Happened the other night, three sets on the flop.

Question: What are odds of three pair being dealt in a 8 person game?

Question 2: What are odds of everypair hitting a set on the flop?

Thanks.

I started calculating the answer to question 1, but, as BruceZ pointed out, I have some trouble with double counting, so I gave it up. I hope for him or someone else to solve that question.

However, question 1 can be further detailed into including or excluding the instances where two players are dealt the same pocket pair, in which cases the probability of all three players flopping sets equals 0, since all cards of that rank are out of the deck and therefore cannot flop. Therefore, I think we better exclude those instances.

Also, it can be further detailed into including or excluding the instances where more than three players are dealt pocket pairs, in which instances the probability that three players flop sets is bigger. For the sake of simplicity, I think we should exclude those cases too.

So, I'd like to specify question 1 into "what is the probability that exactly three players out of eight are dealt different-ranking pocket pairs", and I hope for someone else to show me how to calculate that probability without falling into the traps of double counting.

As for question 2, given that exactly three players have been dealt different-ranking pocket pairs (and assuming we don't know anything about the remaining five players' hole cards, which is not true, because we know they're not holding pocket pairs (how do we account for this information?)), there are

2^3 = 8

combinations of flop cards that give all three players sets, and there are a total of

C(46,3) = 15180

flop combinations. So the probability that all three will flop sets is

8/15180 = 2/3795 ~ 5.27*10^(-4)

Or about 0.000527%.

Edit: No wait! Thinking... brb
Edit2: Done thinking. I think that should be right.

It's quite laborious to calculate this probability exactly, i.e.,
"the probability that exactly three players out of eight are
dealt different-ranking pocket pairs", but as BruceZ would
point out, three or four decimal places of precision should
suffice.

One idea is to compute the probability that three players each
have different pocket pairs and estimate the probability that
the other five players don't have a pocket pair and this is the
idea that I'll be using. If the game were only three-handed,
the probability of these players each being dealt a unique pair
is then

[(13C(4,2))/C(52,2)][(12C(4,2))/C(50,2)][(11C(4,2))/C(48,2)]
= (1/17)(72/1 225)(11/188) = 198/978 775 ~
0.0002022936834.

This is the same probability that the first three players of an
eight-handed game will be dealt a pocket pair. Now, what is
the probability that the fourth player will not be dealt a pair?

[C(10,2)x4x4+10x4x6+C(3,2)x2x2]/C(46,2) = 972/ 1035 =
108/115 ~ 0.9391304348.

Now one could simply use this number and raise it to the fifth
power for approximating the probability that the other five
players don't have a pocket pair. Of course, it's only an
estimate because the probabilities won't be independent.
Another problem is that this estimate may be a little low since
if the fourth player doesn't have a pocket pair, the fifth
player is more likely (compared to the fourth player) not to
have a pocket pair.

It turns out that for the fifth player not to be dealt a pocket
pair given exactly three of the first four players have been
dealt a pair will be on average, slightly higher. There are
three cases for the nonpair of the fourth player and the
probability that the fifth player doesn't have a pair will vary
depending on the type of nonpair the fourth player has:

a) C(10,2)x4x4 = 720 combinations of a nonpair hand with
ranks different from the pocket pairs. For the fifth player not
to have a pocket pair, of the C(44,2)= 946 combinations he
could have, he has no pair on

C(8,2)x4x4+8x4x12+3x3+6x6+C(3,2)x2x2 = 889

of them or a probability of 889/946.

b) 10x4x6 = 240 combinations of a nonpair hand with one
rank different from the pocket pairs and one rank which is the
same as a rank of a pocket pair. There are now

C(9,2)x4x4+9x4x8+3x5+2x2+1x4 = 887

combinations for the fifth hand to be a nonpair.

c) C(3,2)x2x2 = 12 combinations of nonpairs with both ranks
of the hand being among the ranks of the pairs. There are now

C(10,2)x4x4+10x4x4+2x2+1x1 = 885

combinations for the fifth hand to be a nonpair.


On average, the probability that the fifth hand will not be a pair
is then

[(720/972)(889)+(240/972)(887)+(12/972)(885)]/946
~ 0.9391720826.


I'm confident that the probability of a sixth hand being not a
pair will be even higher (given that exactly three out of the
first five hands are unique pocket pairs). To "make an already
long story short", the probability of the first three players
being dealt unique pocket pairs and the other five players
being dealt "nonpairs" could be approximated by:

[198/978 775](0.9392)^5

and then this is multiplied by C(8,3) (since the three "pairs"
can be distributed in so many ways) to get a probability of
about

0.008279. [ Just to get close to four decimal places! ]


Anyone care to check this?

For the sake of simplicity, I would like to calculate the original question 1 only:

I do not want to exclude the possibilities of instances where more than three players are dealt pocket pairs. I could do it with the inclusion/exclusion principle, but the formula becomes to complex and it was not subject of the original question that exactly three pairs being dealt.

