I had some spare time this evening so I took a crack at the derivation using the definition from my previous post just above this post.
You hold As Ks. 5-card board is dealt and we want to know the probability that your hand "improves":
- A or K on board OR
- You make a spade flush OR
- You make a Broadway straight OR
- You make a wheel straight.
Of course there are C(50,5) = 2,118,760 possible 5-card boards since you hold two cards (As Ks).
Let's first consider each of these categories individually. It will be easiest if we derive the "converse" of each case (the tally for no-improvement). Then at the end we will subtract from the total possible to derive the "improvement" tallies.
Case 1 on its own: No A or K
Clearly this is C(44,5) = 1,086,008
Case 2 on its own: No spade flush
Subcase 2A: 0 spades on board= C(39,5) = 575,757
Subcase 2B: 1 spade on board = C(39,4)*C(11,1) = 904,761
Subcase 2C: 2 spades on board = C(39,3)*C(11,2) = 502,645
Total for Case 2 = 1,983,163
Case 3 on its own: No Broadway straight (no [QJT])
Here we will first derive the converse and then subtract from the total at the end.
Subcase 3A: Exactly 1 of each of QJT = C(4,1)*C(4,1)*C(4,1)*C(38,2) = 44,992
Subcase 3B: 1 Dupe in QJT = C(3,1)*C(4,2)*C(4,1)*C(4,1)*C(38,1) = 10,944
Subcase 3C: 2 Dupes in QJT = C(3,2)*C(4,2)*C(4,2)*C(4,1) = 432
Subcase 3D: 1 Trips and 2 Singletons in QJT = C(3,1)*C(4,3)*C(4,1)*C(4,1) = 192
Subtotal of Case 3 Converse: 56,560
So subtotal of Case 3 = 2,118,760 - 56,560 = 2,062,200
Case 4 on its own: No wheel straight (no [5432] excluding 65432)
Again we will first derive the converse.
Subcase 4A: Exactly 1 of each of 5432 = C(4,1)*C(4,1)*C(4,1)*C(4,1)*C(30,1) = 7,680
Subcase 4B: 1 Dupe in 5432 = C(4,1)*C(4,2)*C(4,1)*C(4,1)*C(4,1) = 1,536
Subtotal of Case 4 Converse = 9,216
So Total of Case 4 = 2,118,760 - 9,216 = 2,109,544
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We will now endeavor to "combine" cases. Let's start with Case 1 and Case 2. So we want to derive the number of 5-card boards that do not contain an A or K and have fewer than 3 spades. This is straightforward.
Case12 A: 0 spades = C(33,5) = 237,336
Case12 B: 1 spade = C(33,4)*C(11,1) = 450,120
Case12 C: 2 spades = C(33,3) * C(11,2) = 300,080
Total of Case12 = 987,536
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Now let's derive the number of 5-card boards satisfying the first three cases: no A or K, no spade flush, and no broadway straight (no [QJT]).
We will start with the Case12 results above and then derive the converse for the QJT case.
Case123 A: 0 spades
Subcase123 A1: Exactly 1 of each of QJT
= C(3,1)*C(3,1)*C(3,1)*C(24,2)
= 7,452
Subcase123 A2: 1 Dupe in QJT
= C(3,1)*C(3,2)*C(3,1)*C(3,1)*C(24,1)
= 1,944
Subcase123 A3: 2 Dupes in QJT
= C(3,2)*C(3,2)*C(3,2)*C(3,1)
= 81
Subcase123 A4: 1 Trips and 2 Singletons in QJT
= C(3,1)*C(3,3)*C(3,1)*C(3,1)
= 27
Subtotal of Case123 A = 9,504
Case123 B: 1 spade
Case123 B1: 1 spade in QJT
Subcase123 B11: Exactly 1 of each of QJT
= C(3,1)*C(1,1)*C(3,1)*C(3,1)*C(24,2)
= 7,452
Subcase123 B12: 1 Dupe in QJT
Subcase123 B12A: Spade is dupe
= C(3,1)*C(1,1)*C(3,1)*C(3,1)*C(3,1)*C(24,1)
= 1,944
Subcase123 B12B: Spade is singleton
= C(3,1)*C(2,1)*C(3,2)*C(3,1)*C(1,1)*C(24,1)
