Introduction: How often is anyone dealt a four card hand containing at least one ace plus at least one deuce?
Let’s say you have a question about the probability of being dealt two cards of two particular ranks, for example at least one ace and at least one deuce.
For our purposes, there are three types of cards involved here, “aces,” “deuces,” and “other cards,” or simply “others,” Note that if we cannot see any cards yet, there are four aces, four deuces, and 44 other cards.
First let’s do a “balance,” which involves listing all possible four-card hands from the perspective of aces, deuces, and others. Included within this “balance” will be all the hands with one or more aces plus one or more deuces.
Following are all the possible hands. Note that the order in which we list the cards in an Omaha-8 hand dealt to us, (or anyone), doesn’t matter. In other words, what we list first, second or third doesn’t matter.
AAAA
2222
XXXX
AAA2
AAAX
222A
222X
XXXA
XXX2
AA22
AAXX
22XX
AA2X
A22X
A2XX
The list was made very systematically so as not to omit anything. First, all the possibilities that included four cards of only one type (aces, deuced, and others) were listed. Next, all the possibilities with two types were listed. Finally, all the possibilities with all three types were listed.
You should verify for yourself that the above listing includes all the possibilities. In other words, you should make sure you cannot think of any other possible combinations of ace, deuce, or other. Again, note that the order the cards are listed doesn’t matter when dealing with “combinations.”
Looking at it slightly differently:
First listed were four cards of each type. (4,0,0)
Second listed were three cards of one type plus one card of another. (3,1,0)
Third listed were two cards of one type plus two cards of another. (2,2,0)
Fourth listed were two cards of one type plus one of a second type plus one of the third type. (2,1,1)
Now we need to calculate how many ways we can make put the same four cards together to make a four card hand.
Filling in the chart
The first two listings in the chart, AAAA and 2222, are really easy. Since there are only four aces in a standard deck, there’s only one four card hand that can contain all four aces. Stated slightly differently, there is only one way to choose four aces for a hand, if the order in which they’re chosen doesn’t matter to us. And the same is true of four deuces.
AAAA1 way
22221 way
Four “others” is more complicated.
There are 44 ways to choose the first “other.” Then after having selected that first card, there are 43 ways to choose the second “other.” Then after having selected those two cards, there are 42 ways to choose the third “other.” Then after having selected those three cards, there are 41 ways to choose the fourth “other.”
The product of those four numbers, 44*43*42*41=3,258,024, is the number of “permutations” in four cards after the aces and deuces are removed from a 52 card deck. However, the actual number of "permutations" doesn’t matter to us because we're going to be interested in the number of "combinations" instead. And in order to find the number of "combinations," we’re immediately going to divide the number of "permutations" by the number of ways we could have chosen the four cards.
Once we get a particular four cards together in a hand, the order in which we selected them isn’t important. We could just as well have chosen any of the four cards first, and then we would have had three cards left. And we could have chosen any of the three next, and then we would have had two cards left, and then we could have chosen either of the two cards next, and there would be only one card left to occupy the fourth slot in the hand.
Thus, once we have selected the four cards to be in the hand, there are 4 ways to choose the first “other.” Then after having selected that card, there are 3 ways to choose the second “other.” Then after having selected those two cards, there are 2 ways to choose the third “other.” Then after having selected those three cards, there is 1 ways to choose the fourth “other.”
That total is 4*3*2*1=24.
Thus there are 24 ways we could have selected the 3,258,024 permutations.
3,258,024/24=135,751. That number, 135,751, is the total number of combinations and is important because it’s the result that goes into our chart. That total, 135,751, is the actual number of different hands containing four “other cards.”
XXXX135,751 ways … from 44*43*42*41/4/3/2/1
Continuing, in order to get the number of different possible four card hands consisting of three aces plus one deuce, there are four ways to remove one ace and four ways to choose a deuce, a total of 4*4=16 ways. And it’s the same number of ways to choose three deuces plus one ace.
AAA216 ways … from 4*4
222A16 ways … from 4*4
To get the number of different possible four card hands consisting of three aces plus one other, there are four ways to remove one ace and forty four ways to choose an “other”, a total of 4*44=176 ways. And it’s the same number of ways to choose three deuces plus one other.
AAAX176 ways … from 4*44
222X176 ways … from 4*44
To get the number of different possible four card hands consisting of three others plus one ace, there are 4 ways to select one ace and then there are 44 ways to choose the first, other, 43 ways to choose, the second other, 42 ways to choose the third other, and once we’ve chosen the three others, there are 3*2*1 ways to order them. And it’s the same number of ways to choose three others plus one deuce. (Dividing by “3*2*1” can be written as “/3/2/1.”
