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Old 11-12-2016, 12:30 AM   #1
piranha
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Badugi Probability Question

Been working on this for hours and hours and hours and can't figure out how to do it.

In Badugi, I'm trying to figure out the odds of being dealt a specific 2 card hand like A2. There are twelve combos of A2 where the Ace and Deuce must be 2 different suits. The remaining 50 cards can't make a tri or Badugi. In other words out of the remaining 50 cards, they must be a pair or of the same suit.

Examples: Ac2s9s8s Ac2s5c8s Ac2sAhKs

With a 2 card hand like 54 for example, again there are 12 combos. This time however, the remaining cards that make up the 4 card hand must be of the same suit, a pair, or a higher rank. In other words, you couldn't have 5c4h3h because that would then be counted as a 53 instead.

Examples: 5h4d9h8h, 5s4c4hQh, 5d4sJdJs

Some additional info:

-The odds of being dealt a 2 card hand in Badugi, and this encompasses all 78 two card hands from A2 to QK, is 35.55%.

-There are 270,725 combinations of possible 4 card poker hands. This means there are 96,242 combos of two card hands. I'm actually looking for the answer above to be in terms of combinations but percentage would work as well.

For some really good discussion regarding how to calculate 3 card hands, particularly the post by David Sklanski, see here:

https://forumserver.twoplustwo.com/25...99/?highlight=

And another good discussion on a similar topic here, in particular Bruce Z's response:

http://math.stackexchange.com/questi...cards-out-of-4
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Old 11-12-2016, 02:14 AM   #2
whosnext
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Re: Badugi Probability Question

I'll give it a crack. I broke the problem down to cases. Each case is straightforward to calculate. At the end we add the results from each case. As I often say in these situations, sometimes a first attempt can grease the skids even if it is wrong (a la Dr. Watson and Sherlock Holmes).

Call the two ranks in question X and Y and without loss of generality, say X<Y. They have to be in different suits and it will make what follows easier if we say X is a Heart and Y is a Club (of course, we will generalize to all pairs of suits at the end).

Case 1: Hand has 4 distinct ranks. Call then X, Y, W, and Z.

Then W and Z must each be either a Heart higher than X (not Y) or a Club higher than Y. This is a total of (12-X)+(13-Y) cards from which you must have two. Using the Choice function this would be written as C([(12-X)+(13-Y)],2).

Case 2: Hand has 3 distinct ranks X, X, Y, and Z.

Case 2a) Other X is a Club. Then 4th card must be any Club higher than Y or any Heart higher than Y. This is a total of 2*(13-Y) cards from which to choose one.

Case 2b) Other X is a Spade or Diamond. Then 4th card must be a Club higher than Y. This is a total of (13-Y) cards and we have a choice of two X's, so total hands in this subcase is 2*(13-Y).

Case 3. Hand has 3 distinct ranks X, Y, Y, and Z.

Case 3a) Other Y is a Heart. Then 4th card must be either a Heart higher than Y or a Club higher than Y. So this is a total of 2*(13-Y) cards from which we must choose one.

Case 3b) Other Y is a Spade of Diamond. Then 4th card must be a Heart higher than X but not equal to Y. This is (12-X) cards. And we had a choice of two Y's. So there are a total of 2*(12-X) hands in this subcase.

Case 4. Hand has 3 distinct ranks X, Y, Z, Z.

The only hands that work here are if both Z's are higher than Y and a Heart and a Club respectively. So there are (13-Y) hands in this case.

Case 5. Hand has 2 distinct ranks X, X, X, Y.

Any other pair of X's work here. So there are C(3,2)=3 hands in this case.

Case 6. Hand has 2 distinct ranks X, X, Y, Y.

And other X's or Y's work here. So there are C(3,1)*C(3,1) = 3*3 = 9 hands in this case.

Case 7: Hand has 2 distinct ranks X, Y, Y, Y.

Any other pair of Y's work here. So there are C(3,2)=3 hands in this case.

If I did these correctly, now we need to sum the cases. Again, hoping that I did the simple summing correctly, and using the fact that there are 12 possible X-Y combos with two different ranks and two different suits, I get that there are:

12 * [C([25-(X+Y)],2) + 7(13-Y) + 2(12-X) + 15] four-card hands that give a 2-card XY.

Let's try a few X-Y combos:

A2 -> X=1 and Y=2. The formula above says that there are 12*345=4140 hands.

75 -> X=5 and Y=7. The formula above says that there are 12*149=1788 hands.

KQ -> X=12 and Y=13. The formula above says that there are 12*15=180 hands.

I probably won't have time to review this until late on the weekend.

