I read somewhere that the poker hand AK suited has a 51% chance of hitting top pair, a flush or a straight, by the River, but I have no idea whether this is true or not or how to determine it.

I assume we would first determine the odds of NOT catching any of these hands by the River.

I don’t know how it all interrelated with one another.

For instance, you can’t River a flush if you don’t get at least 1 card to your suit on the flop. So we could determine the odds of NOT flopping ANY heart, when holding 2 hearts.

You can’t River a straight with AK unless you flop at least a Queen a Jack OR a Ten on the flop. So, what are the odds your flop will contain no Heart, no Queen, no Jack and no Ten? Something like this to start with?

Then I guess you want to NOT flop an Ace or. King also, aaaasrgh
Brain explosion at this point,

If you can give me the general format for the equation I can try myself and check results,

Thanks.
Magnum

So for instance, there are 50 cards remaining, but you CAN'T flop any A,K,Q,J,T or Heart.

You hold 1 Ace and 1 King, so there are 3 of those cards remaining. = 6 cards.

There are 3 non-Heart Queens, Jacks and Tens = 9 cards.

There are 11 Hearts that we need to dodge as well (which also includes the QJT of hearts). = 11 Hearts.

So that is 26 cards we have to dodge on the flop to NOT make Top pair, a Flush or a straight by the River. That, I guess is a 52% chance of NOT flopping a good hand.

If we DON'T flop any of those cards then we are done with the idea of making a Flush or a Straight by the River, but we can still hit top pair on either the Turn or the River.

There are now 47 cards remaining and 6 of them are Aces and Kings. So we have a 87.3% chance of NOT hitting an Ace or a King on the Turn.

I guess pretty similar for the River except we have 1 fewer card to draw from (46).

-------------------

This is kind of how I am looking at it but I have no idea how to put all of this together and get a final solution.

I know .... it is a mess.

I don't think there is any magical easy way to do this.

One way I might approach it is to write down mutually exclusive cases which "qualify" as AKs "making a hand" by the river. Maybe something like:

1. Four of a kind of A or K

2. Full house (using A and/or K; not quads)

3. Flush (using A and/or K; includes royal flush/straight flush)

4. Straight (using A and/or K; not a flush)

5. Three of a kind of A or K (not quads, not full house, not flush, not straight)

6. Two pair (using A and/or K, not quads, not full house, not flush, not straight, not three of a kind)

7. One pair of A or K (not quads, not full house, not flush, not straight, not three of a kind, not two pair)

Each one of these categories could then be tallied using combinatorics and logic.

Another way is to simply brute force it. Write a computer program to run over all possible boards and tally how often AKs "makes a hand".

A computer simulation would also be possible, though in this case the brute-force method seems better (it would be fast/easy for a computer to run all possible boards).

To start I think the question should be better defined. What exactly counts as making a hand? Does 23456 count as a straight? Does 33345 count as trips? What about 2222K?

My approach would be to look at the entire board rather than go street by street. Finding the probability that 5 cards contain at least one Ax or Kx isn't too hard, for example. But then, that approach would count 2222K.

browni3141 raises a very good point about the definition of the question.

Also, as you two have intimated, it may be easier to tally the converse of "making a hand" in this case.

So you would need 5 cards, none of which is an A or K, no three of which are spades (say you have AK of spades), no three of which are QJT, and no four of which are 5432 (excluding 65432).

Maybe constructing mutually exclusive categories of this converse is feasible. Or the Principle of Inclusion-Exclusion could potentially be used to take into account overlaps in the above categories.

As mentioned above this is very easy for a computer (once the exact definition of what we are looking for is nailed down), so if an answer would be helpful to aid in derivation or checking the results of a derivation, this could be readily provided.

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.

WOW!!!!!!!!!!

Thank you very much for the time you put into this. This is amazing. I am very grateful, not just for the ultimate answer but for the framework of the process.

Magnum