Math question.

[Regs]

How man 21's are possible in a 52 card deck?

Using rules of black jack. One ace and a 10-value card is 21.

So there are 4 aces, which leads to 4 2-card possibilities. The rest...x,y,z...been a long time.

 

[Colt45]

42

 

[MagnusTheBrewer]

Oh! I thought this was a math problem. Your talkin' probability which has the same relation to math that accounting does ie. none to speak of.

 

[Regs]

Ok I'm sticking with 18.

 

[OdiN]

It gets more complex as aces can be both 11 or 1

 

[Tiamat]

Originally posted by: OdiN
It gets more complex as aces can be both 11 or 1

qft

 

[yankeesfan]

Shouldn't be that hard to figure it out.

 

[mugs]

Originally posted by: Regs
So there are 4 aces, which leads to 4 2-card possibilities. The rest...x,y,z...been a long time.

I take it you're not allowing reuse of any cards? Then figure out the most you can get with one suit and multiply by 4. (edit: Oh wait, that's not right... it's possible that you'd have some cards left over after doing that and they could be combined leftover cards from other suits to make 21)

Although I suspect that's not really what the question was asking...

 

[TuxDave]

Are you asking how many 21's are possible with 2 cards from a 52 card deck?

or

How many 21's are possible with any combination of cards from a 52 card deck?



To answer the first one:
There's 4 aces and 16 cards with a value of 10. So there's 4x16 combinations if order doesn't matter and 4x16x2 if order does matter.

 

[Regs]

Originally posted by: TuxDave
Are you asking how many 21's are possible with 2 cards from a 52 card deck?

or

How many 21's are possible with any combination of cards from a 52 card deck?



To answer the first one:
There's 4 aces and 16 cards with a value of 10. So there's 4x16 combinations if order doesn't matter and 4x16x2 if order does matter.

Thanks Tux.

I'm looking for #2, the more difficult one of course.

 

[jman19]

Originally posted by: MagnusTheBrewer
Oh! I thought this was a math problem. Your talkin' probability which has the same relation to math that accounting does ie. none to speak of.

Not true.

 

[DrPizza]

2 cards: 4C1 * 16C1
3 cards 10 + 10 + 1: 16C2 * 4C1
3 cards: A + A + 9: 4C2 * 4C1

3 cards: 2 + 10 + 9: 4C1 * 16C1 * 4C1
3 cards: 3 + 10 + 8: 4C1 * 16C1 * 4C1
3 cards: 4 + 10 + 7: 4C1 * 16C1 * 4C1
3 cards: 5 + 10 + 6: 4C1 * 16C1 * 4C1

3 cards: 3 + 9 + 9: 4C1 * 4C2
3 cards: 4 + 9 + 8: 4C1 * 4C1 * 4C1
3 cards: 5 + 9 + 7: 4C1 * 4C1 * 4C1
3 cards: 6 + 9 + 6: 4C2 * 4C1

3 cards: 5 + 8 + 8: 4C2 * 4C1
3 cards: 6 + 8 + 7: 4C1 * 4C1 * 4C1

Then, there's a shitload of ways to get 21 with 4 cards, and another shitload of ways to get 21 with 5 cards.
(starting with the first ace counting as 11)
AAA8
AA27
AA36
AA45
A226
A235
A244
A334
(10)AA9
(10)A28
(10)A37
(10)A46
(10)A55
(10)227
(10)236
(10)245
(10)335
(10)344
9AA(10) <---- do not count this one. If you're looking for my pattern, the last digit has to be less than or equal to the first digit.
9A29
9A38
9A47
9A56
9228
9237
9246
9255
9336
9345
9444
8238
8247
8256
8337
8346
8355
7347
7356
7446
7455
etc.
Have fun.

 

[Argo]

with 2 cards - 64

 

[chuckywang]

Originally posted by: Argo
with 2 cards - 64

Yeah, 4 aces and 16 "10" cards.

 

[vrbaba]

Originally posted by: chuckywang

Originally posted by: Argo
with 2 cards - 64

Yeah, 4 aces and 16 "10" cards.

lol, i saw the topic as "Math Question" and told myself, chuckywang must have a reply in it - before I saw the thread.

 

[QED]

I don't know why, but this problem intrigued me.

This problem is very similiar to counting the number of partitions for the number 21, which is quite simple.

But there are several factors which makes this a much more complicated problem, and make it almost impossible to
exhaustively list and count every possibility (at least by hand, that is):

1) An ace can count for one point or eleven points.
2) There are 4 suits of cards, so there are 4 2's, 4 3's, 4 aces, etc.
3) Within each suit, 4 cards (10, Jack, Queen, King) have identical value (10 points).

One way to try to count all the possibilites is to first count the number of possibilities which use only two cards,
then count the number which use exactly 3 cards, and so on. The problem with this method, as Dr. Pizza has shown,
is that the number of possibilities grows rather unwiedly as you increase the number of cards to use.