I apologize in advance if there are some glaring and obvious errors in the above. If there are glaring and obvious errors, they should be easily found and fixed.

Last edited by whosnext; 11-12-2016 at 02:24 AM.
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Old 11-12-2016, 09:30 AM   #3
Buzz
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Re: Badugi Probability Question

Quote:
Originally Posted by piranha View Post
In Badugi, I'm trying to figure out the odds of being dealt a specific 2 card hand like A2. There are twelve combos of A2 where the Ace and Deuce must be 2 different suits. The remaining 50 cards can't make a tri or Badugi. In other words out of the remaining 50 cards, they must be a pair or of the same suit.
I think there are 139,308 possible ways to be dealt a specific 2 card hand like A2. My work is shown below:

specific 2 cards Total possible for XY
Ac2sXY 378
Ac3sXY 351
Ac4sXY 325
Ac5sXY 300
Ac6sXY 276
Ac7sXY 253
Ac8sXY 231
Ac9sXY 210
AcTsXY 190
AcJsXY 171
AcQsXY 153
AcKsXY 136
2c3sXY 325
2c4sXY 300
2c5sXY 276
2c6sXY 253
2c7sXY 231
2c8sXY 210
2c9sXY 190
2cTsXY 171
2cJsXY 153
2cQsXY 136
2cKsXY 120
3c4sXY 276
3c5sXY 253
3c6sXY 231
3c7sXY 210
3c8sXY 190
3c9sXY 171
3cTsXY 153
3cJsXY 136
3cQsXY 120
3cKsXY 105
4c5sXY 231
4c6sXY 210
4c7sXY 190
4c8sXY 171
4c9sXY 153
4cTsXY 136
4cJsXY 120
4cQsXY 105
4cKsXY 91
5c6sXY 190
5c7sXY 171
5c8sXY 153
5c9sXY 136
5cTsXY 120
5cJsXY 105
5cQsXY 91
5cKsXY 78
6c7sXY 153
6c8sXY 136
6c9sXY 120
6cTsXY 105
6cJsXY 91
6cQsXY 78
6cKsXY 66
7c8sXY 120
7c9sXY 105
7cTsXY 91
7cJsXY 78
7cQsXY 66
7cKsXY 55
8c9sXY 91
8cTsXY 78
8cJsXY 66
8cQsXY 55
8cKsXY 45
9cTsXY 66
9cJsXY 55
9cQsXY 45
9cKsXY 36
TcJsXY 45
TcQsXY 36
TcKsXY 28
JcQsXY 28
JcKsXY 21
QcKsXY 15
total 11609

11609*12=139.308

Buzz

Last edited by whosnext; 11-12-2016 at 07:02 PM. Reason: removing the spoiler tags
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Old 11-12-2016, 02:00 PM   #4
piranha
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Re: Badugi Probability Question

Using whosnext method above, I get 114,912 combos. I definitely could of made an error doing each combo manually but I'm confident it would be a small one - I double checked everything last night.

If you look at the links in my op, multiple people (a lot smarter than me), came up with the total number of 3 card hands being 154,440 combos.

It's universally accepted Badugis make up 6.34% of hands or 17,630 combos.

I think we all agree there are 270,725 combos of 4 card hands.

So subtracting 154,440 and 17,630 (#of 3 card+#of 4 card, aka tris and badugis) from the total number of 4 card combos (270,725), that leaves 99,124 combos for 2 card and 1 card hands.

Therefore I don't think it's possible there are more than 100K 2 card hands.

Last edited by piranha; 11-12-2016 at 02:06 PM.
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Old 11-12-2016, 06:58 PM   #5
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Re: Badugi Probability Question

Quote:
Originally Posted by piranha View Post
Using whosnext method above, I get 114,912 combos. I definitely could of made an error doing each combo manually but I'm confident it would be a small one - I double checked everything last night.
I easily could have made an error too.

Quote:
It's universally accepted Badugis make up 6.34% of hands or 17,630 combos.
I hate it when my result is different from what is universally accepted. But I make it 17,160 (as shown below).

Basically my logic is take away a rank and a suit for each successive card chosen. Thus:
Choose any of 13 ranks and 4 suits for the first card.
Choose any of 12 ranks and 3 suits for the second card.
Choose any of 11 ranks and 2 suits for the third card.
Choose any of 10 ranks and 1 suit for the fourth card.

