I have looked on this forum, founf a topic about the odds of hitting one pair with an unpaired hand, but can't find the answer to this question: what are roughly the odds of hitting exactly top pair on an XYZ board with two unpaired cards in the hole? For instance, if you have JT, how often would you hit a board like Jxy where x< J and y<J or Txy where x< T and y<T where x< T and y<T. I see that the odds are lower for a hand like 76 than for a hand like AK, but maybe you guys can give me a rough idea.
Otherwise, e.g. on a Q93 flop, what are the odds my opponent, supposed his vpip is 100%, holds exactly Qx with a pair of queens.
I'm sorry, i'm no great mathematician.

Thanks!

I think you are over thinking this haha. So, your odds of hitting a pair on the flop are roughly 27% (correct any of the numbers if I'm wrong someone). So lets divide that by two because you are talking about a specific cards chances and not two. So one card has a 13.5% chance of making a pair on the flop. Now, how high the card is depends on how likely the odds are that it will be the highest card showing on the flop. For example, an eight has to worry about half of the deck, but an Ace is guaranteed to be top pair.

Using the odds of flopping exactly one pair (28.6% I believe), you then just need to calculate the odds of the other two cards on the flop being lower than your pair. This is actually fairly easy to do, but obviously varies depending on how high your cards are. All you have to do is add up all the cards lower than your pair and find the odds of those cards hitting the flop. Using JT has an example to hit EXACTLY top pair, you have 17 cards that will either be higher or give you better than one pair, and 32 cards that are lower. Since one of the cards is 'set' for the example, you can calculate the odds of the other two cards both being lower by going 1-(32/49*31/48)= 57.8%. Now multiply this by the chance of flopping a pair, 28.6% (.286*.578)= 16.5% chance of flopping exactly top pair with JT.

Actually, forgot to mention my example doesn't take into account instances where you flop a straight on a 789 flop, or the other two cards also forming a pair, such as a flop of J66. It would be significantly more complex to eliminate those possibilities..

For J T, if "exactly" top pair eliminates quads, trips and two pair (either from a board pair or pairing both J and T), the probability is 13.71%. Also, you can't flop both top pair and a straight.

If R1 is the higher rank, you can show that there are 48*C(R1-3,2) combos that meet the criterion, R1>4.

For the lower rank, say R2, the number of combos is 48*C(R2-2),2), R2>3.

Note that for J T, each rank has the same number of combos, namely 1344.

For holding R1,R2, to get the probability, add the combos together and divide by C(50,3)

Clever, I'm still in the learn as I go stage in probability. Trouble is I would never have thought of doing that and so searching for how to do it would be problematic.

Too often my approach is:

solve it, find a better way, find another better way....

Is there some 80/20 resource you could recommend that can provide a lot of such examples that I could reference?

So let's walk it all the way through. I can see my hand and it contains two unpaired cards, we'll call them V and W. There are 50C3 = 19600 flops. Their flavours are:

2 VVV/WWW
44 XXX
18 VVW/VWW
264 VVX/WWX
396 VXX/WXX
2640 XXY
396 VWX
5280 VXY/WXY
10560 XYZ

Since we only want to count top-pair XYZ flops, we're looking at (a subset of) the 5280 flops that are VXY or WXY, where there are naturally 2640 of each. The following table counts the number of VXY flops with no overcards, depending on how many unoccupied ranks are above your card. For instance if you have K8, there is 1 rank (A) above the king and 5 (A,Q,J,T,9) above the 8. Just add up the two percentages to get your answer, so for K8 it's 11.02 + 3.67 = 14.69%.

ranks -- count -- %
0 -- 2640 -- 13.47
1 -- 2160 -- 11.02
2 -- 1728 -- 8.81
3 -- 1344 -- 6.86
4 -- 1008 -- 5.14
5 -- 720 -- 3.67
6 -- 480 -- 2.45
7 -- 288 -- 1.47
8 -- 144 -- 0.73
9 -- 48 -- 0.24
10,11 -- 0 -- 0.00

The second question---If the flop is XYZ, what's the probability that Villain flopped top pair---is actually easier. There are 3 cards (in your example, the three queens) that match the top flop card, and 40 that miss completely, so 120 possible hands give Villain top pair. This is out of 49C2 = 1176 hands for a percentage of 10.2%.

But that's only if you haven't looked at your hand, and given that there's a flop, you probably have. This affects the percentage depending on whether, and what, you've hit.