Instead, we should try to expoit some of the symetries of the problem, and break the problem down into two smaller, but related problems:

1) For each number 0 through 21, find out how many possible
card combinations from a single suit of cards add up to that number.

and

2) Find all possible ways to split 21 points amongst the four suits.


How does this help? We are basically categorizing each possible card combination that sums to 21 by the sum of the cards within each suit. By counting the number of possible card combinations in each of these categories, and then summing this over all of the categories we will have our final answer. Problem #1 will tell us how many combinations are in a specific category, while Problem #2 will give us a list of each category.

We start with Problem #1: for each number 0 through 21, how many possible card combinations from a single suit of cards add up that number? For each number n from 0 through 21, let's call this number P(n).

For simplicity in counting, let's label each card combination (hand) with a number, where each digit represents a card's value, by ordering the cards from highest to lowest in value. I use a '1' to designate an ace being used as a one, an 'A' to designate an ace with a point value of 11, and 'T' to designate any one of the four cards (ten,jack,queen, king) with a point-value of 10.

Hence, the card combination labeled "A53" represents an ace (with value 11), a 5, and a 3, all of one suit. This hand totals 19.

Now the question is, how many combinations of a single suit add up to 0? Obviously, only a hand with no cards-- so just one possibility. So P(0) = 1.

How many combinations add up to 1? Only the hand "1" (i.e. the ace being used as a one), so the answer is also 1. So P(1) = 1.

How many add up to 2? Only the hand "2" does, so the answer is again 1. So P(2) = 1.

How many add up to 3? There are two: "3" and "21". So P(3) = 2.

How many add up to 4? There are two: "4", and "31". Note we cannot use "22" as there is only one 2 card in the suit. So P(4)= 4.

For 5, there are 3 hands ("5", "41", and "32"). For 6, there are 4 hands ("6", "51", "42", and "321").

Using the greedy algorithm, it is a relatively trivial process to list and count all hands of a single suit that sums to a particular number from 0 through 21. You can even re-use some of your early work later on, as listing all of the one-suit hands that total to 18 with a largest card value of 7 is equivalent to listing all hands that total to 11 with a largest card value less than 7. Just remember to not count any hand that has an ace being used as both a one and an eleven, any hand labeled with a single "T" actually represents 4 hands, and any hand label with two "T"'s actually represents 4 * 3 / 2 = 6 hands (since each T can actually be a ten, jack, queen, or king).

We find that P(7) = 5, P(8) = 6, P(9) = 8, P(10) = 13, P(11) = 15, P(12) = 16, P(13) = 22, P(14) = 24, P(15) = 30, P(16) = 36, P(17) = 41, P(18) = 48, P(19) = 56, P(20) = 69, and P(21) = 77.



Now for Problem #2-- list each distinct way to split 21 points among the 4 suits. This is also relatively easy to do using the greedy algorithm.

For simplicity, we group each possible way by ordering the point totals from each suit from highest to lowest, which is then used to label the group. Hence the label is irrespective of the actual suits used.

For instance, one possible way is for all of the hearts to total 12, all of the diamonds to total 6, all of the clubs to total 2, and all of the spades to total 1. We label this category "12,6,2,1". Note that, another possible way is for all of the diamonds to total 12, all of the spades to total 6, all of the hearts to total 2, and all of the clubs to total 1. This way would also be labeled "12,6,2,1"-- in fact, any way in which one hand totals 12, another totals 6, another 2, and the remaning hand totals 1 would be so labeled. Hence the label "12,6,2,1" actually represents 4 * 3 * 2 * 1 = 24 distinct ways or categories.

Similiarily, the label "15,5,5,1" actually represents 4 * 3 * 2 * 1 / 2 = 12 distinct ways or categories.
The label "12,3,3,3" actually represents 4 * 3 * 2 * 1 / ( 3 * 2 * 1) = 4 distinct ways or categories.

It turns out there are 118 groups. The obvious group

"21,0,0,0" represents all categories in which 21 points come from a single suit, and 0 points come from each of the other 3. There are 4 such categories. From Problem #1 above, we know for a single suit there are P(21), or
77, ways for cards of that suit to total 21, and P(0) = 1 way for cards to total 0. Hence, the group "21,0,0,0" represents 4 * 77 * 1 * 1 * 1 = 308 distinct card hands that total 21.

The next group is

"20,1,0,0" and represents all categories in which 20 points come from a single suit, 1 point comes from a second suit, and 0 points come from the other 2. There are 12 such categories. From Problem #1, we know for a single suit there are P(20) = 69 ways for cards of that suit to total 20, P(1) = 1 way for cards to total 1, and P(0) = 1 way for cards to total 0. Hence, the group "20,1,0,0" represents 12 * 69 * 1 * 1 * 1 = 828 distinct card hands that total 21.

