: 4 | Accepted answer

# TLDR:

The probability is about 8%.

# Distribution of diamonds for any deal

With no restrictions at all then dealing a pack between four hands results in the following probabilities for distributions of diamonds:

```
(0, 0, 0, 13): 0.00% or ~1/158753389900
(0, 0, 1, 12): 0.00% or ~1/313123057
(0, 0, 2, 11): 0.00% or ~1/8697863
(0, 0, 3, 10): 0.00% or ~1/646948
(0, 0, 4, 9): 0.00% or ~1/103512
(0, 0, 5, 8): 0.00% or ~1/31948
(0, 0, 6, 7): 0.01% or ~1/17971
(0, 1, 1, 11): 0.00% or ~1/4014398
(0, 1, 2, 10): 0.00% or ~1/91236
(0, 1, 3, 9): 0.01% or ~1/9953
(0, 1, 4, 8): 0.05% or ~1/2212
(0, 1, 5, 7): 0.11% or ~1/922
(0, 1, 6, 6): 0.07% or ~1/1382
(0, 2, 2, 9): 0.01% or ~1/12165
(0, 2, 3, 8): 0.11% or ~1/922
(0, 2, 4, 7): 0.36% or ~1/276
(0, 2, 5, 6): 0.65% or ~1/154
(0, 3, 3, 7): 0.27% or ~1/377
(0, 3, 4, 6): 1.33% or ~1/75
(0, 3, 5, 5): 0.90% or ~1/112
(0, 4, 4, 5): 1.24% or ~1/80
(1, 1, 1, 10): 0.00% or ~1/252654
(1, 1, 2, 9): 0.02% or ~1/5615
(1, 1, 3, 8): 0.12% or ~1/851
(1, 1, 4, 7): 0.39% or ~1/255
(1, 1, 5, 6): 0.71% or ~1/142
(1, 2, 2, 8): 0.19% or ~1/520
(1, 2, 3, 7): 1.88% or ~1/53
(1, 2, 4, 6): 4.70% or ~1/21
(1, 2, 5, 5): 3.17% or ~1/32
(1, 3, 3, 6): 3.45% or ~1/29
(1, 3, 4, 5): 12.93% or ~1/8
(1, 4, 4, 4): 2.99% or ~1/33
(2, 2, 2, 7): 0.51% or ~1/195
(2, 2, 3, 6): 5.64% or ~1/18
(2, 2, 4, 5): 10.58% or ~1/9
(2, 3, 3, 5): 15.52% or ~1/6
(2, 3, 4, 4): 21.55% or ~1/5
(3, 3, 3, 4): 10.54% or ~1/9

```

# Distribution of diamonds given I have four of them

Assuming instead that I can see four of the diamonds in my hand then the distribution for the other three hands becomes:

```
(0, 0, 9): 0.00% or ~1/98795
(0, 1, 8): 0.05% or ~1/2111
(0, 2, 7): 0.38% or ~1/264
(0, 3, 6): 1.39% or ~1/72
(0, 4, 5): 2.61% or ~1/38
(1, 1, 7): 0.41% or ~1/244
(1, 2, 6): 4.93% or ~1/20
(1, 3, 5): 13.55% or ~1/7
(1, 4, 4): 9.41% or ~1/11
(2, 2, 5): 11.08% or ~1/9
(2, 3, 4): 45.16% or ~1/2
(3, 3, 3): 11.04% or ~1/9

```

# Distribution of diamonds given I have four and treating my partner differently

Treating my partner's hand as distinct from my two opponents results in the following probabilities:

```
(0, (0, 9)): 0.00% or ~1/148192
(0, (1, 8)): 0.02% or ~1/6333
(0, (2, 7)): 0.13% or ~1/792
(0, (3, 6)): 0.46% or ~1/216
(0, (4, 5)): 0.87% or ~1/115
(1, (0, 8)): 0.02% or ~1/6333
(1, (1, 7)): 0.27% or ~1/365
(1, (2, 6)): 1.64% or ~1/61
(1, (3, 5)): 4.52% or ~1/22
(1, (4, 4)): 3.14% or ~1/32
(2, (0, 7)): 0.13% or ~1/792
(2, (1, 6)): 1.64% or ~1/61
(2, (2, 5)): 7.39% or ~1/14
(2, (3, 4)): 15.05% or ~1/7
(3, (0, 6)): 0.46% or ~1/216
(3, (1, 5)): 4.52% or ~1/22
(3, (2, 4)): 15.05% or ~1/7
(3, (3, 3)): 11.04% or ~1/9
(4, (0, 5)): 0.87% or ~1/115
(4, (1, 4)): 6.27% or ~1/16
(4, (2, 3)): 15.05% or ~1/7
(5, (0, 4)): 0.87% or ~1/115
(5, (1, 3)): 4.52% or ~1/22
(5, (2, 2)): 3.69% or ~1/27
(6, (0, 3)): 0.46% or ~1/216
(6, (1, 2)): 1.64% or ~1/61
(7, (0, 2)): 0.13% or ~1/792
(7, (1, 1)): 0.14% or ~1/731
(8, (0, 1)): 0.02% or ~1/6333
(9, (0, 0)): 0.00% or ~1/296385

```

From this we can see that the probability of my partner and one opponent having at least four diamonds each is given by these rows:

```
(4, (0, 5)): 0.87% or ~1/115
(4, (1, 4)): 6.27% or ~1/16
(5, (0, 4)): 0.87% or ~1/115

```

Adding these together gives a probability of about 8%.

> **Comments:**
> **username_2**: This looks a bit high to me. My calculation for (4, (1, 4)) is about 4.5%  
> **username_3**: I dealt with the partner then the two opponents. I get P(Partner gets exactly 4 of 9 diamonds) = 0.22 and P(Opponent gets 1 or 4 diamonds from 5 remaining) = 0.14 + 0.14 = 0.28.  Working this out gives P((4, (1, 4)) = 0.22 * 0.28 = 0.06.  
> **username_2**: Okay, my calc is wrong. [Casual Enthusiast](http://www.occasionalenthusiast.com/bridge-partnership-two-hands-probability-analysis/) agrees with you on the odds of partner having exactly 4, and the 28% odds for a 4-1 break of the opponents' five cards is a well known value.  

---

