In Spades, what is the probability a suit is distributed in a way, given I hold n cards from that suit?

Question

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 :( .

Answer 1

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%.