A list of all groups, the number of categories it represents, and the total number of distinct card hands that total 21 in that group follow in the code section below. Summing that all up, we have our answer.






Cliffs:

There are 187,332 distinct card combinations that total 21... unless I made a mistake somewhere, which is quite likely.





EDIT: The code box is stripping the carriage returns, so I'll list the 118 groupings of 21-point distributions amongst the 4 suits below. Next to each grouping is how many different suit combinations are represented in the group, and then how many total card hands fall into that group.

EDIT 2: Added two more groupings I initially missed. The total is the latest and greatest.

{21,0,0,0} 4 308
{20,1,0,0} 12 828
{19,2,0,0} 12 672
{19,1,1,0} 12 672
{18,3,0,0} 12 1,152
{18,2,1,0} 24 1,152
{18,1,1,1} 4 192
{17,4,0,0} 12 984
{17,3,1,0} 24 1,968
{17,2,2,0} 12 492
{17,2,1,1} 12 492
{16,5,0,0} 12 1,296
{16,4,1,0} 24 1,728
{16,3,2,0} 24 1,728
{16,3,1,1} 12 864
{16,2,2,1} 12 432
{15,6,0,0} 12 1,440
{15,5,1,0} 24 2,160
{15,4,2,0} 24 1,440
{15,4,1,1} 12 720
{15,3,3,0} 12 1,440
{15,3,2,1} 24 1,440
{15,2,2,2} 4 120
{14,7,0,0} 12 1,440
{14,6,1,0} 24 2,304
{14,5,2,0} 24 1,728
{14,5,1,1} 12 864
{14,4,3,0} 24 2,304
{14,4,2,1} 24 1,152
{14,3,3,1} 12 1,152
{14,3,2,2} 12 576
{13,8,0,0} 12 1,584
{13,7,1,0} 24 2,640
{13,6,2,0} 24 2,112
{13,6,1,1} 12 1,056
{13,5,3,0} 24 3,168
{13,5,2,1} 24 1,584
{13,4,4,0} 12 1,056
{13,4,3,1} 24 2,112
{13,4,2,2} 12 528
{13,3,3,2} 12 1,056
{12,9,0,0} 12 1,536
{12,8,1,0} 24 2,304
{12,7,2,0} 24 1,920
{12,7,1,1} 12 960
{12,6,3,0} 24 3,072
{12,6,2,1} 24 1,536
{12,5,4,0} 24 2,304
{12,5,3,1} 24 2,304
{12,5,2,2} 12 576
{12,4,4,1} 12 768
{12,4,3,2} 24 1,536
{12,3,3,3} 4 512
{11,10,0,0} 12 2,340
{11,9,1,0} 24 2,880
{11,8,2,0} 24 2,160
{11,8,1,1} 12 1,080
{11,7,3,0} 24 3,600
{11,7,2,1} 24 1,800
{11,6,4,0} 24 2,880
{11,6,3,1} 24 2,880
{11,6,2,2} 12 720
{11,5,5,0} 12 1,620
{11,5,4,1} 24 2,160
{11,5,3,2} 24 2,160
{11,4,4,2} 12 720
{11,4,3,3} 12 1,440
{10,10,1,0} 12 2,028
{10,9,2,0} 24 2,496
{10,9,1,1} 12 1,248
{10,8,3,0} 24 3,744
{10,8,2,1} 24 1,872
{10,7,4,0} 24 3,120
{10,7,3,1} 24 3,120
{10,7,2,2} 12 780
{10,6,5,0} 24 3,744
{10,6,4,1} 24 2,496
{10,6,3,2} 24 2,496
{10,5,5,1} 12 1,404
{10,5,4,2} 24 1,872
{10,5,3,3} 12 1,872
{10,4,4,3} 12 1,248
{9,9,3,0} 12 1,536
{9,9,2,1} 12 768
{9,8,4,0} 24 2,304
{9,8,3,1} 24 2,304
{9,8,2,2} 12 576
{9,7,5,0} 24 2,880
{9,7,4,1} 24 1,920
{9,7,3,2} 24 1,920
{9,6,6,0} 12 1,536
{9,6,5,1} 24 2,304
{9,6,4,2} 24 1,536
{9,6,3,3} 12 1,536
{9,5,5,2} 12 864
{9,5,4,3} 24 2,304
{9,4,4,4} 4 256
{8,8,5,0} 12 1,296
{8,8,4,1} 12 864
{8,8,3,2} 12 864
{8,7,6,0} 24 2,880
{8,7,5,1} 24 2,160
{8,7,4,2} 24 1,440
{8,7,3,3} 12 1,440
{8,6,6,1} 12 1,152
{8,6,5,2} 24 1,728
{8,6,4,3} 24 2,304
{8,5,5,3} 12 1,296
{8,5,4,4} 12 864
{7,7,7,0} 4 500
{7,7,6,1} 12 1,200
{7,7,5,2} 12 900
{7,7,4,3} 12 1,200
{7,6,6,2} 12 960
{7,6,5,3} 24 2,880
{7,6,4,4} 12 960
{7,5,5,4} 12 1,080
{6,6,6,3} 4 512
{6,6,5,4} 12 1,152
{6,5,5,5} 4 432
--------
TOTAL: 188,052

 

[Regs]

Wow thats great work. I'll have a look through Dr.Pizza and QED's method as soon as I finish my coffee.