= 1,296
Subcase123 B13: 2 Dupes in QJT
Subcase123 B13A: Spade is dupe
= C(3,2)*C(2,1)*C(1,1)*C(3,1)*C(3,2)*C(3,1)
= 162
Subcase123 B13B: Spade is singleton
= C(3,1)*C(3,2)*C(3,2)*C(1,1)
= 27
Subcase123 B14: 1 Trips and 2 Singletons in QJT
Subcase123 B14A: Spade is trips
= C(3,1)*C(1,1)*C(3,2)*C(3,1)*C(3,1)
= 81
Subcase123 B14B: Spade is singleton
= C(3,1)*C(2,1)*C(3,3)*C(3,1)*C(1,1)
= 18
Subtotal of Case123 B1 = 10,980
Case123 B2: No spade in QJT
Subcase123 B21: Exactly 1 of each of QJT
Subcase123 B21A: Other two cards are singletons
= C(3,1)*C(3,1)*C(3,1)*C(8,2)*C(2,1)*C(1,1)*C(3,1)
= 4,536
Subcase123 B21B: Other two cards are a pair
= C(3,1)*C(3,1)*C(3,1)*C(8,1)*C(1,1)*C(3,1)
= 648
Subcase123 B22: 1 Dupe in QJT
= C(3,1)*C(3,2)*C(3,1)*C(3,1)*C(8,1)
= 648
Subcase123 B23: 2 Dupes in QJT
= Impossible
= 0
Subcase123 B24: 1 Trips and 2 Singletons in QJT
= Impossible
= 0
Subtotal of Case123 B2: 5,832
So Subtotal of Case123 B = 10,980 + 5,832 = 16,812
Case123 C: 2 spades
Case123 C1: 2 spades in QJT
Subcase123 C11: Exactly 1 of each of QJT
= C(3,2)*C(1,1)*C(1,1)*C(3,1)*C(24,2)
= 2,484
Subcase123 C12: 1 Dupe in QJT
Subcase123 C12A: Spades are dupe & singleton
= C(3,1)*C(2,1)*C(1,1)*C(3,1)*C(3,1)*C(1,1)*C(24,1)
= 1,296
Subcase123 C12B: Spades are singletons
= C(3,1)*C(3,2)*C(1,1)*C(1,1)*C(24,1)
= 216
Subcase123 C13: 2 Dupes in QJT
Subcase123 C13A: Spades are dupes
= C(3,2)*C(1,1)*C(3,1)*C(1,1)*C(3,1)*C(3,1)
= 81
Subcase123 C13B: Spades are dupe & singleton
= C(3,2)*C(2,1)*C(1,1)*C(3,1)*C(3,2)*C(1,1)
= 54
Subcase123 C14: 1 Trips and 2 Singletons in QJT
Subcase123 C14A: Spades are trips & singleton
= C(3,1)*C(2,1)*C(1,1)*C(3,2)*C(3,1)*C(1,1)
= 54
Subcase123 C14B: Spades are singletons
= C(3,1)*C(3,3)*C(1,1)*C(1,1)
= 3
Subtotal of Subcase123 C1: 4,188
Case123 C2: 1 spade in QJT
Subcase123 C21: Exactly 1 of each of QJT
Subcase123 C21A: Other two cards are singletons
= C(3,1)*C(1,1)*C(3,1)*C(3,1)*C(8,2)*C(2,1)*C(1,1)*C (3,1)
= 4,536
Subcase123 C21B: Other two cards are a pair
= C(3,1)*C(1,1)*C(3,1)*C(3,1)*C(8,1)*C(1,1)*C(3,1)
= 648
Subcase123 C22: 1 Dupe in QJT
Subcase123 C22A: Spade is dupe
= C(3,1)*C(1,1)*C(3,1)*C(3,1)*C(3,1)*C(8,1)
= 648
Subcase123 C22B: Spade is singleton
= C(3,1)*C(2,1)*C(3,2)*C(1,1)*C(3,1)*C(8,1)
= 432
Subcase123 C23: 2 Dupes in QJT
= Impossible
= 0
Subcase123 C24: 1 Trips and 2 Singletons in QJT
= Impossible
= 0
Subtotal of Subcase123 C2: 6,264
Case123 C3: 0 spades in QJT
Subcase123 C31: Exactly 1 of each of QJT
Subcase123 C31A: Other two cards are singletons
= C(3,1)*C(3,1)*C(3,1)*C(8,2)*C(1,1)*C(1,1)
= 756
Subcase123 C31B: Other two cards are a pair
= Impossible
= 0
Subcase123 C32: 1 Dupe in QJT
= Impossible
= 0
Subcase123 C33: 2 Dupes in QJT
= Impossible
= 0
Subcase123 C34: 1 Trips & 2 Singletons in QJT
= Impossible
= 0
Subtotal for Subcase123 C3: 756
So Subtotal of Subcase123 C = 4,188 + 6,264 + 756 = 11,208
Grand Total for Subcase123 Converse = 9,504 + 16,812 + 11,208 = 37,524
So Grand Total of Subcase123 = 987,536 - 37,524 = 950,012
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Now let's fold in the 4th Category (no wheel straight) so that we would have a tally of all 4 categories combined: no A or K, no spade flush, no broadway straight, and no wheel straight.
