Live poker probability question

Hello, in Texas hold ‘em before the deal what’s the probability I will I end up with Aces full of Tens or better by the River? Both hole cards must be used. I guessing it’s over 1 in 1000 but I don’t understand how to work the math. Any help would be great.

I'm not entirely sure there is an easy way to "do the math" with the "must use both hole cards" restriction. It sounds like something you solve with a computer, by going through all 1326 possible sets of hole cards, then, for each one, checking all sets of 2,118,760 possible boards and counting how many can make A,A,A,10,10 or better using both hole cards. Despite the fact that this is about 2.8 billion deals, it wouldn't take too long to count.

Thanks for that, is there a website for that kind of thing? Or some software I could buy that would do that kind of calculation?

Quote: 1sock

Hello, in Texas hold ‘em before the deal what’s the probability I will I end up with Aces full of Tens or better by the River? Both hole cards must be used. I guessing it’s over 1 in 1000 but I don’t understand how to work the math. Any help would be great.
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This seems tough, but I think I can maybe do it.

1.) We need Aces full of tens by the river. Both of our hole cards must be used in making the final hand. Because of that, the hand must either be a 4OaK that uses both of our cards or an applicable Full House.

2.) The key starting hands we can have are AA, A10 and 1010. I almost disregarded 1010, but quad tens would still qualify and you could also end up with three Aces on the board. I guess there could also be a FH with a better inside pair on the board, but you never specified that you need to win the hand and that's pretty unlikely not to be a winning hand anyway.

3.) Other inside pairs are also key because the other two of those would make quads, and we would be using both of our cards.

4.) We could get Royal Flushes or Straight Flushes. Believe it or not, I think Royal Flushes and Straight Flushes are actually going to be relatively easy to figure out.

5.) I guess we could have something like KJ, get like the other three Kings (or Jacks) and our card also outkicks the board.

We do know we need both hole cards, so let's start there:

AA:

(4/52) * (3/51) = 0.00452488687

1010:

(4/52) * (3/51) = 0.00452488687

A10:

(4/52 * 4/51) = 0.0060331825

I suppose we could also start with KK, QQ or JJ and end up getting three aces on the board. We could also start with a pair of nines, or lower, and end up with quads.

The easiest case to actually do is starting with a Pair of Nines, or lower, and ending up with Quads. The reason why is because we either get the other two of our pair in the five community cards, or we don't.

99:

(4/52) * (3/51) = 0.00452488687

nCr(48,3)*nCr(2,2)/nCr(50,5) = 0.0081632653061224

Above is the probability of getting two of the same card in five cards, with two of them already being removed, amongst the 50 cards that we don't know. We must combine that with the probability of starting with any of the eight (2-9) applicable pairs in the first place:

(0.00452488687 * 8) * 0.0081632653061224 = 0.000295502815999998226983104---Lower Pair Than 10 Probability

Inside the parenthesis is the probability of starting with any specific inside pair in the first place multiplied by the eight applicable inside pairs. There's no possible way to arrive at any qualifying hand that is not quads, using both of our cards, if we start with a pair of nines, or lower. We multiply that by eight because there are eight pairs (2-9) for which this is true. After that, we multiply by the probability of getting the other two cards we would need, in five of fifty cards, which would be the same for all pairs.

The next thing that we will do is look at 10/10.

We are already aware that the probability that we start with a pair of tens is 0.00452488687. The two ways that we could accomplish a final hand of Aces full of tens, or better, is simply to have either three aces come out or the other two tens.

From above, we already know the probability of the other two tens coming out is 0.0081632653061224 in five cards.

With that, we simply need to know what the probability of getting three aces on the board is:

nCr(46,2)*nCr(4,3)/nCr(50,5) = 0.0019539730785931

I don't think that you could have all four aces come out, otherwise, you wouldn't be using both tens anymore.

We combine our probabilities of this happening IF we have a pair of tens with starting with a Pair of Tens in the first place.

