Four of a kind VS Royal Flush

[imcastleman]

In Omaha, what is the probability that Player one gets (T T x x)where one of the tens is not the Th and Player two gets (two of AKQJ hearts, x x) and the board is (2 of AKQJ hearts, T?, Th, x)? Can someone help explain how to calculate this? If not, just the correct result would be satisfactory.

By the way, this happened to me in a PokerStars tournament yesterday.

Thanks,
Doug

[bigpooch]

[ QUOTE ]

In Omaha, what is the probability that Player one gets (T T x x)where one of the tens is not the Th and Player two gets (two of AKQJ hearts, x x) and the board is (2 of AKQJ hearts, T?, Th, x)? Can someone help explain how to calculate this? If not, just the correct result would be satisfactory.


[/ QUOTE ]

You're quite specific! /images/graemlins/smile.gif

First of all, C(m,n)= the number of combinations of m
objects chosen n at a time; for example, C(4,2)=6 is the
number of possible combinations of a pair of aces (suits
being significant). It is computed as

C(m,n) = m x ... x (m-n+1)/ [1 x ... x n] or
m!/[(m-n)! x n!] where k! = 1 x ... x k.

1) Player One will get dealt an Omaha hand with exactly two
tens excluding the ten of hearts with a probability that can
be computed as follows:

There are a total C(52,4) = 270725 Omaha hands. A pair of
tens without the ten of hearts gives C(3,2)=3 combinations.
The other two cards that make up the Omaha hand must not be
a ten nor a heart higher than a ten: altogether, that's a
total of 4+4=8 cards that can't be chosen so that there are
C(52-8,2) = C(44,2)=946 combinations of the other cards.
Thus, the total possible combinations is the product
3 x 946 = 2838. This is the number of combinations out of a
total of 270725 or the probability is 2838/270725.

2) For Player Two to get two heart honors not including the
ten and two other cards, neither of which is a ten or a
heart honor, there are C(4,2) combinations of the heart
honors and now since there are already two non-tens taken up
by Player One, there are now only C(42,2) combinations of
the remaining cards. The total number of possible hands is
now C(48,4) as there are four cards taken up by Player One's
hand. Thus, the number of hands is 6 x 861 = 5166 and this
is out of 194580 possible hands. This probability given the
event of Player One getting a hand with your specifications
is then 5166/194580.

3) This is the easy part: the number of possible
combinations of board cards is just C(44,5) = 1086008 and
for the cards on board to include the remaining heart honors
and the other ten takes up four cards and there is only one
combination of that! What's left of the deck? There are
four other cards (other than a ten or heart honor) taken up
in the other hands so there are only 52-4-4-4 = 40 cards to
chose from, therefore there are only 40 combinations of
board cards. Thus, this probability is 40/1086008. Of
course, it's another conditional probability based on the
players getting their specific kind of hands.

The chances for all of this to happen is just the product of
the probabilities but bear in mind that Player One and
Player Two could have their hands interchanged and the same
bad beat would occur!

For the more general case then, since there are twenty
possible key cards (the key card in this case is the ten of
hearts that gives a royal flush and quads) that could yield
this kind of bad beat, you would multiply this by 20 for
any quads versus a royal flush. Together with the factor
of two from the previous paragraph, this would be a factor
of 40.

The factor of twenty above is only an approximation, as
there is the unusual but possible combination that Player
One could make two sets of quads and Player Two makes a
royal flush! So, you would have to make a small adjustment
and the calculation is not difficult and I'll leave that as
an exercise.

Okay, now how would that translate to a ring game with ten
players? Well, there are C(10,2) = 45 choices of the two
players and then a good approximation for this to occur in
a ring game would be to multiply this out by 45. The only
reason that it's merely an approximation, although an
extremely good one, is that there is a slim chance that
there will be a royal flush and two players making quads.
Of course, if that were to happen, someone might think there
was a mechanic at the table or at least a cold deck was
dealt!
/images/graemlins/smile.gif

[imcastleman]

Thank you for your reply. I have done some calculation surrounding the probability in a heads up situation(which was mine) where one player gets a 4-of a kind(Tens or bigger) and the other player gets a royal flush. Could you tell me if you get the same result?

Board must look like:
Board A:
Quad card, Quad card, RF card, RF card, RF card (IMPOSSIBLE WITH Quads being TENS or Bigger)

OR
Board B:
Quad card, Quad/RF shared card, RF card, RF card, and off card.

Board A:
Total number of boards that meet the given criteria:
(5 possible ranks for quads) * C(4,2)(of the 4 quads choose 2 of them) * (4 possible suits for royal) * C(5,3)(of the 5 royal cards choose 3 of them)

= 5*6*4*10 = 30*40 = 1,200 boards.(I JUST REALIZED THIS IS IMPOSSIBLE)

Board B:
Total number of boards that meet the given criteria:
(5 possible ranks for quads) * C(4,2)(of the 4 quads choose 2 of them) * (2 possible suits for royal) * C(4,2)(of the 4 remaining royal cards choose 2 of them) * 44(the off card)

= 5*6*2*6*44 = 30*12*44 = 15,840 boards.

(just verifying that my logic in determining the possible boards is right)
(4 possible suits for royal) * C(5,3)(of the 5 royal cards choose 3 of them) * (3 possible ranks for quads) * C(3,1)(of the 3 remaining quad cards choose 1 of them) * 44(the off card)

= 4*10*3*3*44 = 40*3*132=120*132 = 15,840 boards.


Possibilities:

Player one has quads and player 2 has RF

Player One(has quads):
Total number of combinations where players’ 4-card hand contains the 2 quad cards:
(1 possible rank for quads)*43*42

= 1,806 ways to have this hand.

Player Two(has RF):
Total number of combinations where players’ 4-card hand contains the 2 remaining RF cards:
(1 possible way to get 2 remaining RF cards)*41*40

= 1,640 ways to have this hand.

Player one has RF and player 2 has quads

Player One(has RF):
Total number of combinations where players’ 4-card hand contains the 2 remaining RF cards:
(1 possible way to get 2 remaining RF cards)*43*42

= 1,806 ways to have this hand.

Player Two(has quads):
Total number of combinations where players’ 4-card hand contains the 2 remaining quad cards:
(1 possible way to get 2 remaining quad cards)*41*40

= 1,640 ways to have this hand.


P(Player one has quads and player 2 has RF)

= [15,840/C(52,5)] * [1,806/C(47,4)] * [1,640/C(43,4)]
= 469,155,456 / 572,083,715,843,640 ~ 1/1,219,390

P(Player one has RF and player 2 has RF)

is the same as previous calculation.

Therefore, P(one of the 2 situations) is:

~ 1/609,695
Am I correct?