3 players flop the nuts

[Toopskees]

What are the odds against this occurrence, and how is the calculation done?

6-handed home game. Flop is 8-9-J rainbow. Lots of action. At showdown, 3 players each had Q-T.

[jgca]

Depends what you're asking. Given your title, I guess you want to consider only the 3-way nuts, which can only be straights (with one very rare exception, that can't occur on the flop anyway). Do you want just Q-T with all of its straights (all of its straights are nut straights), or nut straights from all hands, or just ones that 3 people might be playing multiway? (If you told me that 3 out of 6 people played 74o and flopped 653, the odds would seem almost zero on it being true.)

Anyway, here's a start:

Part I:

Total combinations of 6 hands: (52!)/(40!) / 2^6 =
1,544,626,032,777,828,000

Combinations that have 3 QT hands and 3 non:

20 (6 choose 3; that is, groups of 3 players it could be) *
16 * 9 * 4 *
843,013,710 (combinations of 3 hands from the remaining 46 cards) =
9,711,517,939,200.

About 1 in 159,000, just for 3 people out of 6 having QT, with a slight error because I didn't adjust for the tiny chance of 4 people having QT. Of course, when you consider the number of playable hands that might have produced the story, you get a ratio around 1 in 9,000. So 3 identical hands 6-way is gonna come up for you, more than once.

[bigpooch]

First, you would have to classify those flops that would
allow a three-way tie, namely those flops that yield a
possible straight without a possible flush. There are
the combinations

AKQ, KQJ, ..., 32A (12)
AKJ, KQT, ..., 43A (11)
AQJ, KJT, ..., 42A (11)
AQT, KJ9, ..., 53A (10)

so altogether, there are 44 combinations of ranks that allow
for this to happen. Each rank can have one of four suits,
so excluding the case of all of the ranks on board being of
the same suit, there are 44 x (4x4x4 - 1) = 2772
combinations out of the possible C(52,3) = 22100 flops that
yield a possible straight and no possible flush.
2272/22100 = 693/5525 is the ratio of flops for which a
three-way tie can occur.


Second, given that the flop comes down as above, what are
the chances that precisely three players out of six have
the required combination giving them each "the nuts"? If it
were three-handed, the probability would then be p =
16*9*4/(C(49,2)*C(47,2)*C(45,2)) = 4/8 739 885. There are
six players, so there are now about C(6,3)=20 as many ways
of having this occur. On the other hand, there is an
adjustment that must be made because of "quadruple counting"
because if players A,B,C and D all have the "nuts", we've
counted it four times (ABC, ABD, ACD, BCD). Thus, we have
to subtract three times the chances that there are four
players that flop "the nuts". The chances that four players
have the nuts is then
C(6,4)*16*9*4*1/(C(49,2)*C(47,2)*C(45,2)*C(43,2)) which is
just C(6,4)/C(43,2) = 15/903 = 5/301 times p above.

Thus, the probability that precisely three of six players
have the requisite cards is (20-3*5/301)*p or about
9.130630947 x 10^(-6). Including the requirement for the
flop, the probability is about 1.1452538 x 10^(-6) or
the odds are about 873167.9 to 1 against.