 

[DrPizza]

I believe QED missed the hands where there are pairs or 3 of a kinds

 

[QED]

Originally posted by: DrPizza
I believe QED missed the hands where there are pairs or 3 of a kinds

You cannot have a pair or a three-of-a-kinds in a single suit, which is why there weren't counted in Problem #1.

A pair, or a three-of-a-kind, would contribute points to two (or three) different suits, which IS accounted for in Problem #2.

For instance, the possible hand of Four of Hearts, Four of Clubs, Four of Diamonds, and 9 of Spades would be accounted for in the category group labeled {9,4,4,4}-- the set of all hands where one suit has 9 total points, and the remaining 3 each have point total 4.

 

[Kyteland]

Originally posted by: QED
TOTAL: 187,332

I get 190624 by exhaustive search. I'll make a list like yours in a second. Edit: I see my mistake. I'm treating 1 and 11 as separate cards, which isn't right. New answer is 186184.

{ 1 1 1 1 2 2 2 2 3 3 3 } 4
{ 1 1 1 1 2 2 2 2 3 6 } 16
{ 1 1 1 1 2 2 2 2 4 5 } 16
{ 1 1 1 1 2 2 2 2 9 } 4
{ 1 1 1 1 2 2 2 3 3 5 } 96
{ 1 1 1 1 2 2 2 3 4 4 } 96
{ 1 1 1 1 2 2 2 3 8 } 64
{ 1 1 1 1 2 2 2 4 7 } 64
{ 1 1 1 1 2 2 2 5 6 } 64
{ 1 1 1 1 2 2 3 } 24
{ 1 1 1 1 2 2 3 3 3 4 } 96
{ 1 1 1 1 2 2 3 3 7 } 144
{ 1 1 1 1 2 2 3 4 6 } 384
{ 1 1 1 1 2 2 3 5 5 } 144
{ 1 1 1 1 2 2 3 10 } 384
{ 1 1 1 1 2 2 4 4 5 } 144
{ 1 1 1 1 2 2 4 9 } 96
{ 1 1 1 1 2 2 5 8 } 96
{ 1 1 1 1 2 2 6 7 } 96
{ 1 1 1 1 2 3 3 3 6 } 64
{ 1 1 1 1 2 3 3 4 5 } 384
{ 1 1 1 1 2 3 3 9 } 96
{ 1 1 1 1 2 3 4 4 4 } 64
{ 1 1 1 1 2 3 4 8 } 256
{ 1 1 1 1 2 3 5 7 } 256
{ 1 1 1 1 2 3 6 6 } 96
{ 1 1 1 1 2 4 4 7 } 96
{ 1 1 1 1 2 4 5 6 } 256
{ 1 1 1 1 2 5 } 16
{ 1 1 1 1 2 5 5 5 } 16
{ 1 1 1 1 2 5 10 } 256
{ 1 1 1 1 2 6 9 } 64
{ 1 1 1 1 2 7 8 } 64
{ 1 1 1 1 3 3 3 3 5 } 4
{ 1 1 1 1 3 3 3 4 4 } 24
{ 1 1 1 1 3 3 3 8 } 16
{ 1 1 1 1 3 3 4 7 } 96
{ 1 1 1 1 3 3 5 6 } 96
{ 1 1 1 1 3 4 } 16
{ 1 1 1 1 3 4 4 6 } 96
{ 1 1 1 1 3 4 5 5 } 96
{ 1 1 1 1 3 4 10 } 256
{ 1 1 1 1 3 5 9 } 64
{ 1 1 1 1 3 6 8 } 64
{ 1 1 1 1 3 7 7 } 24
{ 1 1 1 1 4 4 4 5 } 16
{ 1 1 1 1 4 4 9 } 24
{ 1 1 1 1 4 5 8 } 64
{ 1 1 1 1 4 6 7 } 64
{ 1 1 1 1 5 5 7 } 24
{ 1 1 1 1 5 6 6 } 24
{ 1 1 1 1 7 } 4
{ 1 1 1 1 7 10 } 64
{ 1 1 1 1 8 9 } 16
{ 1 1 1 2 2 2 2 } 4
{ 1 1 1 2 2 2 2 3 3 4 } 96
{ 1 1 1 2 2 2 2 3 7 } 64
{ 1 1 1 2 2 2 2 4 6 } 64
{ 1 1 1 2 2 2 2 5 5 } 24