0.00452488687 * (0.0019539730785931 + 0.0081632653061224) = 0.000045779359127659174635485

The interesting thing about this is it will also be true of Jacks, Queens and Kings. We couldn't do this for other pairs because three Aces in the community cards would not have mattered as the FH would not have been sufficiently good. Thus:

0.000045779359127659174635485 * 4 = 0.00018311743651063669854194---Pair 10's-K's Probability

Finally, for pairs, we have to look at starting with a Pair of Aces. If we start with a Pair of Aces, then we can either end up with Quad Aces, or alternatively, get ONE more Ace but then we need, of the other four cards, a pair of tens, or better.

We already know the probability of starting with a Pair of Aces and the probability of getting the other two:

0.00452488687 * 0.0081632653061224 = 0.000036937851999999778372888---Quad Aces, Started with Pair of Aces

Aces full and Quad Aces are the only possible hands that help us. Even other quads are bad because we'd only use one of our Aces, if that happened.

For that reason, we want EXACTLY one more Aces (otherwise we have Quad Aces anyway) and we want a pair of tens, or better.

The first thing that we are going to do is determine the probability of having exactly one other ace.

nCr(48,4)*nCr(2,1)/nCr(50,5) = 0.1836734693877551

I wouldn't have guessed that high. I guess the simple probability of none to be an Ace is:

(48/50) * (47/49) * (46/48) * (45/47) * (44/46) = 0.8081632653061224 -1 = -0.1918367346938776. Then roughly .0081633 for two aces to come out. Hmm. I guess that's right. Would you have guessed you end up with at least Trips 18.367% of the time starting with an inside pair? I don't think I would have.

For this, we're taking for granted that one card MUST be an Ace, which is why we did the above step. What that leaves us with is the following:

49 Cards, of which:

1.) We select four.

2.) We must select a pair.

3.) We must not select the only remaining Ace.

4.) The pair must be tens, or better.

There are two ways that this could happen. The first is that we get a pair and the other two cards are unpaired. The second way is that we get two pairs with the other four cards. This is going to get a little tricky. We can begin to simplify this though.

The first thing that we are going to start off with is the fact that we ONLY care about pairs that are a pair of tens, or better. Other than for who takes down the main pot, other pairs don't really matter to us at all. We also care that the other cards are not the remaining ace.

Let's figure out the probability of getting only one specific pair before we do anything else. As you know, the remaining card composition consists completely of collected quads of cards not counting aces.

How can we do this? Well, we're going to start with our first card. We want the first card to not be an Ace:

(48/49) = 0.9795918367346939

Our first card has successfully not been the other ace. Now, we want our next card to not be an Ace AND not be the same as our first card:

(44/48) = 0.9166666666666667

Our second card has not been an Ace and has also not been the first card. We want our third card not to be our first card, second card or an Ace.

(40/47) = 0.851063829787234

Our third card has not been the first or second and has also not been an Ace. We want our fourth card not to be the same rank as our first three, but also, to not be an Ace.

(36/46) = 0.78260869565

Okay, if we multiply all of these results together, then we will know the probability of, in our four non-Ace cards, that there is NOT a pair, or two Pair.

0.9795918367346939 * 0.9166666666666667 * 0.851063829787234 * 0.78260869565 = 0.598085672750325667730181804436241307808970030033264991473

In other words, roughly 59.808563% of the time, these other four non-Ace cards have nothing to do with each other.

Remember, this is with one of the five cards that come out already being an Ace, but we will factor that back in later.

With that, 1 - 0.598085672750325667730181804436241307808970030033264991473 = 0.401914327249674332269818195563758692191029969966735008527 and will be one of the following things:

1.) One Pair, Two Off Cards, but neither an Ace.

2.) Two Pairs

3.) Trips.