(any of 13 ranks and 4 suits) = 13*4=52
(any of 12 ranks and 3 suits) = 12*3=36
(any of 11 ranks and 2 suits) = 11*2=22
(any of 10 ranks and 1 suit) = 10*1=10

52*36*22*10/1/2/3/4=17160

Quote:
multiple people (a lot smarter than me), came up with the total number of 3 card hands being 154,440 combos.
If 10 cards of the fourth suit make a badugi, then of the 49 remaining cards (after three cards are already chosen), 39 must not. Therefore
52*36*22*39/1/2/3/4=66,924 is my guess at the number of three-card-badugis.

Quote:
I think we all agree there are 270,725 combos of 4 card hands.
52*51*50*49/1/2/3/4=270,725.

Buzz
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Old 11-12-2016, 08:42 PM   #6
piranha
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Re: Badugi Probability Question

Quote:
Originally Posted by Buzz View Post
I easily could have made an error too.


I hate it when my result is different from what is universally accepted. But I make it 17,160 (as shown below).

Basically my logic is take away a rank and a suit for each successive card chosen. Thus:
Choose any of 13 ranks and 4 suits for the first card.
Choose any of 12 ranks and 3 suits for the second card.
Choose any of 11 ranks and 2 suits for the third card.
Choose any of 10 ranks and 1 suit for the fourth card.

(any of 13 ranks and 4 suits) = 13*4=52
(any of 12 ranks and 3 suits) = 12*3=36
(any of 11 ranks and 2 suits) = 11*2=22
(any of 10 ranks and 1 suit) = 10*1=10

52*36*22*10/1/2/3/4=17160

If 10 cards of the fourth suit make a badugi, then of the 49 remaining cards (after three cards are already chosen), 39 must not. Therefore
52*36*22*39/1/2/3/4=66,924 is my guess at the number of three-card-badugis.

52*51*50*49/1/2/3/4=270,725.

Buzz
The 17630 was a typo on my part, you are absolutely correct it's 17160.

So if we're in agreement on the number of Badugi combos as being 17160 and Tris as being 154,440 combos, then the sum of those two is 171,600 combos.

We both agree 270,725 combos as being the total number of 4 card hands.

So that leaves: 270,725-171,600 combos for 1 and 2 card hands or
~99,125 combos (This number could be +/- 1 due to rounding).

Note: Both DougZ and David Sklanski indirectly arrived at the ~171,600 combos as the number of tris using different methods here:

http://math.stackexchange.com/questi...cards-out-of-4

I'm content to accept this number as accurate given their credibility and their documented work/methods to back it up.

Bruce Z took it a step further and said the 2 card/1 card breakdown is:

2 Cards: ~35.55%
1 Cards: ~1.06%

So the point I'm trying to make is that if we're arriving at 100K+ combos when summing up all our individual 2 card hands whether it be my method, your method, or anyone else's, we're making an error somewhere.

Lastly and a bit off topic, I don't think I've thanked you for the amount of time you've put into this. Although we haven't found the correct answer, you've done more than your share to help find it, not only in my threads but older ones on the same topic. Thank you.

Tony
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Old 11-13-2016, 12:05 AM   #7
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Re: Badugi Probability Question

Just a quick pop-in. Not sure it has been mentioned, but the number of 1-card badugi hands is very easy to calculate. So there is no great mystery on that front.
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Old 11-13-2016, 01:19 AM   #8
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Re: Badugi Probability Question

Quote:
Originally Posted by Buzz View Post
I think there are 139,308 possible ways to be dealt a specific 2 card hand like A2. My work is shown below:

specific 2 cards Total possible for XY
2cKsXY 120
3cKsXY 105
 

11609*12=139.308

Buzz
i thought these 2 are of a similar case, (as are all the kings)and i thought this case was the easiest to count, and so i started here.

i'm wondering how you arrived at the numbers you posted in the table because i would think 110*12, and 90*12 and then if we are supposed to add the the non-badugi 33kk,3kkk,333k thats 54 hands to add to each.
(so if i'm right all the Kings-case are overcounted) ( and if we are to count the 2pair and trip non-badugis, then globally multiplying by 12 would seem suspect)

Last edited by ngFTW; 11-13-2016 at 01:29 AM. Reason: language
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Old 11-13-2016, 01:29 AM   #9
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Re: Badugi Probability Question

I have found one error in what I posted earlier. Unfortunately, correcting the error gives more 2-card badugis rather than fewer.

I will highlight the correction below.

-----

I'll give it a crack. I broke the problem down to cases. Each case is straightforward to calculate. At the end we add the results from each case. As I often say in these situations, sometimes a first attempt can grease the skids even if it is wrong (a la Dr. Watson and Sherlock Holmes).

Call the two ranks in question X and Y and without loss of generality, say X<Y. They have to be in different suits and it will make what follows easier if we say X is a Heart and Y is a Club (of course, we will generalize to all pairs of suits at the end).