{ 1 1 1 2 2 2 2 10 } 64
{ 1 1 1 2 2 2 3 3 3 3 } 16
{ 1 1 1 2 2 2 3 3 6 } 384
{ 1 1 1 2 2 2 3 4 5 } 1024
{ 1 1 1 2 2 2 3 9 } 256
{ 1 1 1 2 2 2 4 4 4 } 64
{ 1 1 1 2 2 2 4 8 } 256
{ 1 1 1 2 2 2 5 7 } 256
{ 1 1 1 2 2 2 6 6 } 96
{ 1 1 1 2 2 3 3 3 5 } 384
{ 1 1 1 2 2 3 3 4 4 } 864
{ 1 1 1 2 2 3 3 8 } 576
{ 1 1 1 2 2 3 4 7 } 1536
{ 1 1 1 2 2 3 5 6 } 1536
{ 1 1 1 2 2 4 } 96
{ 1 1 1 2 2 4 4 6 } 576
{ 1 1 1 2 2 4 5 5 } 576
{ 1 1 1 2 2 4 10 } 1536
{ 1 1 1 2 2 5 9 } 384
{ 1 1 1 2 2 6 8 } 384
{ 1 1 1 2 2 7 7 } 144
{ 1 1 1 2 3 3 } 96
{ 1 1 1 2 3 3 3 3 4 } 64
{ 1 1 1 2 3 3 3 7 } 256
{ 1 1 1 2 3 3 4 6 } 1536
{ 1 1 1 2 3 3 5 5 } 576
{ 1 1 1 2 3 3 10 } 1536
{ 1 1 1 2 3 4 4 5 } 1536
{ 1 1 1 2 3 4 9 } 1024
{ 1 1 1 2 3 5 8 } 1024
{ 1 1 1 2 3 6 7 } 1024
{ 1 1 1 2 4 4 4 4 } 16
{ 1 1 1 2 4 4 8 } 384
{ 1 1 1 2 4 5 7 } 1024
{ 1 1 1 2 4 6 6 } 384
{ 1 1 1 2 5 5 6 } 384
{ 1 1 1 2 6 } 64
{ 1 1 1 2 6 10 } 1024
{ 1 1 1 2 7 9 } 256
{ 1 1 1 2 8 8 } 96
{ 1 1 1 3 3 3 3 6 } 16
{ 1 1 1 3 3 3 4 5 } 256
{ 1 1 1 3 3 3 9 } 64
{ 1 1 1 3 3 4 4 4 } 96
{ 1 1 1 3 3 4 8 } 384
{ 1 1 1 3 3 5 7 } 384
{ 1 1 1 3 3 6 6 } 144
{ 1 1 1 3 4 4 7 } 384
{ 1 1 1 3 4 5 6 } 1024
{ 1 1 1 3 5 } 64
{ 1 1 1 3 5 5 5 } 64
{ 1 1 1 3 5 10 } 1024
{ 1 1 1 3 6 9 } 256
{ 1 1 1 3 7 8 } 256
{ 1 1 1 4 4 } 24
{ 1 1 1 4 4 4 6 } 64
{ 1 1 1 4 4 5 5 } 144
{ 1 1 1 4 4 10 } 384
{ 1 1 1 4 5 9 } 256
{ 1 1 1 4 6 8 } 256
{ 1 1 1 4 7 7 } 96
{ 1 1 1 5 5 8 } 96
{ 1 1 1 5 6 7 } 256
{ 1 1 1 6 6 6 } 16
{ 1 1 1 8 } 16
{ 1 1 1 8 10 } 256
{ 1 1 1 9 9 } 24
{ 1 1 2 2 2 2 3 3 5 } 144
{ 1 1 2 2 2 2 3 4 4 } 144
{ 1 1 2 2 2 2 3 8 } 96
{ 1 1 2 2 2 2 4 7 } 96
{ 1 1 2 2 2 2 5 6 } 96
{ 1 1 2 2 2 3 } 96
{ 1 1 2 2 2 3 3 3 4 } 384
{ 1 1 2 2 2 3 3 7 } 576
{ 1 1 2 2 2 3 4 6 } 1536
{ 1 1 2 2 2 3 5 5 } 576
{ 1 1 2 2 2 3 10 } 1536
{ 1 1 2 2 2 4 4 5 } 576
{ 1 1 2 2 2 4 9 } 384
{ 1 1 2 2 2 5 8 } 384
{ 1 1 2 2 2 6 7 } 384
{ 1 1 2 2 3 3 3 6 } 576
{ 1 1 2 2 3 3 4 5 } 3456
{ 1 1 2 2 3 3 9 } 864
{ 1 1 2 2 3 4 4 4 } 576
{ 1 1 2 2 3 4 8 } 2304
{ 1 1 2 2 3 5 7 } 2304
{ 1 1 2 2 3 6 6 } 864
{ 1 1 2 2 4 4 7 } 864
{ 1 1 2 2 4 5 6 } 2304
{ 1 1 2 2 5 } 144
{ 1 1 2 2 5 5 5 } 144
{ 1 1 2 2 5 10 } 2304
{ 1 1 2 2 6 9 } 576
{ 1 1 2 2 7 8 } 576
{ 1 1 2 3 3 3 3 5 } 96
{ 1 1 2 3 3 3 4 4 } 576
{ 1 1 2 3 3 3 8 } 384
{ 1 1 2 3 3 4 7 } 2304
{ 1 1 2 3 3 5 6 } 2304
{ 1 1 2 3 4 } 384
{ 1 1 2 3 4 4 6 } 2304
{ 1 1 2 3 4 5 5 } 2304
{ 1 1 2 3 4 10 } 6144
