## Flop Analysis Part 3 – Pair Bias

Dec 18th, 2009 | Posted by spadebidder

Here we examine the card-removal effect that results from players tending to see flops when they hold paired hole cards.  This part is a little tedious but it’s necessary to predict what we should see when analyzing an enormous sample of community cards.  If you aren’t interested in the math then just skip to the bottom for the result, and then jump to the actual data in Part 7.  That’s why you’re here anyway, right?

—————

From a full deck there are 78 possible pairs that can be dealt. Each rank of four cards can make 6 different pair combinations, which are            , and with 13 ranks we have 6*13 for 78 pairs.  And there are 1248 non-pairs that can be formed, for a total of 1326 hole card combinations, from C(52,2)=1326.  When we remove one pair from the deck, we have left only one other possible pair in that rank, eliminating 5 possible pairs and leaving 73.  At the same time we have eliminated 96 possible non-pairs from the deck, as each card could have been matched with any of the 48 cards in other ranks.  So the chance for paired hole cards  goes from 78/1326 or 5.882% initially, to 73/1225 or 5.959% after removing one pair.

This means that when you hold a pair, the chance another player was dealt a pair goes up slightly.  It also means that the chance of another pair being dealt to the board goes up, so let’s calculate that difference.

Dealing from a full deck (or the deck stub when no hole cards are known) the chance for a paired flop is
(52*3*48)*3 / (52*51*50) = 16.94117647%

This is the chance for a flop that is paired but not triplets, since we didn’t count that combination (which would be 52*3*50 in the numerator).  The second *3 in the numerator is for the three ways the pair can be formed from 3 cards,  using the first and second card, first and third, or second and third.

Now let’s look at scenarios where players see the flop holding paired and unpaired hole cards and how that might affect the board distribution.  Note that all the calculations shown were confirmed by full enumeration of the boards.

Scenario 1 - two players see a flop with one holding a pair and one holding a non-pair in other ranks.  We calculate the probability of them seeing a paired flop (but not triplets):

10 ranks (40*3*44)*3
2 ranks (6*2*45)*3
1 rank, i.e. he flopped quads = (2*1*46)*3
(15840 + 1620 + 276) / (48*47*46) = 17.090656799% chance for paired flop.

As expected, when someone holds a pair the chance for a paired flop increases. The increase is (17.090656799 – 16.94117647) / 16.94117647 or nearly 1% more paired flops than a random board.

Scenario 2 - Two players hold non-pairs but they have one matching rank between them.  This is equivalent to #1 and also has a 17.090656799% chance for paired flop

Scenario 3 – two players both holding pairs (in different ranks) see the flop.  We calculate the probability of them seeing a paired flop (but not triplets):

11 ranks (44*3*44)*3
2 ranks, i.e. somebody hit quads = (4*1*46)*3
= (17424 + 552) / (48*47*46) = 17.321924144% chance for paired flop.

So as expected, the chance rises a bit more. Now we have 2.2% more paired flops than a random board.

Scenario 4 – two players both holding non-pairs with no ranks in common see the flop.  We calculate the probability of them seeing a paired flop (but not triplets):

9 ranks (36*3*44)*3
4 ranks (12*2*45)*3
= (14256 + 3240) / (48*47*46) = 16.859389454% chance for paired flop.

The chance for a paired flop goes down when no one holds a pocket pair.

Scenario 5 – two non-paired players hold the same ranks for both cards.  This is surprisingly common at a 9-player table, and it’s similar to the birthday problem.   We’ll calculate the frequency when we do the scenario weighting.    This scenario is equivalent to #3 and also has a 17.321924144% chance for paired flop.

All other flops - for all times when more than 2 players see the flop, the effects will tend to cancel out since 3 or more players holding pairs is rare, and paired and non-paired hole cards have opposite but similar effects on the chance of a paired flop.  We’re also disregarding the rare case when two players see the flop holding the same pair, since the weighting would be so small.   So we’ll count all other flops together as having the random expectation of being paired, which was 16.94117647%.

Now we need to attempt to weight our scenarios.

This means making some assumptions about player behavior, but I’ll try to use reasonable averages that might fit most games, and try to stay out of wild guess range.

