# Pocket pair starting hands vs. probability of opponent having a pocket pair

According to Wikipedia, the probability of being dealt any pocket pair as the starting hand is ~6% (78/1326=0.0588), but yet the probability of facing another player holding a larger pair when holding a pocket pair yourself in pre-flop is significantly high especially when you're playing with 6 or more players. This is strange to me given the likelihood of getting a pocket pair for anyone is only 6%.

The following table from Wikipedia shows the probability that before the flop another player has a larger pocket pair when there are one to nine other players in the hand. Note the figures when you're playing against 6 or more players is ridiculously high. If anyone has a 6% chance of getting a pocket pair as starting hand, when you look at the probabilities in the against 6-9 players columns for comparison it simply doesn't make sense, to me anyway. How is this explained?

ps: will need to x100 to get percentage for the probabilities.

Edit, this is a more accurate table:

• can I suggest a more descriptive post title. I think a lot of calculations are strange. Maybe something to do with starting hands and pocket pairs. Commented Mar 11, 2013 at 10:36
• By the way, that table has been shown to have errors, especially for lower ranked pairs and a large number of players. For example for 22 with 9 players it shows 36.3% while the correct value is 41.9%. A more accurate table can be found here.
– Theo
Commented Mar 11, 2013 at 14:46
• If the current table has errors, perhaps replace it with the one that doesn't?! Commented Mar 11, 2013 at 23:01
• If you are wondering why the wiki entries are wrong then I think it is because the probabilities were treated as independent instead of dependently. If I have time I will try to post an answer showing how you would actually calculate these entries. It is worth noting that the table you provide is just from a simulation of 1 million hands and so not exact (although it is accurate enough for any purposes) Commented Mar 12, 2013 at 17:39
• And I thought the toughest part of Texas Hold'um was not giving any tale tale signs away. You guys are talking Enstine! Commented Nov 24, 2021 at 9:14

First off, I believe it is easier to understand the probability of getting dealt a pocket pair this way:

Your first card can be any of 52 cards, so it is not relevant in the calculation. The 2nd card must be one of the 3 cards that match the value of your other card. There are 51 cards left in the deck now, so the probability of being dealt a pocket pair is 3/51 = 1/17 = 5.88%.

Another way to think about this is that you should be dealt a pocket pair, on average, once every 17 hands.

It's difficult to answer your question as to why the probability "seems too high", since the numbers won't seem high to someone who lives and breathes probabilities. But perhaps this will help: At a 10 person table you're getting 10 tries at a 1/17 shot. The chance of at least 1 person getting a pocket pair is 45%. So maybe now it's easier to see that the likelihood of 2 or more getting a pocket pair is less than 45%, but not "much" less.

• I don't think the OP is asking about "the likelihood of 2 or more [players] getting a pair", so not sure how this answers the question? Commented Dec 29, 2016 at 12:26
• @3N1GM4 - this question is exactly about 2 or more players getting dealt a pocket pair. I'm not sure what you mean?
– TTT
Commented Dec 29, 2016 at 17:31
• @3N1GM4 - Ahah! I'm guessing you thought I meant having a pair on the river? I added the word "pocket" to clarify. ;)
– TTT
Commented Dec 29, 2016 at 17:48
• As I'm reading it, the question is about the chance of one or more players having a pair, not two or more? Commented Dec 29, 2016 at 18:00
• Oh wait, I see my misunderstanding, two or more players (in total), including Hero. I was thinking of "one other player", already assuming Hero has a pair as per the question. Sorry for the confusion! Commented Dec 29, 2016 at 18:02

A poker math geek gave me the following formula to use as an estimation.

(Edit: The credit for this shortcut goes to Phil Gordon and is known as the Gordon Pair Principle.)

Let's say you have `55`:

1. Work out how many pairs are higher than your pair - `66`, `77`, `88`, `99`, `10`, `JJ`, `QQ`, `KK`, `AA` = 9 pairs left

2. Count how many players are left = 8

3. Multiply `higher pairs` by `players left` and divide by 2

So in the above example, with `55` and 8 players left:

``````(9 x 8) / 2 = 36% chance of at least one of the remaining players having a higher pocket pair
``````

Say you have `99` on the button and it's folded to you:

1. 5 pairs left
2. 2 players left
3. `(5 x 2) / 2 = 5%` chance of at least one of the remaining players having a higher pocket pair
• This is actually a pretty good approximation based on my own calculations - it gives results within a 2% margin of the actual result, which is definitely good enough to make decisions at the table. Commented Dec 29, 2016 at 12:31

There are 10 players at the table, any one of them has a 6% chance of having a pocket pair.