Case 1: Hand has 4 distinct ranks. Call then X, Y, W, and Z.

Then W and Z must each be either a Heart higher than X (not Y) or a Club higher than Y. This is a total of (12-X)+(13-Y) cards from which you must have two. Using the Choice function this would be written as C([(12-X)+(13-Y)],2).

Case 2: Hand has 3 distinct ranks X, X, Y, and Z.

Case 2a) Other X is a Club. Then 4th card must be any Club higher than Y or any Heart higher than Y. This is a total of 2*(13-Y) cards from which to choose one.

Case 2b) Other X is a Spade or Diamond. Then 4th card must be a Club higher than Y. This is a total of (13-Y) cards and we have a choice of two X's, so total hands in this subcase is 2*(13-Y).

Case 3. Hand has 3 distinct ranks X, Y, Y, and Z.

Case 3a) Other Y is a Heart. Then 4th card must be either a Heart higher than X (not Y) or a Club higher than Y. So this is a total of (12-X) + (13-Y) cards from which we must choose one.

Case 3b) Other Y is a Spade of Diamond. Then 4th card must be a Heart higher than X but not equal to Y. This is (12-X) cards. And we had a choice of two Y's. So there are a total of 2*(12-X) hands in this subcase.

Case 4. Hand has 3 distinct ranks X, Y, Z, Z.

The only hands that work here are if both Z's are higher than Y and a Heart and a Club respectively. So there are (13-Y) hands in this case.

Case 5. Hand has 2 distinct ranks X, X, X, Y.

Any other pair of X's work here. So there are C(3,2)=3 hands in this case.

Case 6. Hand has 2 distinct ranks X, X, Y, Y.

And other X's or Y's work here. So there are C(3,1)*C(3,1) = 3*3 = 9 hands in this case.

Case 7: Hand has 2 distinct ranks X, Y, Y, Y.

Any other pair of Y's work here. So there are C(3,2)=3 hands in this case.

If I did these correctly, now we need to sum the cases. Again, hoping that I did the simple summing correctly, and using the fact that there are 12 possible X-Y combos with two different ranks and two different suits, I get that there are:

12 * [C([25-(X+Y)],2) + 6(13-Y) + 3(12-X) + 15] four-card hands that give a 2-card XY.

Edit: From the above, summing over all the possible X-Y combos yields 118,716 2-card badugis.

This is inconsistent (too high) with the prevailing wisdom on how many 1-card (known with certainty), 3-card, and 4-card (known with certainty) badugis.

I am extremely busy this weekend so cannot do anything further until later.
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Old 11-13-2016, 02:08 AM   #10
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Re: Badugi Probability Question

JFC, I just figured out what the problem was. Several of the cases involved double-counting. I don't have the time to go through this carefully now, but I will amend my just-amended post (ahem). I will bold and italicize these new fixes.

-----

I'll give it a crack. I broke the problem down to cases. Each case is straightforward to calculate. At the end we add the results from each case. As I often say in these situations, sometimes a first attempt can grease the skids even if it is wrong (a la Dr. Watson and Sherlock Holmes).

Call the two ranks in question X and Y and without loss of generality, say X<Y. They have to be in different suits and it will make what follows easier if we say X is a Heart and Y is a Club (of course, we will generalize to all pairs of suits at the end).

Case 1: Hand has 4 distinct ranks. Call then X, Y, W, and Z.

Then W and Z must each be either a Heart higher than X (not Y) or a Club higher than Y. This is a total of (12-X)+(13-Y) cards from which you must have two. Using the Choice function this would be written as C([(12-X)+(13-Y)],2). Multiplying by 12 is correct here.

Case 2: Hand has 3 distinct ranks X, X, Y, and Z.

Case 2a) Other X is a Club. Then 4th card must be any Club higher than Y or any Heart higher than Y. This is a total of 2*(13-Y) cards from which to choose one. Multiplying by 12 is correct here?

Case 2b) Other X is a Spade or Diamond. Then 4th card must be a Club higher than Y. This is a total of (13-Y) cards and we have a choice of two X's, so total hands in this subcase is 2*(13-Y). Double-counting means that multiplying by 6 here is correct.

Case 3. Hand has 3 distinct ranks X, Y, Y, and Z.

Case 3a) Other Y is a Heart. Then 4th card must be either a Heart higher than X (not Y) or a Club higher than Y. So this is a total of (12-X) + (13-Y) cards from which we must choose one. Multiplying by 12 is correct here.

Case 3b) Other Y is a Spade of Diamond. Then 4th card must be a Heart higher than X but not equal to Y. This is (12-X) cards. And we had a choice of two Y's. So there are a total of 2*(12-X) hands in this subcase. Double-counting means that multiplying by 6 here is correct.