Scenario 1 Weight - Some player at 9-player table gets a pair in the hole approximately
1 – ( (1 – (1/17))^9) = 42.052%.
We’re assuming independence in this approximation but the actual lengthy calculation is very close, within a fraction of a percent.  We’ll now assume players see the flop 60% of the time when they are paired (certainly debatable within a reasonable range), so 25.2312% of the hands some player sees the flop holding a pair.   Now we need the chance that a single opponent is not paired.  If we assume the typical cash player sees the flop 20% of the time, and he has a pair 5.882%*60% = 3.5292%, he is unpaired (.20 – .03592) / .20  or 82.09% of the time.   So we have (.8209 * .252312) = 20.7123% of flops.   Since this is an estimate based on an assumption, we’ll round and say that when 2 players see the flop, 21% of the time they will have a pair and a non-pair with no matching ranks.  That sounds reasonable to me.
Scenario 1 = 21%.

Scenario 2 Weight - For the equivalent case of two non-paired players with a matching rank between them, we know that when a player sees the flop with unpaired cards and faces one opponent, the opponent will hold at least one of the same rank [1 - (44/50 * 43/49)] = 22.7755% of the time.  So we need to multiply that factor by how often two players with unpaired cards will see a flop.   Using the same assumption as in #1,  a 20% flop player sees a flop with no pair in 16.4708% of the hands he is dealt.   And he can expect to have a potential opponent with non-paired holdings approximately (1-(1-.164708)^8) or 76.3023% of the time, thus (.763023 * .164708) = 12.5676%.   Now multiply by the chance the opponent has a matching rank, and we have (.125676% * 22.7755%) = 2.862% of flops.   Again we’ll round and say that when 2 players see the flop, 3% of the time they will have a pair and a non-pair with a matching rank.  Can’t be too far off.
Scenario 2 = 3%.

Scenario 3 Weight - When one player is paired, the chance for a second player to be paired is then not 1/17, but is 73/1225 or .0595918, so we approximate 1 – ( (1 – .0595918)^8) = 38.831%  again assuming independence around the table, accepting a tiny error.  So two players at a 9-player table get a different pair in the hole 42.052 * 38.831% = 16.3292%  and we’ll say they both see the flop 30% of the time this happens (debatable within a reasonable range), so 4.8988% of flops.
Scenario 3 = 5%.

Scenario 4 Weight – using the calculations in #2 for two opponents with neither holding a pair, we had 12.5676%, and then deducting the 2.862% with matching ranks, we’re left with 9.7056%.
Scenario 4 = 10%.

Scenario 5 Weight -  This calculation turns out to be very difficult, but I think we have a pretty good estimate from this thread on 2+2:
http://forumserver.twoplustwo.com/25/probability/31-7-chance-someone-table-has-same-ranks-630649/ (thanks to BruceZ).

So I’m going to go with 23%, which should be within a percent or two.   Now, how often will these two specific players see the  flop together?  If we just go with a 20% flop rate for both, then we get 0.2^2 or 4% of the time.   So now we have 23% * 4% to the flop, or a .92% weight for this scenario.  Note also that we’ve not differentiated the rare case of holding the same pairs, which has a greater effect on the chance of seeing a paired flop.  It’s infrequent enough to just leave it in this group.
Scenario 5 = 1%.

All others -  The rest of the flops would be  1 – (.21 + .03 + .05 + .10 + .01) = 60% of flops.

Add up our weightings and we determine that paired flops with these assumptions will occur:

(.21*.17090656799) + (.03*.17090656799) + (.05*.17321924133) + (.10*.16859389454) + (.01 * .17321924133) +  (.60*.1694117647)
= .0358904 + .0051272 + .0086610 + .0168594 + .0017322 + .1016471

= 16.99173% , vs. 16.94117647% if there were no removal effect. So that’s a 0.05% increase in paired flops overall, using our assumptions. Therefore:

In full ring NLHE we expect to see around 50 / 100,000 more paired flops than a random distribution.

That doesn’t seem like much, but in a large sample it is very significant. In 100 million flops this is an offset of over 13 standard deviations from the mean of a random distribution.  In a smaller sample of 2 million flops, the offset would be about 2 standard deviations.  And in a single player sample of 200,000 flops the offset would only be 0.6 standard deviations, and thus would never be noticed.  This illustrates how a consistent offset becomes progressively more significant as the sample size increases, and how a relatively minor effect like this takes a huge sample size to detect.

In this exercise we’ve  also derived some statistics that are interesting on their own:

• Someone at 9-player table has a pocket pair approximately 42% of the time.
• Two players at a 9-player table have a different pocket pair approximately 16% of the time.
• When you have a pocket pair, the chance any opponent has a pair goes up slightly.
• Someone at a 9-player table is dealt the exact same two ranks as another player ~23% of the time (such as AK AK or T2 T2).

—————