You've hit -- Villain %
top set -- 3.70
other set or bottom 2 pair -- 11.10
other 2 pair -- 7.40
pocket pair or air -- 10.55
top pair -- 7.22
lower pair -- 10.82

Hmm interesting. How does having a lower pair give a higher chance of the opp having top pair, compared to if you had air though?

Let’s do a specific case. Suppose the flop was A J 3.

You hold T 4, an air hand relative to the flop (actually, an air hand – period.)

For villain to have exactly top pair, he has to have Ax, x <> A, J, or 3

The deck has 47 remaining cards, 3 of which are aces of which V has to have exactly one. Since no pair is allowed for the other card (x), a J or 3 cannot be in his hand. That leaves 47 -3 -6 = 38 allowable cards. Therefore the number of successful combos is 3*38 that results in the stated 10.55%

Now, instead of your air hand, you hit a pair, say by having J 4. That means villain can have any of 47 -3 – 5 = 39 allowable cards, for now there are only 3 “bad for villain” jacks instead of 4 since you hold one and it was already excluded. Therefore the probability of V hitting exactly top pair is 3*38/C(50,2) = 10.82%

This is another version of the somewhat non-intuitive fact that if you have a pocket pair, it increases the chance that an opponent also has a pocket pair.

thanks for your responses guys

This is a method I use so that it is easy to rationalize what is being done. I'm going to figure out the chances of A9 hitting top pair.

50*49*48 = 117600 total combos
(You have 2 of the 52 possible cards so it goes 50 for the first card and once less for each proceeding card.)

Ace: 3*44*43*3 = 17028
Nine: 3*28*27*3 = 6804
(First number is number of other cards it can pair. The second and third numbers are the other cards that are possible that don't pair or is lower. The fourth number is for rearrangements of the order. The fourth number is always 6, 3, or 1. In this particular case I figured out for both the nine and the ace seperately.)

Ace + Nine = 23832 which is our top-pair-hitting flops
(I have to add them together in order to get all the possible flop combos which work.)

top-pair-hitting flops/total combos


Which is this 23832/117600 = 20.27% chance of flopping top pair.

I've used this method for awhile. I get confused with all the C and P things with factorials and such so I tend to stick to this. Is my method correct? The hardest part is trying to figure out the fourth number when deciding rearrangements of the order. I have a hard time figuring out if it is 6, 3, or 1.

That's total permutations (which is fine). If you wanted combos you'd divide by 3!.
What you did is correct. It was a 3 here because there were 3 possibilities for which flop card was the A (or the 9). The total flop orders should come to 3!. Before the *3 at the end, only 2! orders were accounted for. After *3, there were 3!.

You might be wondering how it's known that 2! orders were accounted for before the *3.

Let's revisit the ace: 3*44*43.

3 chooses from the set of aces.
44 chooses from the set of non-aces.
43 again chooses from the set of non-aces.

44*43 counts 2! orders because it chooses one element from the same set twice (separately). Choosing from the same set twice counts order. Had you chosen two cards in one step by saying C(44,2), that wouldn't count order.

Earlier I pointed out that 50*49*48 is total permutations. That's because you're choosing from the same set three times, so it counts all three-card orderings.

So whenever you see choosing from the same set multiple times, you know order is being counted. How many orders depends on how many repeated choosings there were, and how many cards were being chosen in each choosing.

If you understand all that, here's another example. How many ways to deal all hearts on the flop and turn, if order of flop&turn matters but not order of the flop cards?
C(13,3)*10
We chose from the same set (hearts) twice. We chose 3 cards and 1 card, a total of 4, so we counted C(4,3) or C(4,1) orders. So another way to write that solution would have been to say
C(13,4) * 4

All hearts on flop, turn and river = C(13,3)*10*9
I'll let you take a stab now: how many orders does that count?

Going back to the 4th factor you asked about, on the flop it would be a *6 only if you're choosing from 3 different sets. Like if you want the flop to be JQK that's 4*4*4*6 = 384. That didn't choose from the same set twice anywhere so it counted no order, so to make it a permutation we multiply by 3!. But for that example, if I wanted perms, I'd say 12*8*4 = 384. You could say that chooses from the same set of 3 ranks three times.

Finally, it's *1 if you chose from the same set three times. Permutations of all-heart flops = 13*12*11. That already takes care of all 3! orders so no factor (or only a *1) is needed.