Case 4. Hand has 3 distinct ranks X, Y, Z, Z.

The only hands that work here are if both Z's are higher than Y and a Heart and a Club respectively. So there are (13-Y) hands in this case. Multiplying by 12 is correct here.

Case 5. Hand has 2 distinct ranks X, X, X, Y. There are clearly C(4,3)*3=12 hands in this case.

Case 6. Hand has 2 distinct ranks X, X, Y, Y. There are clearly C(4,2)*C(4,2)=6*6=36 hands in this case.

Case 7: Hand has 2 distinct ranks X, Y, Y, Y. There are clearly C(4,3)*3=4*3=12 hands in this case.


If I did these correctly, now we need to sum the cases. Again, hoping that I did the simple summing correctly, I get that there are:

12*C([25-(X+Y)],2) + 60*(13-Y) + 24(12-X) + 60 four-card hands that give a 2-card XY.

Edit: From the above, summing over all the possible X-Y combos yields 99,060 2-card badugis.

This is inconsistent (too high) with the prevailing wisdom on how many 1-card (known with certainty), 3-card, and 4-card (known with certainty) badugis.

I am extremely busy this weekend so cannot do anything further until later.
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Old 11-13-2016, 12:34 PM   #11
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135,876 is my new revised total

Quote:
Originally Posted by ngFTW View Post
i'm wondering how you arrived at the numbers you posted in the table
New revised sub totals are shown in middle column in table below, method is shown in right hand column.

Quote:
because i would think 110*12, and 90*12 and then if we are supposed to add the the non-badugi 33kk,3kkk,333k thats 54 hands to add to each.
(so if i'm right all the Kings-case are overcounted) ( and if we are to count the 2pair and trip non-badugis, then globally multiplying by 12 would seem suspect)
I'm doing it a little bit differently. I'll try to explain my logic for 3cKsXY.

First, all red cards except 3, 3, K, K are excluded from being X or Y.
Next the A and 2 are excluded.
Next all spades are excluded. That leaves a group of 14 cards.
Then we go back and include the 3, but only with the four red cards plus the K, making a group of 5 cards.

Then the "method" to find the number of possibilities for XY is
14*13/2+1*5=96 possible combinations of XY cards to fit in 3cKsXY.

Finally there are actually 12 ways to use a trey and a king from two different suits. So I summed the table and then multiplied by 12.
(Is "summed" a word?)

If I did it right, each entry in the table excludes all specific combos above it that are already counted. Thus the total for XY for Ac5sXY excludes combos already counted in Ac2sXY, Ac3sXY, and Ac4sXY.

specific 2 cards Total possible for XY method
Ac2sXY 378. 28*27/2
Ac3sXY 350. 26*25/2+25
Ac4sXY 323. 25*24/2+23
Ac5sXY 297. 24*23/2+21
Ac6sXY 272. 23*22/2+19
Ac7sXY 248. 22*21/2+17
Ac8sXY 225. 21*20/2+15
Ac9sXY 203. 20*19/2+13
AcTsXY 182. 19*18/2+11
AcJsXY 162. 18*17/2+9
AcQsXY 143. 17*16/2+7
AcKsXY 125. 16*15/2+5
2c3sXY 325. 26*25/2
2c4sXY 299. 24*23/2+23
2c5sXY 274. 23*22/2+21
2c6sXY 250. 22*21/2+19
2c7sXY 227. 21*20/2+17
2c8sXY 205. 20*19/2+15
2c9sXY 184. 19*18/2+13
2cTsXY 164. 18*17/2+11
2cJsXY 145. 17*16/2+9
2cQsXY 127. 16*15/2+7
2cKsXY 110. 15*14/2+5
3c4sXY 276. 24*23/2
3c5sXY 252. 22*21/2+21
3c6sXY 229. 21*20/2+19
3c7sXY 207. 20*19/2+17
3c8sXY 186. 19*18/2+15
3c9sXY 166. 18*17/2+13
3cTsXY 147. 17*16/2+11
3cJsXY 129. 16*15/2+9
3cQsXY 112. 15*14/2+7
3cKsXY 96. 14*13/2+5
4c5sXY 231. 22*21/2
4c6sXY 209. 20*19/2+19
4c7sXY 188. 19*18/2+17
4c8sXY 168. 18*17/2+15
4c9sXY 149. 17*16/2+13
4cTsXY 131. 16*15/2+11
4cJsXY 114. 15*14/2+9
4cQsXY 98. 14*13/2+7
4cKsXY 83. 13*12/2+5
5c6sXY 190. 20*19/2
5c7sXY 170. 18*17/2+17
5c8sXY 151. 17*16/2+15
5c9sXY 133. 16*15/2+13
5cTsXY 116. 15*14/2+11
5cJsXY 100. 14*13/2+9
5cQsXY 85. 13*12/2+7
5cKsXY 71. 12*11/2+5
6c7sXY 153. 18*17/2
6c8sXY 135. 16*15/2+15
6c9sXY 118. 15*14/2+13
6cTsXY 102. 14*13/2+11
6cJsXY 87. 13*12/2+9
6cQsXY 73. 12*11/2+7
6cKsXY 60. 11*10/2+5
7c8sXY 120. 16*15/2
7c9sXY 104. 14*13/2+13
7cTsXY 89. 13*12/2+11
7cJsXY 75. 12*11/2+9
7cQsXY 62. 11*10/2+7
7cKsXY 50. 10*9/2+5
8c9sXY 91. 14*13/2
8cTsXY 77. 12*11/2+11
8cJsXY 64. 11*10/2+9
8cQsXY 52. 10*9/2+7
8cKsXY 41. 9*8/2+5
9cTsXY 66. 12*11/2
9cJsXY 54. 10*9/2+9
9cQsXY 43. 9*8/2+7
9cKsXY 33. 8*7/2+5
TcJsXY 45. 10*9/2
TcQsXY 35. 8*7/2+7
TcKsXY 26. 7*6/2+5
JcQsXY 28. 8*7/2
JcKsXY 20. 6*5/2+5
QcKsXY 15. 6*5/2
11323  
Then 11323*12=135,876

