Odds that someone makes or holds a pair on the flop, based on the # of players

Question

QUESTION: In Hold ’Em, what are the combined odds after the flop that at least one player at the table either has a pair in the hole or has flopped at least a pair, depending on the number of players?

DISCUSSION:

I’ve seen the odds of flopping a pair with any two unpaired hole cards pegged at a little less than 1/3, at 32.5%.

I’ve also seen the odds of being dealt a pair pegged at 5.88%, or about 1 in every 17 hands.

So presumably my chance of having either a pocket pair or making a pair on the flop is a combined ~38.38%, or slightly worse than 2 out of 5 times.

It would be useful to have a rule of thumb to use at the table for the likelihood that if I missed the flop, the rest of the table did, too, according to how many people stayed in to see the flop. What are the odds that one of the players is either holding a pair or has flopped a pair (the board matches one of their hole cards), based on the number of players?

Secondarily, how about the chances of someone flopping top pair, based upon the number of players? (This may require leaving aside the fact that most players see flops with Broadway cards than lower ones; I would be more than happy with an answer that just dealt with random holdings.)

A simple chart would be great but my statistics skills are quite rusty. My inclination is to take the chances for one player, subtract it from 1, and multiply that by the number of players, then subtract for the possibility of cross-holdings. But with this speculation I am mainly just exposing my ignorance, no doubt.

Any help would be much appreciated. (NOTE: I mistakenly posted this in an existing thread some time ago, and only just got around to posting it as a separate question.)

Answer 1

If we assume that any player seeing the flop is equally likely to have any possible hand and that we do not take our own specific hand into account, we can provide an answer.

Taking the simple case first, of a single opponent seeing the flop, the chances of that player having a pair is indeed 5.88% as you state:

[Chance of second card in hand being same rank as first]
= [chance of a random first card] + {[cards in desk matching first card] / [unknown cards remaining in deck]}
= 1 + [3 / 51]
= 0.05882 (5.88%)

The chance of them having at least a pair after the flop is a little more complex. Firstly, the chance of the player having an unpaired hand is 94.12% (simply 100% - 5.88% chance of having a pair). The chances of making exactly a pair, 2pair, trips, a full house of quads with an unpaired hand is:

[chance of flopping a pair] + [chance of flopping 2 pair on paired flop] + [chance of flopping 2 pair on paired flop] + [chance of flopping trips] + [chance of flopping full house] + [chance of flopping quads]
= [{(6 * 44 * 40) / (50 * 49 * 48)} * 3] + [{(6 * 3 * 44) / (50 * 49 * 48)} * 3] + [{(3 * 3 * 44) / (50 * 49 * 48)} * 6] + [{(6 * 2 * 44) / (50 * 49 * 48)} * 3] + [{(6 * 2 * 3) / (50 * 49 * 48)} * 3] + [{(6 * 2 * 1) / (50 * 49 * 48)} * 3]
= [0.2694] + [0.0202] + [0.0202] + [0.0135] + [0.00092] + [0.00010]
= 0.3243 (32.43%)

The chance of being dealt a hand which can flop a straight (any connector or 1/2/3 gapper*) is 24.74% and such a hand will flop a straight 0.64% of the time. The chance of being dealt a suited hand is 23.53% and such a hand will flop a flush 0.84% of the time.

Putting this all together to get your final answer:

[chance of  a player either having a pair or having flopped a pair or better with an unpaired hand]
= [chance of having a pair] + {[chance of having an unpaired hand] * [chance of making a pair or better with an unpaired hand]}
= 0.05882 + { 0.9412 * [0.3243 + {(0.2474/0.9412)*0.0064} + {(0.2353/0.9412)*0.0084}]}
= 0.05882 + { 0.9412 * [0.3243 + {0.2628 * 0.0064} + {0.25 * 0.0084}]}
= 0.05882 + { 0.9412 * [0.3243 + 0.001682 + 0.0021]}
= 0.05882 + { 0.9412 * 0.3281}
= 0.05882 + { 0.9412 * 0.3281}
= 0.05882 + 0.3088
= 0.3676 (36.76%)

Or conversely, a 63.24% chance that a player misses the flop entirely and does not hold a pocket pair.

From this we can answer your original question, for any given number of players seeing the flop, what is the chance that at least one of them either holds a pocket pair or has flopped one pair or better:

/--------------------\
| # Players |   %    |
|-----------+--------|
|     1     | 36.76% |
|     2     | 60.01% |
|     3     | 74.71% |
|     4     | 84.01% |
|     5     | 89.89% |
|     6     | 93.60% |
|     7     | 95.95% |
|     8     | 97.44% |
\--------------------/

These figures were arrived at by taking the inverse of the chance of none of the players either holding a pair or hitting the flop:

1 - (0.6324 ^ NumPlayers)

As has already been discussed, this doesn't take into account the fact that not all hands are equally likely to see a flop - to incorporate this accuracy we would need to establish the probability of someone seeing a flop with each class of hand and factor this into the calculation.

*i.e. A2, 32, 43, 54, 65, 76, 87, 98, T9, JT, QJ, KQ, AK, A3, 42, 53, 64, 75, 86, 97, T8, J9, QT, KJ, AQ, A4, 52, 63, 74, 85, 96, T7, J8, Q9, KT, AJ, A5, 62, 73, 84, 95, T6, J7, Q8, K9 or AT.

Answer 2

Let's just call it 1/3 chance that a player makes a pair on the flop. Change it into an and so you can multiply. The chance of not having a pair is 2/3. So if three saw the flop 1 - (2/3 * 2/3 * 2/3) = 1 - 8/27 = 19/27. So about 2/3 chance at least one player is paired. But that is a random hand - a hand that will see a flop is not random (e.g. pairs rarely fold pre flop).

If the combined number is 2/5 then
1 - (3/5 * 3/5 * 3/5) = 1 - (27 / 125) = (125 - 27) / 125 = 98 / 125 = 0.784