One more thing: if you do the problem using combos instead of perms, you never need a factor at the end. For instance, chance of flopping top pair of A's or A's up = 3*C(44,2) / C(50,3)

What heehaw said is fine. The only problem is that TG’s answer is wrong if you are looking for EXACTLY top pair. The term 3*44*43*3 he used for an ace flop includes the case of a pair on the flop excluding aces and nines. For example, he includes a flop of A88, which would give hero two pair. The correct term for excluding such events using permutations is 3*(11*10/2)*4*4*6. That’s 3 ways for the ace, 11*10/2 ways for a pair of non-ace, non-9 ranks, 4 ways for a card from each of those ranks and 6 ways to order the ace and non –ace ranks.

Then a flop for exactly one ace and no pair has 3*55*4*4*6 = 15840 ways for a probability of 15,840/117,600 = 13.47%, the same result StMisbehavin and I got.

My formula for the higher pair flop was 48*C(Rank-3, 2)/C(50,3). Ace has rank 14. Therefore,

Prob =48* C(11,2)/C(50,3) = 48*11*5*6/(50*49*48) =13.47%.

Oh wow, I have learned a lot from this. Thank you! A light bulb just went on in my head.

With C(13,3)*10*9 you are counting 3 orders.

The C(13,3) is selecting 3 cards discounting order. The 10*9 will allow for cards like 5 2 to also show up like 2 5 so they count as 2. So add them together and we get 2+1=3.

In order to make it five orders I would make it 13*12*11*10*9 or to do one order I could just do C(13,5).

For the hitting top pair problem, I originally was thinking I was eliminating more and more cards from all the total permutations. When selecting a random card you can have another random card but one less. So I considered the whole deck thinking it got smaller going down and that the 44*43 would not allow the same card to show up. I was wrong, because this can happen:

A 7 Q
A Q 7

The A doesn't move around, but the other cards can according to the calculation. This is the 2! that you were talking about. Which can be simplified to just 2, but to grasp the concept I understand why you put 2! now.


I've done some further research on the internet to figure out P(x,y) and C(x,y) and let me know if I got this right.

P(44,3) = 44*43*42
C(44,3) = 44*43*42 / (3*2*1)

The last division part is what eliminates other permutations of the same thing. In your example you said C(4,3) and C(4,1), because they are symmetrical and get bigger as the last number gets closer to the most middle number possible.

Now I am a fan of C-ing and P-ing like all the other probability masters. Woohoo! That is if I got this right.

Everything quoted above is right. In general, C(n,r) = C(n, n-r). That's a useful identity to know.

C(13,3)*C(10,1)*C(9,1) is three draws from the same set three times. But the extra wrinkle is that not each draw is of one card. One simple way to get the answer is to divide all that by C(13,5). The difference between that an C(13,5) has to be the number of orderings counted. If you do that, you see that it's 20 times as large as C(13,5), so it counts 20 orderings.

Here's how to understand why it's 20. We're treating the 3 flop cards as indistinguishable. It's really like how to arrange the letters FFFTR. There are C(5,3) ways to distribute the F's among the other letters, then 2! ways to order the T & R. C(5,3)*2=20. Or you can start the calculation with any letter you want: there are 5 places to put the T, then 4 to put the R, then the F's spots are forced.

Or without the letter analogy, you can just say, there are C(5,3) possibilities for which 3 cards are on the flop, then 2 possibilities for which one is on the turn, then the river is forced. (Or again, you can start that calculation with whichever street you want. E.g. 5 possibilities for which card is the river, etc.)
They make life easier for sure.

@statmanhal:

Okay so you did it in a way that puts rank in a category so you can do multiple cards at once in a program or by following an equation.

To eliminate pairs I can also do this:

Ace: 3*44*40*3
Nine: 3*28*24*3
Ace+Nine / P(50,3)


I can also do this:

Ace: 3*44*40
Nine: 3*28*24
Ace+Nine / C(50,3)

This gives a total of 18.61% for the whole thing excluding XYY flops. Is this right?

My thinking is that you simplified by doing:
3*4(Rank-3)*4(Rank-4)
48*(Rank-3)(Rank-4)
48*C(Rank-3,2)

and...

48*(Rank-2,2) for the second card (the Nine in this case) you take off 4, because you don't have to worry about the higher card pairing since it was already eliminated.

But wouldn't (Rank-3)(Rank-4) simplify into P(Rank-3,2) instead of C(Rank-3,2).

I think I may be confusing myself again somehow.

Those aren't the same. The first way is right. The second way counts 2! orders when the denominator counts no order, so it gets an answer of 37.22% (twice as much as the first answer).