{ 1 1 2 3 5 9 } 1536
{ 1 1 2 3 6 8 } 1536
{ 1 1 2 3 7 7 } 576
{ 1 1 2 4 4 4 5 } 384
{ 1 1 2 4 4 9 } 576
{ 1 1 2 4 5 8 } 1536
{ 1 1 2 4 6 7 } 1536
{ 1 1 2 5 5 7 } 576
{ 1 1 2 5 6 6 } 576
{ 1 1 2 7 } 96
{ 1 1 2 7 10 } 1536
{ 1 1 2 8 9 } 384
{ 1 1 3 3 3 } 24
{ 1 1 3 3 3 3 7 } 24
{ 1 1 3 3 3 4 6 } 384
{ 1 1 3 3 3 5 5 } 144
{ 1 1 3 3 3 10 } 384
{ 1 1 3 3 4 4 5 } 864
{ 1 1 3 3 4 9 } 576
{ 1 1 3 3 5 8 } 576
{ 1 1 3 3 6 7 } 576
{ 1 1 3 4 4 4 4 } 24
{ 1 1 3 4 4 8 } 576
{ 1 1 3 4 5 7 } 1536
{ 1 1 3 4 6 6 } 576
{ 1 1 3 5 5 6 } 576
{ 1 1 3 6 } 96
{ 1 1 3 6 10 } 1536
{ 1 1 3 7 9 } 384
{ 1 1 3 8 8 } 144
{ 1 1 4 4 4 7 } 96
{ 1 1 4 4 5 6 } 576
{ 1 1 4 5 } 96
{ 1 1 4 5 5 5 } 96
{ 1 1 4 5 10 } 1536
{ 1 1 4 6 9 } 384
{ 1 1 4 7 8 } 384
{ 1 1 5 5 9 } 144
{ 1 1 5 6 8 } 384
{ 1 1 5 7 7 } 144
{ 1 1 6 6 7 } 144
{ 1 1 9 } 24
{ 1 1 9 10 } 384
{ 1 2 2 2 2 3 3 3 3 } 4
{ 1 2 2 2 2 3 3 6 } 96
{ 1 2 2 2 2 3 4 5 } 256
{ 1 2 2 2 2 3 9 } 64
{ 1 2 2 2 2 4 4 4 } 16
{ 1 2 2 2 2 4 8 } 64
{ 1 2 2 2 2 5 7 } 64
{ 1 2 2 2 2 6 6 } 24
{ 1 2 2 2 3 3 3 5 } 256
{ 1 2 2 2 3 3 4 4 } 576
{ 1 2 2 2 3 3 8 } 384
{ 1 2 2 2 3 4 7 } 1024
{ 1 2 2 2 3 5 6 } 1024
{ 1 2 2 2 4 } 64
{ 1 2 2 2 4 4 6 } 384
{ 1 2 2 2 4 5 5 } 384
{ 1 2 2 2 4 10 } 1024
{ 1 2 2 2 5 9 } 256
{ 1 2 2 2 6 8 } 256
{ 1 2 2 2 7 7 } 96
{ 1 2 2 3 3 } 144
{ 1 2 2 3 3 3 3 4 } 96
{ 1 2 2 3 3 3 7 } 384
{ 1 2 2 3 3 4 6 } 2304
{ 1 2 2 3 3 5 5 } 864
{ 1 2 2 3 3 10 } 2304
{ 1 2 2 3 4 4 5 } 2304
{ 1 2 2 3 4 9 } 1536
{ 1 2 2 3 5 8 } 1536
{ 1 2 2 3 6 7 } 1536
{ 1 2 2 4 4 4 4 } 24
{ 1 2 2 4 4 8 } 576
{ 1 2 2 4 5 7 } 1536
{ 1 2 2 4 6 6 } 576
{ 1 2 2 5 5 6 } 576
{ 1 2 2 6 } 96
{ 1 2 2 6 10 } 1536
{ 1 2 2 7 9 } 384
{ 1 2 2 8 8 } 144
{ 1 2 3 3 3 3 6 } 64
{ 1 2 3 3 3 4 5 } 1024
{ 1 2 3 3 3 9 } 256
{ 1 2 3 3 4 4 4 } 384
{ 1 2 3 3 4 8 } 1536
{ 1 2 3 3 5 7 } 1536
{ 1 2 3 3 6 6 } 576
{ 1 2 3 4 4 7 } 1536
{ 1 2 3 4 5 6 } 4096
{ 1 2 3 5 } 256
{ 1 2 3 5 5 5 } 256
{ 1 2 3 5 10 } 4096
{ 1 2 3 6 9 } 1024
{ 1 2 3 7 8 } 1024
{ 1 2 4 4 } 96
{ 1 2 4 4 4 6 } 256
{ 1 2 4 4 5 5 } 576
{ 1 2 4 4 10 } 1536
{ 1 2 4 5 9 } 1024
{ 1 2 4 6 8 } 1024
{ 1 2 4 7 7 } 384
{ 1 2 5 5 8 } 384
{ 1 2 5 6 7 } 1024
{ 1 2 6 6 6 } 64
{ 1 2 8 } 64
{ 1 2 8 10 } 1024
{ 1 2 9 9 } 96
{ 1 3 3 3 3 4 4 } 24
{ 1 3 3 3 3 8 } 16
{ 1 3 3 3 4 7 } 256
{ 1 3 3 3 5 6 } 256
{ 1 3 3 4 } 96
{ 1 3 3 4 4 6 } 576
{ 1 3 3 4 5 5 } 576
{ 1 3 3 4 10 } 1536
