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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.

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Edit, this is a more accurate table:

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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. –  hmmmm Mar 11 '13 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 Mar 11 '13 at 14:46
    
If the current table has errors, perhaps replace it with the one that doesn't?! –  Toby Booth Mar 11 '13 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) –  hmmmm Mar 12 '13 at 17:39

2 Answers 2

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 pair is 45%. So maybe now it's easier to see that the likelihood of 2 or more getting a pair is less than 45%, but not "much" less.

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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

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What relation does this answer have to the question? –  hmmmm Mar 11 '13 at 20:15
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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: {11,12,13,14,15,16,21,22,23,24,25,26,31,32,33,34,35,36,41,42,43,44,45,46,51,52,5‌​3,54,55,56,61,62,63,64,65,66} 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 –  hmmmm Mar 11 '13 at 20:19

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