44*40 counts 2! orders because you're picking from the same set of 11 ranks twice. It distinguishes between say AJQ and AQJ. So you have to divide by 2!.
So, Ace = 3*44*40/2 = 880

Or you could have said 3 * C(11,2) * C(4,1)^2 = 880

I decided to post this chart on here, because I may use it in the future. Instead of having it on a random document that may get lost on my computer, I have decided to post it up on here where I can easily access it.

Order by card rank: Order by percentage:
AK = 26.94% - - - - 26.94% = AK
AQ = 24.49% - - - - 24.49% = AQ
AJ = 22.29% - - - - 22.29% = AJ
AT = 20.33% - - - - 22.04% = KQ
A9 = 18.61% - - - - 20.33% = AT
A8 = 17.14% - - - - 19.84% = KJ
A7 = 15.92% - - - - 18.61% = A9
A6 = 14.94% - - - - 17.88% = KT
A5 = 14.20% - - - - 17.63% = QJ
A4 = 13.71% - - - - 17.14% = A8
A3 = 13.47% - - - - 16.16% = K9
A2 = 13.47% - - - - 15.92% = A7
KQ = 22.04% - - - - 15.67% = QT
KJ = 19.84% - - - - 14.94% = A6
KT = 17.88% - - - - 14.69% = K8
K9 = 16.16% - - - - 14.20% = A5
K8 = 14.69% - - - - 13.96% = Q9
K7 = 13.47% - - - - 13.71% = JT
K6 = 12.49% - - - - 13.71% = A4
K5 = 11.76% - - - - 13.47% = K7
K4 = 11.27% - - - - 13.47% = A3
K3 = 11.02% - - - - 13.47% = A2
K2 = 11.02% - - - - 12.49% = Q8
QJ = 17.63% - - - - 12.49% = K6
QT = 15.67% - - - - 12.00% = J9
Q9 = 13.96% - - - - 11.76% = K5
Q8 = 12.49% - - - - 11.27% = Q7
Q7 = 11.27% - - - - 11.27% = K4
Q6 = 10.29% - - - - 11.02% = K3
Q5 = 9.55%Z - - - - 11.02% = K2
Q4 = 9.06% - - - - 10.53% = J8
Q3 = 8.82% - - - - 10.29% = T9
Q2 = 8.82% - - - - 10.29% = Q6
JT = 13.71% - - - - 9.55% = Q5
J9 = 12.00% - - - - 9.31% = J7
J8 = 10.53% - - - - 9.06% = Q4
J7 = 9.31% - - - - 8.82% = T8
J6 = 8.33% - - - - 8.82% = Q3
J5 = 7.59% - - - - 8.82% = Q2
J4 = 7.10% - - - - 8.33% = J6
J3 = 6.86% - - - - 7.59% = T7
J2 = 6.86% - - - - 7.59% = J5
T9 = 10.29% - - - - 7.35% = 98
T8 = 8.82% - - - - 7.10% = J4
T7 = 7.59% - - - - 6.86% = J3
T6 = 6.61% - - - - 6.86% = J2
T5 = 5.88% - - - - 6.61% = T6
T4 = 5.39% - - - - 6.12% = 97
T3 = 5.14% - - - - 5.88% = T5
T2 = 5.14% - - - - 5.39% = T4
98 = 7.35% - - - - 5.14% = 96
97 = 6.12% - - - - 5.14% = T3
96 = 5.14% - - - - 5.14% = T2
95 = 4.41% - - - - 4.90% = 87
94 = 3.92% - - - - 4.41% = 95
93 = 3.67% - - - - 3.92% = 86
92 = 3.67% - - - - 3.92% = 94
87 = 4.90% - - - - 3.67% = 93
86 = 3.92% - - - - 3.67% = 92
85 = 3.18% - - - - 3.18% = 85
84 = 2.69% - - - - 2.94% = 76
83 = 2.45% - - - - 2.69% = 84
82 = 2.45% - - - - 2.45% = 83
76 = 2.94% - - - - 2.45% = 82
75 = 2.20% - - - - 2.20% = 75
74 = 1.71% - - - - 1.71% = 74
73 = 1.47% - - - - 1.47% = 65
72 = 1.47% - - - - 1.47% = 73
65 = 1.47% - - - - 1.47% = 72
64 = 0.98% - - - - 0.98% = 64
63 = 0.73% - - - - 0.73% = 63
62 = 0.73% - - - - 0.73% = 62
54 = 0.49% - - - - 0.49% = 54
53 = 0.24% - - - - 0.24% = 53
52 = 0.24% - - - - 0.24% = 52
43 = 0% - - - - 0% = 43
42 = 0% - - - - 0% = 42
32 = 0% - - - - 0% = 32