{ 1 3 3 5 9 } 384
{ 1 3 3 6 8 } 384
{ 1 3 3 7 7 } 144
{ 1 3 4 4 4 5 } 256
{ 1 3 4 4 9 } 384
{ 1 3 4 5 8 } 1024
{ 1 3 4 6 7 } 1024
{ 1 3 5 5 7 } 384
{ 1 3 5 6 6 } 384
{ 1 3 7 } 64
{ 1 3 7 10 } 1024
{ 1 3 8 9 } 256
{ 1 4 4 4 8 } 64
{ 1 4 4 5 7 } 384
{ 1 4 4 6 6 } 144
{ 1 4 5 5 6 } 384
{ 1 4 6 } 64
{ 1 4 6 10 } 1024
{ 1 4 7 9 } 256
{ 1 4 8 8 } 96
{ 1 5 5 } 24
{ 1 5 5 5 5 } 4
{ 1 5 5 10 } 384
{ 1 5 6 9 } 256
{ 1 5 7 8 } 256
{ 1 6 6 8 } 96
{ 1 6 7 7 } 96
{ 1 10 } 64
{ 1 10 10 } 480
{ 2 2 2 2 3 3 3 4 } 16
{ 2 2 2 2 3 3 7 } 24
{ 2 2 2 2 3 4 6 } 64
{ 2 2 2 2 3 5 5 } 24
{ 2 2 2 2 3 10 } 64
{ 2 2 2 2 4 4 5 } 24
{ 2 2 2 2 4 9 } 16
{ 2 2 2 2 5 8 } 16
{ 2 2 2 2 6 7 } 16
{ 2 2 2 3 3 3 6 } 64
{ 2 2 2 3 3 4 5 } 384
{ 2 2 2 3 3 9 } 96
{ 2 2 2 3 4 4 4 } 64
{ 2 2 2 3 4 8 } 256
{ 2 2 2 3 5 7 } 256
{ 2 2 2 3 6 6 } 96
{ 2 2 2 4 4 7 } 96
{ 2 2 2 4 5 6 } 256
{ 2 2 2 5 5 5 } 16
{ 2 2 2 5 10 } 256
{ 2 2 2 6 9 } 64
{ 2 2 2 7 8 } 64
{ 2 2 3 3 3 3 5 } 24
{ 2 2 3 3 3 4 4 } 144
{ 2 2 3 3 3 8 } 96
{ 2 2 3 3 4 7 } 576
{ 2 2 3 3 5 6 } 576
{ 2 2 3 4 4 6 } 576
{ 2 2 3 4 5 5 } 576
{ 2 2 3 4 10 } 1536
{ 2 2 3 5 9 } 384
{ 2 2 3 6 8 } 384
{ 2 2 3 7 7 } 144
{ 2 2 4 4 4 5 } 96
{ 2 2 4 4 9 } 144
{ 2 2 4 5 8 } 384
{ 2 2 4 6 7 } 384
{ 2 2 5 5 7 } 144
{ 2 2 5 6 6 } 144
{ 2 2 7 10 } 384
{ 2 2 8 9 } 96
{ 2 3 3 3 3 7 } 16
{ 2 3 3 3 4 6 } 256
{ 2 3 3 3 5 5 } 96
{ 2 3 3 3 10 } 256
{ 2 3 3 4 4 5 } 576
{ 2 3 3 4 9 } 384
{ 2 3 3 5 8 } 384
{ 2 3 3 6 7 } 384
{ 2 3 4 4 4 4 } 16
{ 2 3 4 4 8 } 384
{ 2 3 4 5 7 } 1024
{ 2 3 4 6 6 } 384
{ 2 3 5 5 6 } 384
{ 2 3 6 10 } 1024
{ 2 3 7 9 } 256
{ 2 3 8 8 } 96
{ 2 4 4 4 7 } 64
{ 2 4 4 5 6 } 384
{ 2 4 5 5 5 } 64
{ 2 4 5 10 } 1024
{ 2 4 6 9 } 256
{ 2 4 7 8 } 256
{ 2 5 5 9 } 96
{ 2 5 6 8 } 256
{ 2 5 7 7 } 96
{ 2 6 6 7 } 96
{ 2 9 10 } 256
{ 3 3 3 3 4 5 } 16
{ 3 3 3 3 9 } 4
{ 3 3 3 4 4 4 } 16
{ 3 3 3 4 8 } 64
{ 3 3 3 5 7 } 64
{ 3 3 3 6 6 } 24
{ 3 3 4 4 7 } 144
{ 3 3 4 5 6 } 384
{ 3 3 5 5 5 } 24
{ 3 3 5 10 } 384
{ 3 3 6 9 } 96
{ 3 3 7 8 } 96
{ 3 4 4 4 6 } 64
{ 3 4 4 5 5 } 144
{ 3 4 4 10 } 384
{ 3 4 5 9 } 256
{ 3 4 6 8 } 256
{ 3 4 7 7 } 96
{ 3 5 5 8 } 96
{ 3 5 6 7 } 256
{ 3 6 6 6 } 16
{ 3 8 10 } 256
{ 3 9 9 } 24
{ 4 4 4 4 5 } 4
{ 4 4 4 9 } 16
{ 4 4 5 8 } 96
{ 4 4 6 7 } 96
{ 4 5 5 7 } 96
{ 4 5 6 6 } 96
{ 4 7 10 } 256
{ 4 8 9 } 64
{ 5 5 5 6 } 16
{ 5 6 10 } 256
{ 5 7 9 } 64
{ 5 8 8 } 24
{ 6 6 9 } 24
{ 6 7 8 } 64
{ 7 7 7 } 4

