This should be fairly simple to a sharp math brain!
1. I am certain villain has AAxx
2. Flop comes 234ss

What percentage of the time will villain have the spades (or any single particular suit, spades used in this example) suited to the ace, not taking into account any other information other than the raw probability?

If this isn't phrased properly or whatever let me know but I think what I'm trying to ask makes sense!

Thanks for the help!

FWIW I estimate (based on no actual math) that its probably between 21-23%

I think I have followed your assumptions, but I am not 100% sure.

I have assumed board comes our 2s, 3s, 4c, and we are concerned about villain having the nut flush draw in spades. I assume that AAxx means that xx can be any two cards (including aces).

Under these assumptions, the total number of AAxx hands is C(4,2) * C(47,2) = 6 * 1081 = 6486.

How many of these give villain nut flush draw?

Case 1: 2 aces including the As and exactly one other spade = 3*10*37 = 1110

Case 2: 2 aces including the As and exactly two other spades = 3*C(10,2) = 3*45 = 135

Case 3: 3 aces including the As and exactly one other spade = 3*10 = 30

Total nut flush draws = 1110 + 135 + 30 = 1275

Pct = 1275/6486 = 19.6577%

I think the assumptions you followed are exactly what I was looking for and the answer seems to make perfect sense to me. Thanks so much for your help buddy, I appreciate you taking the time to do that!

This seems to be wrong. You are counting AAAA six times and AAAy (where y is a non-A non-234s) three times. You have to check each case separately (or use inclusion-exclusion).

The number of AAxx hands is given by:

AAAA + AAAy + AAyy = 1 + 4*45 + C(4,2)*C(45,2) = 6121

When you write C(4,2) * C(47,2) you say that for every choice of the two aces you have C(47,2) choices of the other two cards and each time you reach a different combo. This isn't true, since, for instance, you may choose AsAd for the aces and then AhAd for the other two cards one time and another time AsAh and AdAc, reaching the same combo. As you can see, with your method, you build six times AAAA and three times each AAAy. If you subtract 5 and 2*4*45 from your total, you reach my number.

Since we already know Villain has AAxx, we only have to count the xx, which is just C(47,2).

I'll do it as 2 cases: AA includes a spade (case 1), vs doesn't.

Case 1 combos = 1/2 * [9*46 - C(9,2)]
Case 2 combos = 1/2 * 8

probability = total / C(47,2) = 193/1081 =~ 17.85%

Thanks heehaww for the correction.

As veterans of this forum can attest, my contributions to these threads resembles that of Dr. Watson to Sherlock Holmes. Taking a first stab which is often wrong but illuminates the correct path. Which is just another way of saying that I was wrong again.

OP, you asked for the math based on straight probability, and not using any other information. But there will be info that you should use, and that is your own hand. If you are not in the hand, then you don't care, and if you are in the hand, the probabilities are going to be very different depending on whether you have 0, 1, 2, 3 or 4 spades in your own hand. If your plan is to bluff depending on the likelihood of him having the spades, this would be very critical information to use.

I also do not agree with this. While it is certainly true that xx can be C(47,2) combos, not every combo has the same probability. Suppose that a guy picks any AAxx. Now he remove AA from his hand and we can bet to guess the xx. Suppose that I bet on 5h5d and you on AsAd. Who do you think is going to win more often? I'm going to win any time the original hand was AA5h5s. You can win only if the original hand is AAAA and not every time! In fact, it's possible that he removed one of your aces and the two xx are formed by other aces.

Again, I think you need to split by cases. As I showed in my previous post, there are 6121 AAxx. Now we count how many of these hands have As plus at least another spade.

AAAA case: not possible.

AAAy case: there are 3 AAA containing the As and the four card can be any of the 9 remaining spades. 3*10=30 combos.

AAyy case: there are 3 AA containing the As. Now we count amongst the C(45,2) yy how many does not contain any spade. They are C(35,2). So we got 3*(C(45,2)-C(35,2)) hands with As and at least another spade in this group.

In total we have:

(30+3*(C(45,2)-C(35,2)))/6121 ~ 19.84%

I may have made some math mistake, but I guess the reasoning is sound. If only BruceZ could jump and state who is in the right! I hope someday he could join 2+2 and this forum again.

Thanks nickthegeek for the correction. (I overlooked you in the original mea culpa.)

I just ran a simple program to brute force the tallies.

I get that there are 6121 total AAxx cases (given the board) and 1215 have the Ace and at least one other spade.

Which is what nickthegeek posted above.

You're right. I'm not sure if how you phrased it in the quote is fully accurate, but that's the gist. There aren't C(47,2) combos, there are some extra ones but with uneven frequencies like you said. When I posted earlier, I thought my way was a shortcut to get the same answer, but no, it was just wrong. I suppose looking at all 4 cards is the way to go.

I'm getting about 19%.

{As, A, A, s} = 3 ways to choose the other two A's, then 9 other spades = 3*9

{As, Ax, s, _} = 3 ways to choose the second ace, then C(45,2)-C(38,2) ways to choose at least one spade without a third ace (since the AAA case was already counted). 45 is the number of non-aces remaining and 38 is the number of non-spades remaining, 47-9.
So it's 3[C(45,2)-C(38,2)]

Probability = 1161/6121 =~ 18.97%

the flop has exactly two spades

in the 2-ace case, there are 52-7 "other" cards and 10 of those are spades, which is why it is C(45,2) - C(35,2)

Shoot, I thought the flop was monotone.

And I typed something wrong in my calculator anyway because what I wrote wouldn't be 1161, but 888.

You know what, that's right. It needn't subtract C(37,2) because C(45,2) already excludes the case of two non-spades involving an ace. So now I'm in agreement with Nick's answer.

Today's not my day, I'll refrain from fouling up more threads til tomorrow

Chance of neither xx being a spade =
(37/47)*(36/46) = 1332 / 2162 = ~.616

So chance of at least one of xx being spade is ~.384

Chance of one of the Aces being spade is 1/2

Chance of having Asxs = .5 * .384 = ~.192

That assumes we know the AA includes a spade. Same mistake I made.

I think my incorrect assumption was that AAxx meant two and only two (not three) Aces.

If AAxx includes AAAx, then I agree with whosnext's answer of 19.66%.

Ok yeah I see you did make that assumption. But you also made my mistake. Whosnext, nick and myself now agree on 19.84%. What I see now is that my mistake and whosnext's are related. In my first post I falsely assumed a monotone flop, but had I done it for two-tone, I'd have got whosnext's answer of 19.66%. My denominators (and now yours) were too big by the same factor as whosnext's (though for different reasons). See nick's reply to my first post.