4.) Quads

The first thing that we will note is that Quads do not actually help us. If quads, then we are not playing one of our Aces.

nCr(45,0)*nCr(4,4)/nCr(49,4) = 0.0000047197417357

The second thing to note is that Trips can help us, but only if they are trip tens or better. There are twelve ranks that can make Trips:

nCr(44,1)*nCr(4,3)*nCr(1,0)/nCr(49,4) = 0.0008306745454889***

That solves for a specific trip, but there are 12: 0.0008306745454889 * 12 = 0.0099680945458668

You'll notice, for this, that the non-trip card is not allowed to be the Ace. We didn't have to do that for Quads because, if Quads, then the Ace couldn't have come out anyway.

0.0099680945458668 * (4/12) = 0.0033226981819556

The reason for the above is that only Trips being 10-K are good enough if we have a pair of Aces and get another Ace. Any pair lower than that and we just have Aces full of something not good enough, like sixes, who likes sixes?

The next thing that we have to do is figure out how many ways there are to make exactly Two Pair. We could just list them:

1.) 2&3
2.) 2&4
3.) 2&5
4.) 2&6
5.) 2&7
6.) 2&8
7.) 2&9
8.) 2&10***
9.) 2&J***
10.) 2&Q***
11.) 2&K***
12.) 3&4
13.) 3&5
14.)3&6
15.) 3&7
16.) 3&8
17.) 3&9
18.) 3&10***
19.) 3&J***
20.) 3&Q***
21.) 3&K***
22.) 4&5
23.) 4&6
24.) 4&7
25.) 4&8
26.) 4&9
27.) 4&10***
28.) 4&J***
29.) 4&Q***
30.) 4&K***
31.) 5&6
32.) 5&7
33.) 5&8
34.) 5&9
35.) 5&10***
36.) 5&J***
37.) 5&Q***
38.) 5&K***
39.) 6&7
40.) 6&8
41.) 6&9
42.) 6&10***
43.) 6&J***
44.) 6&Q***
45.) 6&K***
46.) 7&8
47.) 7&9
48.) 7&10***
49.) 7&J***
50.) 7&Q***
51.) 7&K***
52.) 8&9
53.) 8&10***
54.) 8&J***
55.) 8&Q***
56.) 8&K***
57.) 9&10***
58.) 9&J***
59.) 9&Q***
60.) 9&K***
61.) 10&J***
62.) 10&Q***
63.) 10&K***
64.) J&Q***
65.) J&K***
66.) Q&K***

With that, there are sixty-six different kinds of Two Pair that you can come across. Why does that matter? Well, the next thing we are going to do is figure out the probability of getting a specific Two Pair in four of the remaining 49 cards, and then we will determine which Two Pairs help us and which ones do not.

***As you can see, 28 different types of Two Pairs do not help us and 38 of them help us because they contain at least a Pair of Tens as one of the two.

Any Specific Two Pair:

nCr(41,0)*nCr(4,2)*nCr(4,2)/nCr(49,4) = 0.0001699107024864

Right, so 28 of those do not help us and 38 of them do help us:

0.0001699107024864 * 38 = 0.0064566066944832 (Help)

0.0001699107024864 * 28 = 0.0047574996696192 (Do Not Help)

With that, we go back to our original probability that, of the four cards, there NOT being an Ace and being at least a Pair, Two Pair, Trips or Quads is:

0.401914327249674332269818195563758692191029969966735008527

Okay, we now have to subtract from that all of the ways that can be Quads, Trips or Two Pair and that will leave us with all of the ways that it can be just a single Pair and then two of something else, but not an ace.

(0.401914327249674332269818195563758692191029969966735008527) - (0.0000047197417357 * 12) - (0.0099680945458668) - (0.0001699107024864 * 66) =
0.3806754894388767322698182.