Quote:
i'm wondering how you arrived at the numbers you posted in the table because i would think 110*12, and 90*12
My revised table does have 110 for the XY possibilities for 2cKs (the same as you), but I get 96 (not 90) for the XY possibilities for 2cKs. Maybe I did that one incorrectly. I tried to explain my logic for it up above in this post.

Quote:
and then if we are supposed to add the the non-badugi 33kk,3kkk,333k thats 54 hands to add to each.
I did it differently. I included the red threes and red kings with the clubs that were not excluded. Then the 3 could only be included with the 2 red threes, the 2 red kings, and the K:clubs:.

Quote:
(so if i'm right all the Kings-case are overcounted) ( and if we are to count the 2pair and trip non-badugis, then globally multiplying by 12 would seem suspect)
I don't think we grouped quite the same. Another time I might have done this a different way. The method I chose was rather tedious.

It bothers me when I get a different result from math experts.

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Old 11-13-2016, 05:21 PM   #12
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Re: Badugi Probability Question

Quote:
Originally Posted by whosnext View Post

12*C([25-(X+Y)],2) + 60*(13-Y) + 24(12-X) + 60 four-card hands that give a 2-card XY.

Edit: From the above, summing over all the possible X-Y combos yields 99,060 2-card badugis.
The above formula should yield a much higher total of overall combos than the original one you posted right? Or am I missing something (quite possible)?

I think the original was : 12*C([25-(X+Y)],2) + 7*(13-Y) + 2(12-X) + 15 four-card hands that give a 2-card XY.
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Old 11-14-2016, 05:04 AM   #13
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Re: Badugi Probability Question

Quote:
Originally Posted by piranha View Post
The above formula should yield a much higher total of overall combos than the original one you posted right? Or am I missing something (quite possible)?

I think the original was : 12*C([25-(X+Y)],2) + 7*(13-Y) + 2(12-X) + 15 four-card hands that give a 2-card XY.
You are not noting the difference in the placement of parentheses.

Last edited by whosnext; 11-16-2016 at 05:32 PM.
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Old 11-14-2016, 05:22 AM   #14
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Re: Badugi Probability Question

Just to be clear, there is no doubt as to how many total N-card badugis are possible for N=1,2,3,4. This is easily obtainable by high school math and is well-known and has been reported in many places.

A simple table of tallies of all four-card hands should suffice.

Ranks Suits Number Badugi Class Comment
1
1
0
 
1
2
0
 
1
3
0
 
1
4
13
1
2
1
0
 
2
2
468
2
2
3
3,744
2
2
4
1,092
2
3
1
0
 
3
2
20,592
2
3
3
10,296
2
Unpaired cards are suited
3
3
41,184
3
Unpaired cards are not suited
3
4
10,296
3
4
1
2,860
1
4
2
60,060
2
4
3
102,960
3
4
4
17,160
4
 

Summing the cases immediately shows that there are a total of:

2,873 1-card badugi hands
96,252 2-card badugi hands
154,440 3-card badugi hands
17,160 4-card badugi hands

The only challenge raised in this thread is to further classify the 96,252 2-card badugis into their 2-card buckets. As readily admitted to above, the formula given in post #10 is not quite correct as it adds up to 99,060 hands.