It will be easiest to start from the Case123 results above:
- 0 spades = 237,336 - 9,504 = 227,832
- 1 spades = 450,120 - 16,812 = 433,308
- 2 spades = 300,080 - 11,208 = 288,872
Case1234 A: 0 spades
Throughout this section we will calculate the converse and subtract from the total at the end and we will "exclude" 65432 boards.
Case1234 A1: Exactly 1 of each of 5432
= C(3,1)*C(3,1)*C(3,1)*C(3,1)*C(18,1)
= 1,458
Case1234 A2: 1 Dupe in 5432
= C(4,1)*C(3,2)*C(3,1)*C(3,1)*C(3,1)
= 324
Subtotal of Case1234 A Converse: 1,782
So Subtotal of Case1234 A = 227,832 - 1,782
= 226,050
Case1234 B: 1 spade
Subcase1234 B1: 1 spade in 5432
Subcase1234 B11: Exactly 1 of each of 5432
= C(4,1)*C(1,1)*C(3,1)*C(3,1)*C(3,1)*C(18,1)
= 1,944
Subcase1234 B12: 1 Dupe in 5432
Subcase1234 B12A: Spade is dupe
= C(4,1)*C(1,1)*C(3,1)*C(3,1)*C(3,1)*C(3,1)
= 324
Subcase1234 B12B: Spade is singleton
= C(4,1)*C(3,1)*C(3,2)*C(3,1)*C(3,1)*C(1,1)
= 324
Subtotal of Subcase1234 B1: 2,592
Subcase1234 B2: 1 spade not in 5432
Subcase1234 B21: Exactly 1 of each of 5432
= C(3,1)*C(3,1)*C(3,1)*C(3,1)*C(6,1)*C(1,1)
= 486
Subcase1234 B22: 1 Dupe in 5432
= Impossible
= 0
Subtotal for Subcase1234 B2: 486
Subtotal for Subcase1234 B Converse: 2,592 + 486
= 3,078
So Subtotal for Subcase1234 B: 433,308 - 3,078
= 430,230
Case1234 C: 2 spades
Subcase1234 C1: 2 spades in 5432
Subcase1234 C11: Exactly 1 of each of 5432
= C(4,2)*C(1,1)*C(1,1)*C(3,1)*C(3,1)*C(18,1)
= 972
Subcase1234 C12: 1 Dupe in 5432
Subcase1234 C12A: Spades are dupe & singleton
= C(4,1)*C(3,1)*C(1,1)*C(3,1)*C(3,1)*C(3,1)*C(1,1)
= 324
Subcase1234 C12B: Spades are singletons
= C(4,1)*C(3,2)*C(3,2)*C(3,1)*C(1,1)*C(1,1)
= 108
Subtotal of Subcase1234 C1: 1,404
Subcase1234 C2: 1 spade in 5432
Subcase1234 C21: Exactly 1 of each of 5432
= C(4,1)*C(1,1)*C(3,1)*C(3,1)*C(3,1)*C(6,1)
= 648
Subcase1234 C22: 1 Dupe in 5432
= Impossible
= 0
Subtotal of Subcase1234 C2: 648
Subcase1234 C3: 0 spades in 5432
= Impossible
= 0
Subtotal of Subcase1234 C Converse: 1,404 + 648 + 0
= 2,052
So Subtotal of Subcase1234 C: 288,872 - 2,052
= 286,820
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PUTTING IT ALL TOGETHER
All 4 no-improvement categories:
- No A or K
- No spade flush
- No broadway straight
- No wheel straight
= 226,050 + 430,230 + 286,820
= 943,100
Remember the total number of 5-card boards is C(50,5) = 2,118,760
So the Number of 5-card boards for which As Ks "improves" is given by:
= 2,118,760 - 943,100
= 1,175,660 which is approx 55.49%.
The details of this lengthy difficult-to-follow post are not critical. The idea is that standard combinatorics and logic are sufficient to derive tallies of the sort mentioned in OP.
There are undoubtedly easier and more efficient methods to derive these tallies than presented here. But I leave that to others if they so desire.
Last edited by whosnext; 05-26-2021 at 08:22 PM.
Reason: fix typos on case labels