What that means is, roughly 38.06755% of the time, there is a Pair that comes out in the four cards (which can't include the Ace), but does not result in Two Pair, Trips or Quads. A pair and two different non-Ace cards. With that, we can multiply what remains by 4/12 (10's, J's, Q's and K's and get)

0.3806754894388767322698182 * (4/12) = 0.1268918298129589107566060667

The first thing that we have to do is break these down into categories and make sure we are arriving at 100% of the things that do not include the fourth ace for these other four cards:

No Pair: 0.598085672750325667730181804436241307808970030033264991473
Quads: 0.0000566369008284
Trips That Help: 0.0033226981819556
Trips That Don't Help: 0.0066453963639112
Two Pair That Help: 0.0064566066944832
Two Pair That Don't Help: 0.0047574996696192
Single Pairs That Help: 0.1268918298129589107566060667
Single Pairs That Don't Help: 0.2537836596259178215132121333

We will now sum these:

0.2537836596259178215132121333 + 0.1268918298129589107566060667 + 0.0047574996696192 + 0.0064566066944832 + 0.0066453963639112 + 0.0033226981819556 + 0.0000566369008284 + 0.598085672750325667730181804436241307808970030033264991473 = 1.000000000000000000000000004 or 1

In other words, we have accounted for every possible result that could come from us having wired Aces, ending up with a third Ace in any way, shape or form, and then what the other four cards could be doing a la Pairs, Trips, Two Pair and Quads.

We have already determined this, but let us refresh our memories of the probability of ending up with one Ace, but not two, in the first place:

nCr(48,4)*nCr(2,1)/nCr(50,5) = 0.1836734693877551

With that, what we have to do is multiply the probability of that ever happening in the first place by, assuming it has happened, the probability of that actually helping us:

(0.1836734693877551) * (0.0033226981819556 + 0.0064566066944832 + 0.1268918298129589107566060667) = 0.02510286147356284455392532921143324335386517

As you will recall, all of that assumes that we started with wired aces in the first place, so let's go back to that probability:

(0.02510286147356284455392532921143324335386517) * 0.00452488687 = 0.0001135876082811533674419077291092417367334192714833179

0.0001135876082811533674419077291092417367334192714833179 Is the Probability of Ending Up with Aces Full of Tens, or Better, in any way that is NOT quad Aces)

QUICK SUMMARY

I'm starting to lose my place when doing things, so this seems like a good time for a quick summary of what we have accomplished so far. For this section, we will look at all of the different ways we can end up with Aces Full of Tens, or better, using BOTH of our hole cards.

QUADS (Starting with a pair 2s-9s): 0.000295502815999998226983104+++
QUADS (Aces): 0.000036937851999999778372888
QUADS (10-K) OR FULL HOUSE (10-K) or Aces Full via Three Aces on Board: 0.00018311743651063669854194
FULL HOUSE (We Have Pocket Rockets): 0.0001135876082811533674419077291092417367334192714833179

+++Remember, Quads is the only way this one can happen. Any Three Aces on the board would result in a Full House that is not strong enough.

Straight Flushes and Royal Flushes

The final thing we have to look at is the probability of a Straight Flush or Royal Flush with two of the cards being in your hand. This one is extremely simple as all we need to do is look at seven card poker probabilities:

https://wizardofodds.com/games/poker/

This is the same thing as seven-card stud. We have seven cards, they are fixed and the only difference is we need two of them to be in our hand:

The probability of a Straight Flush, in seven cards, is: 0.00027851

There are seven cards, of which, five of those cards make up the final hand. Those cards are either both in our hand, or they are not; therefore:

nCr(5,3)*nCr(2,2)/nCr(7,5) = 0.4761904761904762

That is how often the two cards would be in our hand, ergo:

0.4761904761904762 * .00027851 = 0.000132623809523809526462 is the probability that we would get a SF this way.

The probability of a Royal Flush, in seven cards, is: 0.00003232

Ergo, 0.4761904761904762 * 0.00003232 = 0.000015390476190476190784

Do you have to also win the hand? I guess you could have a SF, with a Royal on the Board, with 9-8 suited, or a losing straight flush with lower cards, but still using both of your cards, somehow. Seems unlikely.