So the chance that no-one is dealt a pocket pair is 0.94^10 ≈ 0.54, ie chance of someone having a pocket pair at a FR table ≈ 0.46.

Excluding card removal for ease:

You have a pocket pair, what is the chance that out of 9 players none of them has a pocket pair ?

0.94^9 ≈ 0.57, so the chance of one of the other 9 having a pair is ≈ 0.43

(math is approximated in more than 1 way in the examples)

The equation used is a standard way of working out probabilities of multiple events.

If you roll a die, the chance of a 6 is 0.1667

If you roll the die 10 times, what is the chance that you will roll at least one 6 ?

You can't sum the probabilities, 0.1667 added 10 times would give a probability of 1.667 !

You say, on this roll my chance of not rolling a 6 is (1-0.1667). If I roll it again, the chance is the same on that roll, so overall the chances of not having rolled a 6 in either try are

(1-0.1667) * (1-0.1667) ≈ 0.69

so the chances of rolling the die 10 times and never having a 6 are

(1-0.1667)^10 ≈ 0.16

so the chances of rolling a die 10 times and rolling at least one 6 are

1 - 0.16 ≈ 0.84

• What relation does this answer have to the question? Commented Mar 11, 2013 at 20:15
• Also you might want to consider treating the probabilities as discrete distributions as everything is equally likely. This is obviously not possible for even small calculations but it gives better understanding of what is going on. If we use the die example with just 2 rolls we see that the possible combinations are: {11,12,13,14,15,16,21,22,23,24,25,26,31,32,33,34,35,36,41,42,43,44,45,46,51,52,53,54,55,56,61,62,63,64,65,66} Now that is 36 possibilities each as likely as each other and 25 don't have a 6 so there is 25/36=0.694r chance of not having a 6 Commented Mar 11, 2013 at 20:19
• I don't see how this answer deals with the aspect of the question where we're trying to determine the chance of another player holding a higher pair than ours. Commented Dec 29, 2016 at 12:09
• It is not the same as rolling the dice as you pick up the dice and roll again. In poker the 2 cards are down and select the next 2. 1 or the prior 2 could have been an Ace which kill an out. See my answer. Commented Mar 31, 2017 at 20:53

One way to solve this is combination

Will use combin(x,y) for (x/y) as that is what you can use in Excel

How many ways to make a pair? combin(4,2) = 6 (sc, sh, sd, ch, cd, hd)

You have 2 card so how many other starting hands
combin(50,2) = 1225

h = number of other hands
Now number of ways to make a pair is combin(6,1) - the pair is a 1
The calculation for one higher named pair is (E.G. have KK chance of AA)

combin(6, 1) * combin(1225-6, h-1) / combin(1225, h)

combin(1225-6, h-1) is the number of other possible combinations
combin(1225, h) is the total number of combinations

For QQ just double KK as AA and KK are mutually exclusive
For JJ just triple

Running in Excel

``````         1       2       3       4       5       6       7       8       9
KK  0.0049  0.0098  0.0146  0.0194  0.0241  0.0288  0.0335  0.0381  0.0427
QQ  0.0098  0.0195  0.0291  0.0387  0.0482  0.0576  0.0669  0.0761  0.0853
JJ  0.0147  0.0293  0.0437  0.0581  0.0723  0.0864  0.1004  0.1142  0.1280
TT  0.0196  0.0390  0.0583  0.0774  0.0964  0.1152  0.1338  0.1523  0.1706
99  0.0245  0.0488  0.0729  0.0968  0.1205  0.1440  0.1673  0.1904  0.2133
88  0.0294  0.0585  0.0874  0.1161  0.1445  0.1727  0.2007  0.2284  0.2559
77  0.0343  0.0683  0.1020  0.1355  0.1686  0.2015  0.2342  0.2665  0.2986
66  0.0392  0.0780  0.1166  0.1548  0.1927  0.2303  0.2676  0.3046  0.3413
55  0.0441  0.0878  0.1312  0.1742  0.2168  0.2591  0.3011  0.3427  0.3839
44  0.0490  0.0976  0.1457  0.1935  0.2409  0.2879  0.3345  0.3807  0.4266
33  0.0539  0.1073  0.1603  0.2129  0.2650  0.3167  0.3680  0.4188  0.4692
22  0.0588  0.1171  0.1749  0.2322  0.2891  0.3455  0.4014  0.4569  0.5119
``````
• Not enough love for this answer, although it gives different results to my approach - is this due to card removal and the fact I cannot simply multiply to scale up for a larger number of opponents as per the end of my answer, or have I missed something? Commented Dec 29, 2016 at 13:11
• I don't want to edit and make this active but cannot use 1-power((1-x),h) because a single ace in a hand is not a pair and it take away an out for a pair. Commented Mar 31, 2017 at 20:47
• I think this is wrong in that it does not consider two pocket pair. Commented May 9, 2017 at 22:21

