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

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 :( .

• Do you require that the 0 or 1 be your partner's holding? Commented Sep 10, 2019 at 13:42
• @L.ScottJohnson I require it to be one of the opponents Commented Sep 10, 2019 at 13:53
• Clarification: Are you specifically interested in Partnership Spades rather then Cutthroat (ie Solo) Spades? Commented Sep 10, 2019 at 13:57
• In general I am interested in both, In this praticular question I think that the probabilities are the same. Isn't it? Commented Sep 10, 2019 at 14:00
• @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. Commented Sep 10, 2019 at 14:07

# 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%.

• This looks a bit high to me. My calculation for (4, (1, 4)) is about 4.5% Commented Sep 10, 2019 at 16:48
• 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. Commented Sep 10, 2019 at 17:15
• Okay, my calc is wrong. Casual Enthusiast 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. Commented Sep 10, 2019 at 17:33