Last edited by whosnext; 11-16-2016 at 05:32 PM.
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Old 11-14-2016, 01:41 PM   #15
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Re: Badugi Probability Question

Quote:
Originally Posted by whosnext View Post
I am honestly at a loss for words. If I said anything more, I would probably have to ban myself for inappropriate language.
I'm so confused, my last post wasn't meant as being sarcastic at all.

What am I missing? If you want to call me an idiot or any of your other favorite derogatory adjectives, have at it.

The original formula I used from you first post was:

12 * [C([25-(X+Y)],2) + 6(13-Y) + 3(12-X) + 15] four-card hands that give a 2-card XY

I came up with 114,892 combos when I summed all the 2 card hands using this.

You came up with 118,716 combos.

The last formula you listed that you stated got you a lot closer to the real number is:

12*C([25-(X+Y)],2) + 60*(13-Y) + 24(12-X) + 60 four-card hands that give a 2-card XY.

What am I missing here. The choose function has stayed the same right? And then everything else has increased I believe so we should be getting a higher number of overall combos than the initial 118,716 you got.

I'm sure I'm being an idiot and feel free to tell me. I'm just confused.

Last edited by piranha; 11-14-2016 at 01:51 PM.
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Old 11-16-2016, 06:17 PM   #16
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Re: Badugi Probability Question

Here is the correct approach. Break down the cases as shown in post #14. For one, this is simpler. And two, it guarantees the sum of the results will yield the correct tally!

How many ways are there to make a 2-card XY badugi in a four-card hand? In my derivation which follows, I have taken X<Y even though Badugi hands are evaluated from high to low (of course it does not matter since you get the same result either way).

Case 22: 2 Ranks & 2 Suits
Clearly the number of XY hands is the same for all XY. Easy to see that it is C(4,2)=6.

Case 23: 2 Ranks & 3 Suits
Clearly the number of XY hands is the same for all XY. Straightforward to show that it is C(4,2)*2*2*2=48.

Case 24: 2 Ranks & 4 Suits
Clearly the number of XY hands is the same for all XY. Straightforward to show that it is 2^4-2=14.

Case 32A: 3 Ranks & 2 Suits where suit counts are [3,1]
Case 32AL: low rank is paired
Easy to show that this is 12*(13-Y)

Case 32AM: middle rank is paired
Easy to show that this is 12*(13-Y)

Case 32AH: high rank is paired
Easy to show that this is 12*(Y-X-1)

Case 32B: 3 Ranks & 2 Suits where suit counts are [2,2]
There are six subcases to consider depending upon which ranks are paired and the order of the other two cards. In each case, the total number of hands is 6*(13-Y). So the total over all six subcases is 36*(13-Y).

Case 33: 3 Ranks & 3 Suits where the unpaired cards are in the same suit
Case 33A: lower suited card is less than pair
Easy to show that there are 12*(12-X).

Case 33B: lower suited card is higher than pair
Easy to show that there are 12*(13-Y).

Case 42A: 4 Ranks & 2 Suits where suit counts are [3,1]
Case 42A1: lone suited card is less than all three others
Easy to show that this is 6*(13-Y)*(12-Y)

Case 42A2: lone suited card is less than two of the others
Easy to show that this is 6*(13-Y)*(12-Y)

Case 42A3: lone suited card is less than one of the others
Easy to show that this is 12*(Y-X-1)*(13-Y)

Case 42A4: lone suited card is less than none of the others
Easy to show that this is 6*(Y-X-1)*(Y-X-2).
Of course, don't allow this term to be negative (i.e., zero it out if it is negative).

Case 42B: 4 Ranks & 2 Suits where the suit counts are [2,2]
Case 42B1: Domination
Easy to show that it is 12*(Y-X-1)*(13-Y)

Case 42B2: Interwoven
Easy to show that it is 6*(13-Y)*(12-Y)

Case 42B3: Sandwich
Easy to show that it is 6*(13-Y)*(12-Y)

Summing all of these cases yields:

68 + 72*(13-Y) + 12*(12-X) + 12*(Y-X-1) + 24*(13-Y)*(12-Y) + 24*(Y-X-1)*(13-Y) + 6*(Y-X-1)*(Y-X-2) where you zero out this last term if it is negative.

There are undoubtedly ways to "simplify" the above expression by combining a whole bunch of terms. For our purposes, I feel it is better left as is.