This is using binomial distribution
I know it does not agree with other answers but I think this is correct

``````this is for KK - for the rest you just multiply

0 hand          1 hand          2 hand
1   0.9951020408    0.0048979592
2   0.9902280716    0.0097479384    0.0000239900
3   0.9853779750    0.0145502900    0.0000716175
4   0.9805516339    0.0193053644    0.0001425335
5   0.9757489320    0.0240135094    0.0002363922
6   0.9709697535    0.0286750707    0.0003528516
7   0.9662139833    0.0332903915    0.0004915726
8   0.9614815067    0.0378598132    0.0006522199
9   0.9567722095    0.0423836746    0.0008344612

1       2       3       4       5       6       7       8       9
K   0.0049  0.0098  0.0146  0.0194  0.0242  0.0290  0.0338  0.0385  0.0432
Q   0.0098  0.0195  0.0292  0.0389  0.0485  0.0581  0.0676  0.0770  0.0864
J   0.0147  0.0293  0.0439  0.0583  0.0727  0.0871  0.1013  0.1155  0.1297
T   0.0196  0.0391  0.0585  0.0778  0.0970  0.1161  0.1351  0.1540  0.1729
9   0.0245  0.0489  0.0731  0.0972  0.1212  0.1451  0.1689  0.1926  0.2161
8   0.0294  0.0586  0.0877  0.1167  0.1455  0.1742  0.2027  0.2311  0.2593
7   0.0343  0.0684  0.1024  0.1361  0.1697  0.2032  0.2365  0.2696  0.3025
6   0.0392  0.0782  0.1170  0.1556  0.1940  0.2322  0.2703  0.3081  0.3457
5   0.0441  0.0879  0.1316  0.1750  0.2182  0.2613  0.3040  0.3466  0.3890
4   0.0490  0.0977  0.1462  0.1945  0.2425  0.2903  0.3378  0.3851  0.4322
3   0.0539  0.1075  0.1608  0.2139  0.2667  0.3193  0.3716  0.4236  0.4754
2   0.0588  0.1173  0.1755  0.2334  0.2910  0.3483  0.4054  0.4621  0.5186
``````

My approach would be to start with the case of holding `KK` against one player, where the chance of them holding a higher pair (in this case, exactly `AA`) is:

``````(4/50)*(3/49)
``````

or

``````(4*3)/(50*49)
``````

if you'd prefer. This comes to `0.0049` or 0.49%.

From here, we can extrapolate for the number of players by simply multiplying the above by the number of players, so against 2 players, we are twice as likely to face `AA` (this ignore card removal, see below), and so we expect this `0.0049 * 2 = 0.0098` or 0.98% of the time.

Similarly, we can extrapolate our chances of facing a larger pair when we ourselves hold a smaller pair by just multiplying by the number of larger pairs. When we hold `QQ`, the chance of a single opponent holding a higher pair is twice as large (as holding `AA` or `KK` are mutually exclusive). We can show this explicitly:

``````[chance of facing AA or KK when holding QQ]
= (8/50)*(3/49)
= 0.0098 (0.98%)
``````

or by simply doubling our original result.