XXXA52,976 ways … from 4*44*43*42/3/2/1
XXX252,976 ways … from 4*44*43*42/3/2/1
To get the number of different possible four card hands consisting of two aces plus two deuces, there are 4 ways to choose the first ace and 3 ways to choose, the second ace. But once we’ve chosen the two aces, there are two ways we could have chosen either one of them first. 4*3/2/1=6. 6 is the number of ways to choose two aces from all four aces. Similarly, 6 is the number of ways to choose two deuces from all four deuces.
Thus
AA2236 ways … from 6*6
To get the number of different possible four card hands consisting of two aces plus two others, we compute the number of ways to choose two aces, then compute the number of ways to choose two others, and then multiply the two.
There are 6 ways to choose two aces. Next there are 44 ways to choose the first other and then 43 ways to choose the second other, and once we’ve chosen the two others, there are actually two ways we could have chosen either one of them first. Thus there are 44*43/2/1 ways to represent the two others.
And it works a similar way for two others plus two deuces.
Thus
AAXX5,676 ways … from 6*44*43/2/1
22XX5,676 ways … from 6*44*43/2/1
For AA2X, there are 6 ways to choose two aces, 4 ways to choose one deuce, and 44 ways to choose one other. A22X is similar.
Thus
AA2X1,056 ways … from 6*4*44
A22X1,056 ways … from 4*6*44
Finally for A2XX, there are 4 ways to choose an ace, 4 ways to choose a deuce 44 ways to choose the first other, 43 ways to choose the second other, and then 2 ways to select the first of the others, once the two choices have been made.
Thus
A2XX15,136 ways … from 4*4*44*43/2
When we put that all together in a chart, it looks like this:
Hand |
# ways |
Computation method |
AAAA |
1 |
|
2222 |
1 |
|
XXXX |
13,5751 |
44*43*42*41/4/3/2/1 |
AAA2 |
16 |
4*4 |
222A |
16 |
4*4 |
AAAX |
176 |
4*44 |
222X |
176 |
4*44 |
XXXA |
52,976 |
4*44*43*42/3/2/1 |
XXX2 |
52,976 |
4*44*43*42/3/2/1 |
AA22 |
36 |
6*6 |
AAXX |
5,676 |
6*44*43/2/1 |
22XX |
5,676 |
6*44*43/2/1 |
AA2X |
1,056 |
6*4*44 |
A22X |
1,056 |
4*6*44 |
A2XX |
15,136 |
4*4*44*43/2 |
Total |
270,725 |
|
Now we check the total and note that the total number of possible four card hands is 52*51*50*49/4/3/2/1=270,725.
Thus we have a check.
The check doesn’t “prove” anything, but when our total from the table jibes with the total we get by independently calculating the number of possible Omaha-8 hands, we at least don’t think the table is incorrect. (Indeed we think it’s probably correct).
Hands that contain at least one ace plus at least one deuce.
Since we only want to include hands with both at least one ace and at least one deuce, we exclude all the others to get:
Hand |
# ways |
Computation method |
AAA2 |
16 |
4*4 |
222A |
16 |
4*4 |
AA22 |
36 |
6*6 |
AA2X |
1,056 |
6*4*44 |
A22X |
1,056 |
4*6*44 |
A2XX |
15,136 |
4*4*44*43/2 |
Total |
17,316 |
|
We could have gotten to the above chart faster, and normally would, but then we wouldn’t have our all-important check.
Finally, let’s calculate the probability of being dealt a hand containing at least one ace plus at least one deuce.
17,316/270,725=0.06396 or about 0.064
That corresponds to about one time in sixteen (actually one time in 15.63).
If you have no ace or deuce, you can make a similar chart for an opponent. In that case (if you have no ace or deuce), the probability a specific one opponent will be dealt a hand with at least one ace plus at least one deuce becomes 17,316/194,580=0.089 or about one in eleven.
Hands that contain at least one card of one rank plus at least one card of a second rank.
We did the calculation thinking of an ace plus a deuce, but the same idea would hold true for any particular two different ranks. Thus the probability of being dealt a hand containing at least one king plus at least one queen is also about 0.064.
However if you have a hand that contains neither a king nor a queen, that probability an opponent does increases to 17,316/194,580=0.089 or about one in eleven.