Let me know if you have any questions.

Last edited by whosnext; 11-16-2016 at 06:27 PM.
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Old 11-19-2016, 08:57 PM   #17
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Re: Badugi Probability Question

Quote:
Originally Posted by whosnext View Post
Here is the correct approach. Break down the cases as shown in post #14. For one, this is simpler. And two, it guarantees the sum of the results will yield the correct tally!

How many ways are there to make a 2-card XY badugi in a four-card hand? In my derivation which follows, I have taken X<Y even though Badugi hands are evaluated from high to low (of course it does not matter since you get the same result either way).

Case 22: 2 Ranks & 2 Suits
Clearly the number of XY hands is the same for all XY. Easy to see that it is C(4,2)=6.

Case 23: 2 Ranks & 3 Suits
Clearly the number of XY hands is the same for all XY. Straightforward to show that it is C(4,2)*2*2*2=48.

Case 24: 2 Ranks & 4 Suits
Clearly the number of XY hands is the same for all XY. Straightforward to show that it is 2^4-2=14.

Case 32A: 3 Ranks & 2 Suits where suit counts are [3,1]
Case 32AL: low rank is paired
Easy to show that this is 12*(13-Y)

Case 32AM: middle rank is paired
Easy to show that this is 12*(13-Y)

Case 32AH: high rank is paired
Easy to show that this is 12*(Y-X-1)

Case 32B: 3 Ranks & 2 Suits where suit counts are [2,2]
There are six subcases to consider depending upon which ranks are paired and the order of the other two cards. In each case, the total number of hands is 6*(13-Y). So the total over all six subcases is 36*(13-Y).

Case 33: 3 Ranks & 3 Suits where the unpaired cards are in the same suit
Case 33A: lower suited card is less than pair
Easy to show that there are 12*(12-X).

Case 33B: lower suited card is higher than pair
Easy to show that there are 12*(13-Y).

Case 42A: 4 Ranks & 2 Suits where suit counts are [3,1]
Case 42A1: lone suited card is less than all three others
Easy to show that this is 6*(13-Y)*(12-Y)

Case 42A2: lone suited card is less than two of the others
Easy to show that this is 6*(13-Y)*(12-Y)

Case 42A3: lone suited card is less than one of the others
Easy to show that this is 12*(Y-X-1)*(13-Y)

Case 42A4: lone suited card is less than none of the others
Easy to show that this is 6*(Y-X-1)*(Y-X-2).
Of course, don't allow this term to be negative (i.e., zero it out if it is negative).

Case 42B: 4 Ranks & 2 Suits where the suit counts are [2,2]
Case 42B1: Domination
Easy to show that it is 12*(Y-X-1)*(13-Y)

Case 42B2: Interwoven
Easy to show that it is 6*(13-Y)*(12-Y)

Case 42B3: Sandwich
Easy to show that it is 6*(13-Y)*(12-Y)

Summing all of these cases yields:

68 + 72*(13-Y) + 12*(12-X) + 12*(Y-X-1) + 24*(13-Y)*(12-Y) + 24*(Y-X-1)*(13-Y) + 6*(Y-X-1)*(Y-X-2) where you zero out this last term if it is negative.

There are undoubtedly ways to "simplify" the above expression by combining a whole bunch of terms. For our purposes, I feel it is better left as is.

Let me know if you have any questions.
Bingo, that's it! I can't thank you enough.
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Old 07-19-2017, 04:11 PM   #18
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Re: Badugi Probability Question

This weekend I was part of a group that was discussing the great game of badugi and this question arose. So I thought of this thread again.

If anyone is interested, here is a simplified formula for the number of 2-card badugis based upon the value of the two cards. Here we set Y to be the value of the high card and X to be the value of the low card where cards have the following values (King=13, Queen=12, Jack=11, Ten=10, ..., Deuce=2, Ace=1).

Number of YX 2-card badugi hands = 4580 - 342Y - 318X + 6YY + 12XY + 6XX.

To give an example, there are 440 J8 2-card badugi hands where we plugged in J=11 and 8=8 into the formula above for Y and X respectively.

Sum of these over all possible YX cases (14>Y>X>0) is 96,252, the total number of possible 2-card badugi hands.
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Old 07-20-2017, 06:54 PM   #19
piranha
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Re: Badugi Probability Question

Totally coincidental that I logged on today with a 27 question and you posted in this old thread. Thanks a lot for this, I don't know how you guys figure this stuff out. I have the first formula along with all the bi combos in a spreadsheet still - I'll update the formula.
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Old 09-19-2017, 01:54 PM   #20
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Re: Badugi Probability Question

brain explode
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