By this method, we can calculate all of the values:

``````/---------------------------------------------------------------------------------------\
|      |                                    Opponents                                   |
| Hand |    1   |    2   |    3   |    4   |    5   |    6   |    7   |    8   |    9   |
|------+--------+--------+--------+--------+--------+--------+--------+--------+--------|
|  KK  |  0.49% |  0.98% |  1.47% |  1.96% |  2.45% |  2.94% |  3.43% |  3.92% |  4.41% |
|  QQ  |  0.98% |  1.96% |  2.94% |  3.92% |  4.90% |  5.88% |  6.86% |  7.84% |  8.82% |
|  JJ  |  1.47% |  2.94% |  4.41% |  5.88% |  7.35% |  8.82% | 10.29% | 11.76% | 13.22% |
|  TT  |  1.96% |  3.92% |  5.88% |  7.84% |  9.80% | 11.76% | 13.71% | 15.67% | 17.63% |
|  99  |  2.45% |  4.90% |  7.35% |  9.80% | 12.24% | 14.69% | 17.14% | 19.59% | 22.04% |
|  88  |  2.94% |  5.88% |  8.82% | 11.76% | 14.69% | 17.63% | 20.57% | 23.51% | 26.45% |
|  77  |  3.43% |  6.86% | 10.29% | 13.71% | 17.14% | 20.57% | 24.00% | 27.43% | 30.86% |
|  66  |  3.92% |  7.84% | 11.76% | 15.67% | 19.59% | 23.51% | 27.43% | 31.35% | 35.27% |
|  55  |  4.41% |  8.82% | 13.22% | 17.63% | 22.04% | 26.45% | 30.86% | 35.27% | 39.67% |
|  44  |  4.90% |  9.80% | 14.69% | 19.59% | 24.49% | 29.39% | 34.29% | 39.18% | 44.08% |
|  33  |  5.39% | 10.78% | 16.16% | 21.55% | 26.94% | 32.33% | 37.71% | 43.10% | 48.49% |
|  22  |  5.88% | 11.76% | 17.63% | 23.51% | 29.39% | 35.27% | 41.14% | 47.02% | 52.90% |
\---------------------------------------------------------------------------------------/
``````

However, this is not strictly the correct set of values, because the chances of a player making a pair are not mutually exclusive from the chances of another player in the same hand making a pair.

Consider for example a game with 9 players - the chances of a single player getting a pair (without any other information) is `3/52 = 0.0577 (5.77%)`, but what if we already know that some players already have pairs? Let's say for example that we know the hands of 3 other players at the table, who each have 44, 77 and KK respectively. Well now our chances of making a pair are:

``````[chance of making 44, 77 or KK] + [chance of making any other pair]
= [(6/46)*(1/45)] + [(40/46)*(3/45)]
= 0.0609 (6.09%)
``````

We have a greater chance of making a pair because other players have already made pairs, which has polarised the deck with regards to pairs. Taking this to its logical conclusion, think about a situation at a 10-handed table where all 9 of your opponents have pairs (AA x 2, KK x 2, QQ x 2, JJ x 2 and TT). Now our chances of making a pair are:

``````[chance of making JJ+] + [chance of making TT] + [chance of making 22-99]
= [0] + [(2/34)*(1/33)] + [(32/34)*(3/33)]
= 0.0873 (8.73%)
``````

Similarly, sometimes if we know that other players have not made pairs, our chances of making pairs also increases; for example, if we know 3 players hold `AK`, `QJ` and `T9` respectively, our chances of making a pair are:

``````[chance of making 99+] + [chance of making 22-99]
= [(18/46)*(2/45)] + [(32/46)*(3/45)]
= 0.0638 (6.38%)
``````

However, if lots of different ranks are known to be in other players' hands, our chances to make a pair will be decreased. For example, if we are in a 10-handed game where we know 6 of our opponents hold `AK`, `QJ`, `T9`, `87`, `65`, `43`, our chances of making a pair are:

``````[chance of making 22] + [chance of making 33+]
= [(4/40)*(3/39)] + [(36/40)*(2/39)]
= 0.0538 (5.38%)
``````

For this reason, my simplified approach above is not 100% accurate, because it is not correct to say that if the chance of facing an overpair with `KK` against one opponent is 0.49%, then it must be double this chance against two opponents - this is flawed because the chance of that second player having a pair is affected by what the first player held. At least that's my understanding.

In any case, the numbers above will certainly be close enough to the "real" figures to be useful to you, and hopefully this answer has helped show you how these numbers are calculated.

Hmm, so I tried to brush up some of my very rusty math, and came up with this:

Chances that once the flop comes, at least one of the remaining players in the hand has either a pocket pair or flopped a pair (or better), based on the number of players left:

10....... 99%

9......... 98%

8......... 97%

7......... 95%

6......... 92%

5......... 88%

4......... 82%

3......... 73%

2......... 58%

Please, shoot holes in these numbers. I haven't worked out stuff like this in 30 years.

As to how one would compute the odds of someone flopping top pair, I could not even being to think.

• The question has nothing to do with the flop Commented Feb 15, 2016 at 19:52