My solution is:
8*13*6*C(7,2)*((12*6)+1)*(12*6/73+72/73*((11*6)+2))/(C(52,2)*C(50,2)*C(48,2))
=~ 0.0355

or for three different pairs:
8*13*6*C(7,2)*(12*6)*(72/73*11*6)/(C(52,2)*C(50,2)*C(48,2))
=~ 0.0335

This is my solution for Question 2:

~0.0335 * 2^3 / C(46,3) =~ 0.00001766

Question 1: ("three different pair")
8*78*C(7,2)*72*66/(C(52,2)*C(50,2)*C(48,2)) =~ 0.0339

Question 2:
~0.0339 * 2^3 / (C(46,3) =~ 0.00001791



Question 1: exactly three different pairs (with exclusion/inclusion principle):
8*78/1326 * x
x =
C(7,2)*72*66/(C(50,2)*C(48,2))
-C(7,3)*72*66*60/(C(50,2)*C(48,2)*C(46,2))
+C(7,4)*72*66*60*54/(C(50,2)*C(48,2)*C(46,2)*C(44,2))
etc.

Your number for "three different pairs" (0.0335) is much too
high by a factor of about 3. An upper bound for this
probability can be found by the inclusion-exclusion method
with the events A(i,j,k) where i,j,k are three unique indices
in the set {1,2,...,8} where A(i,j,k) denotes the event that
the ith, jth and kth players have a pocket pair:

C(8,3)[13C(4,2)/C(52,2)][12C(4,2)/C(50,2)][11C(4,2)/C(48,2)]
~0.01132844627.

It turns out that the above is a "poor" upper bound, so clearly
at least an "extra term" of a method using inclusion-exclusion
will be needed to get at least three significant figures.


At least everyone is correct about the probability of three
players making a set given these three players have unique
pocket pairs! The estimate I have for three players being
dealt unique pocket pairs and then the flop coming down to
make all three a set is then about

(8/15 180)(0.008279) ~ 0.00004363

or roughly 229 200 to 1 against.

o.k. I think your are right. Then this should be the correct solution for question 1:

C(8;3)*78*72*66/(C(52,2)*C(50,2)*C(48,2)) = 0.0113284462721259000
-C(8;4)*78*72*66*60/(C(52,2)*C(50,2)*C(48,2)*C(46,2))
+C(8;5)*78*72*66*60*54/(C(52,2)*C(50,2)*C(48,2)*C(46,2)*C(44,2))
~=0.010545032

But now I will think again !!!!

1) A quick way to approximate the solution is to assume independence. That is, the probability that 3 players are dealt pocket pairs at an 8-handed table is approximately equal to the probability that you're dealt 3 pocket pairs in "8 independent deals".

C(8,3)*(78/1326)^3*(1248/1326)^5 ~ 0.00841

2) 8c3*78/1326*72/1225*66/1128*2^3/15,180 ~ 0.00000597

At most 3 players can flop sets on the flop and each event where three chosen players flop sets is mutually exclusive to one another. There is no need to use the principle of inclusion/exclusion to work out the other terms since it is not possible for 4 or more players to flop sets.

It is easy calculating the probability of a flop coming down and
three of eight players flopping a set. The exact calculation is:

For an unpaired flop, the probability is [C(13,3)x4^3]/C(52,3).

For every rank on board, somebody has that rank as a pocket
pair and the probability of this happening is
8[3/C(49,2)] x 7[3/C(47,2)] x 6[3/C(45,2)].

Multiplying, the probability is about 0.000005970195664 or the
odds are about 167 497.6979 to 1 against. Your approximation
is very good (accurate to four decimal places).


Question (1) of the OP is vague as "three" could mean either
exactly three or at least three. Of course, the calculation for
the latter is (relatively) easy.

Jay_shark:

Actually, what I stated as "your approximation" is the same
calculation in a different form (now that I looked at it more
carefully!) for Question 2 (where "three" is at least three in
the OP), i.e., it is the exact solution for three or more players
holding a pocket pair.

You are missing a 0 in your decimal expansion.

8/15180*0.008279 ~ 0.000004363

Now it is closer to my answer.

Right. The probability of about 0.000004363 is for exactly
three players out of the eight with a pocket pair and each of
these three players flopping a set. The difference between
this and the other probability seemed smaller than I expected.

So, basically, about 175000-1, or thereabouts?

Yikes!

I witnessed this event at a live table...2s, 4s, and 7s for anyone that cares

I was involved ( on the lossing end) of a set over set over set over 2 pair. 10 person table unopened community pot. Flop comes A 8 5.

If you don't need to find the exact answer you need only to multiply 1/17 cubed times 16/17 to the fifth power times 56 times 6/46 times 4/45 times 2/44.

http://www.youtube.com/watch?v=gb4lS...EAYmTvmPC3npew happened to me yesterday haha 2 hours 30 mins in. im the one with the aces obv lol