I believe that we have accounted for every possible way this can happen. Let's sum them up:

0.000015390476190476190784 + 0.000132623809523809526462 + 0.000295502815999998226983104 + 0.000036937851999999778372888 + 0.00018311743651063669854194 + 0.0001135876082811533674419077291092417367334192714833179 = 0.0007771599985060737885858397291092417367334

Actually, I did forget one thing. I ended up double-counting Full Houses, starting with a pair of 10's-K's, even if you ended up with quads anyway. Let me go ahead and subtract that from our overall probability really fast:

Any Pair: (4/52 * (3/51) = 0.0045248868778281

Pair 10's-K's: 0.0045248868778281 * 4 = 0.0180995475113124

Now, we need the probability of (Nothing) + (Other Two Pair Cards) + (Three Aces) all in the same hand:

nCr(44,0)*nCr(4,3)*nCr(2,2)/nCr(50,5) = 0.0000018878966943

Now, we have to multiply that by the probability of getting one of the applicable pairs to begin with:

0.0000018878966943 * 0.0180995475113124 = 0.00000003417007591493247181459932

Okay, now we can subtract that:

0.0007771599985060737885858397291092417367334 - 0.00000003417007591493247181459932 = 0.0007771258284301588561140251297892417367334

Right. Therefore, playing every single hand out regardless of what you start with, you would end up with:

0.0007771258284301588561140251297892417367334 or 1/0.0007771258284301588561140251297892417367334 = 1286.7929020195615441719929412418480559773006980544

Or, about 1 in 1,286.8 hands.

Of course, that's doing straight flushes and starting inside pairs, but there are other ways this can happen.

I think that's going to be good enough. There are only two other types of cases to account for and both seem pretty unlikely:

1.) You get a 4OaK or Aces Full of Tens or better, starting with a non-pair, such as A-K or A-10.

2.) You start with any two cards and get quads.

For #1---I could probably figure out #1 easily enough because A-2 through A-9 as starting cards would be disqualified as they cannot make a good enough Full House using both of your cards. That means you would have to make a Full House (but not quads) starting with A-10, A-J, A-Q or A-K. The only way it can ever be Aces full is with one of your cards being an Ace because we have already accounted for all possible Aces full if you start with a Pair. If you do not start with an Ace, and you do not start with a pair, then Aces full of anything doesn't help you as you would not be using both of your cards.

You can start here if you want to do that:

nCr(4,1)*nCr(16,1)*nCr(32,0)/nCr(52,2) = 0.0482654600301659

That's the probability of starting with an Ace as well as some other card that is a Ten, Jack, Queen or King.

For #2---No thanks. I can figure out how likely it is that you would start with two different cards and end up with quads, easily enough.

I could actually probably figure out how likely it is that you kicker is good, based on kicker and what the quads are. For example, if the quads are quad aces, and you have one of the aces, then you do not have to outkick an ace...which is impossible. But, what if your non-Ace was a six? How likely is it that one of the other two cards (A-A-A-x-x) beats a six? Pretty darn likely.

I could probably go through and actually figure it out for any two starting cards that are not a pair. Figuring out how likely it is that it becomes quads is actually pretty trivial. The tough part is I would have to figure out a probability for every quads and every kicker for those quads.

I guess what I am saying is I don't want to.

Probably something like 1 in 1,000, or so, all told.

ANOTHER QUICK ESTIMATION

Another quick way of estimating is to take the seven-card stud information from the WizardofOdds page, which I will link again:

https://wizardofodds.com/games/poker/

We can look at:

Royal: 0.00003232
Straight Flush: 0.00027851
Quads: 0.00168067
Full House: 0.02596102

Quad kickers could get messy, as I mentioned above.