 

[oiprocs]

jesus. You guys are insane. There must be at least a thousand hamster wheels spinning in your brains.

 

[Kyteland]

Originally posted by: Oiprocs
jesus. You guys are insane. There must be at least a thousand hamster wheels spinning in your brains.

I do these kind of problems for a living, so this is fun for me.

 

[Atomic Playboy]

Originally posted by: Kyteland

Originally posted by: Oiprocs
jesus. You guys are insane. There must be at least a thousand hamster wheels spinning in your brains.

I do these kind of problems for a living, so this is fun for me.

I do database work for a living, but that doesn't make me want to do it in my spare time...

 

[Kyteland]

Originally posted by: Atomic Playboy

Originally posted by: Kyteland

Originally posted by: Oiprocs
jesus. You guys are insane. There must be at least a thousand hamster wheels spinning in your brains.

I do these kind of problems for a living, so this is fun for me.

I do database work for a living, but that doesn't make me want to do it in my spare time...

I guess I'm one of those lucky people who really likes his job.

 

[QED]

Originally posted by: Kyteland

Originally posted by: QED
TOTAL: 187,332

I get 190624 by exhaustive search. I'll make a list like yours in a second. Edit: I see my mistake. I'm treating 1 and 11 as separate cards, which isn't right. New answer is 186184.

Your post prompted me to check my work again, and so far I found out I missed two more possible
groups of 21-point splits amongst 4 suits: {11,4,3,3} and {6,5,5,5}.

{11,4,3,3} represents 12 unique suit combinations, and there are P(11) * P(4) * P(3) * P(3) =
15 * 2 * 2 * 2 = 120 possible hands for each suit combination , for a total of an extra
12* 120 = 1440 hands.

{6,5,5,3} represents 4 unique suit combinations, and there are P(6) * P(5) * P(5) * P(5) =
4 * 3 * 3 * 3 = 108 possible hands for each suit combination, for a total of an extra
4 * 108 = 432 hands.

EDIT: Also, found that the group {6,6,5,4} represents 12 unique suit combinations, not the 24 I had
mistakenly listed in my original post. Hence there are only 12 * P(6) * P(6) * P(5) * P(4) =
12 * 4 * 4 * 3 * 2 = 1152 total hands in that group, not the 2304 originally listed.


So my total is revised slightly upwards to 188,052.



I'm now off to double-check that P(n) values are correct... since I did all work by hand, it's quite
possible I counted something extra, or forgot to count something.