Of these Full Houses, only 1/13th are Aces full:

0.02596102 * (1/13) = 0.0019970015384615

Of which, only 4/12 are Aces full of Tens, or Better:

0.0019970015384615 * 4/12 = 0.0006656671794871667

In a vacuum, we end up with:

0.0006656671794871667 + 0.00168067 + 0.00027851 + 0.00003232 = 0.0026571671794871667

Which are Aces Full of Tens, or better. We stipulate that two must be in your hand:

nCr(2,2)*nCr(5,3)/nCr(7,5) = 0.4761904761904762

Multiply that by the probability of having a hand like that:

0.4761904761904762 * 0.0026571671794871667 = 0.00126531770451769845387778266178254

Or about 1/0.00126531770451769845387778266178254 or 1 in 790.32

Of course, that ignores four of a kinds (where we have two different ranks) in which our kicker causes us not to qualify due to a better kicker being on the board. I assume that is most of them, because the board gets two cards (if we don't have a pair) to have a better kicker than we do and we only get one. I hope if we have something like AJ and the board is AAAJ9 they say it is our jack that plays.

THE BIG PROBLEM

Of course, the big problem is that I don't think you're going to blindly play out every hand to the end hoping to qualify for whatever prize this is going to give you. Are you going to ride out preflop action on 7-2 off just hoping for three deuces to come out on the board AND for your seven to hold up?

Also, there's also the possibility of other players folding before ever seeing a flop.

Unless, maybe the whole table is just going to collude if the prize is large enough to justify it. Ten players all sitting around just calling to the BB and then checking around the table? That wouldn't look suspicious, I don't think. Happens all the time. Probably 1 in 1000 to 1 in 1100, or so. That's per individual hand dealt, so a full table just checking all the way is going to make it more likely.

Of course, because of the BIG PROBLEM, there's no reason to spend the next eight hours getting it exact. There's really no way to even know for sure unless you're committed to making it to the River regardless of what your hand is, or only folding when such a result is no longer possible.

I did a brute force search, and I get 5,418,688 out of the 2,809,475,760 possible deals, or about 1 in 518.

Quote: ThatDonGuy

I did a brute force search, and I get 5,418,688 out of the 2,809,475,760 possible deals, or about 1 in 518.
link to original post

How does that work? That seems way too frequent imo.

Quote: ThatDonGuy

I did a brute force search, and I get 5,418,688 out of the 2,809,475,760 possible deals, or about 1 in 518.
link to original post

Actually, that number is correct only if you are allowed to take a "lesser hand" that qualifies.

I can find three kinds of hands where you can have Aces full of 10s or better using both hole cards, but you can make a better hand without using both:

Hole cards A K, board A A A K and anything else
This is Aces full of Kings with both hole cards, but Four Aces with just one

Hole cards K K, board A A A A K
Again, you can make Aces full of Kings with both hole cards, but Four Aces without either of them

Hole cards A A, board A K K K K
Aces full of Kings with both hole cards, or Four Kings without them

There are:
For the first type, if the last card is not a King: 4 (ranks for the non-Ace hole card) x 4 (suits for the Ace) x 4 (suits for the non-Ace) x 3 (suits for the non-Ace on the board) x 44 (cards for the non-AK) = 8448
For the first type, if the last card is a King: 4 (ranks for the non-Ace hole card) x 4 (suits for the Ace) x 4 (suits for the non-Ace) x 3 (suits for the non-Ace on the board) x 3 (pairs for the Kings on the board) = 192
For the second type: 4 (ranks for the hole cards) x 6 (pairs of that rank) x 2 (cards remaining for the non-Ace card on the board) = 48
For the third type: 6 (pairs of Aces) x 2 (remaining Aces for the board) x 4 (non-Ace ranks for the four of a kind) = 48
The total = 8736
This means there are 5,409,952 winning hands out of 2,809,475,760, or 1 in 519.

That does seem a little high of a probability for this, but it checks out in simulation.