How do I figure out the probability of being dealt a tri?
Of those combos, how do I figure the probability of being dealt a tri 7 or better?

What is the formula/logic behind the math here?

For a badugi I know it’s 13x12x11x10, but why? Is the formula/logic

13C4 x 4C1 x 3C1 x 2C1?
(13 cards of a suit, choose 4 of them) x
(4 suits, choose 1) x
(3 suits, choose 1) x
(2 suits, choose 1)

Prob any tri is 57%

Prob 7 or better low tri is 12.7%

I calculated many of these things around 5 years ago and had Buzz check my work (RIP buddy), may be able to provide some other guidance later but I’m at the table and my backup is elsewhere

Thanks as always

Nice and very helpful website btw

Thanks man

Actually I’m no longer involved with that website so there will be no more material added. I may be starting another one though in a few months.

That one article on Badugi Part 2 will give you the guidance you are looking for in regards to tris

In no way did I want to link the site anymore more

I went in with a guy who created the site and told him I did not want to create any potential training vids if I’m only getting 50% of revenues from them. It’s a ton of work and I feel like I’m emptying the vault. So he responded by shutting me out of site and is no longer sharing affiliate revenue with me.

Being dealt a pat badugi means being dealt 4 unpaired unsuited cards. The first card is arbitrary so has no probabilistic bearing, so it can be any of the 52 cards in a standard deck. Probability of having a badugi-eligible at this point is 52/52 = 1. The second card can't be the same suit or rank as the first one so there are 12 cards per remaining suit = 36 cards it can be for a probability of 36/51 (51 not 52 since 1 card has already been dealt without replacement). This is the chance of having a hand capable of making a badugi having been dealt 2 cards. Conditional on that, the 3rd card can be either of the 11 cards of the remaining 2 suits, so a 22/50 chance of having a 3 card dugi after 3 cards given the first 2 cards were a 2 card dugi. The final card has to be one of the 10 non-pairing cards in the remaining suit, for a conditional probability given a 3 card dugi of 10/49 to make a 4 card pat badugi. To determine the total probability, you multiply together the conditional probabilities. The total probability therefore is:
(36/51)(22/50)(10/49) ~ 6.3%

I just skipped to the end and the conclusion is wrong, you are dealt a badugi around 6% of the time..

I realized as soon as I submitted my last post that I made grave errors. I have since edited the post you quoted. In the interest of intellectual honesty, it is for the best that my mistake is immortalized in your quote but I felt it important to edit the original lest someone take my mistake for truth. I do arrive at the 6% probability you advertise in my corrected explanation.

Ok great! Lol

The analytical calculation of getting a 3card dugi (or better) in 4 cards is a lot more involved but I just did it anyways

So again the first card is arbitrary.

The 2nd card will be a 2c dugi with probability (w.p.) 36/51 as before. The 3rd card will make a 3c dugi with conditional probability (w.c.p.) 22/50. The 3rd card will not make a 3c dugi w.c.p. 28/50 but for the next step it's important we know what sort of 2c dugi it is. It could be XXsYs (e.g. AA2), XsXoYo (e.g. AA2), or XoYsZs (e.g. A22). Given the 2c dugi we're assuming here of XoYs (e.g. A2), XXsYs occurs w.c.p. 2/50 (onsuit pairing card), XsXoYo occurs w.c.p. 4/50 (offsuit pairing card), and XoYsZs occurs w.c.p. 22/50 (onsuit non-pairing card). Given XXsYs, you make a tri w.c.p. 22/49 (e.g. must be a nonpairing or if you have AA2). Given XsXoYo, you make a tri w.c.p. 11/49 (e.g. must be a nonpairing if you have AA2). Given XoYsZs, you make a tri w.c.p. 24/49 (e.g. must be a or given A23 but can pair one of the spades).

2 dealt cards will not be a 2dugi w.p. 15/51, but again it depends what type of 2dugi it is for the next step. The 2 cards will be suited w.p. 12/51 or paired w.p. 3/51. If suited, then the 3rd card will make a 2dugi w.c.p. 39/50 (any unsuited card works, even if it pairs). If paired, then the 3rd card will make a 2dugi w.c.p. 48/50 (any non-pairing card works). However, it again matters what type of 3 card is made. If the first 2 cards were suited XsYs (e.g. A2), then the 3rd card either pairs and makes XXsYs w.c.p. 6/50 or doesn’t pair and makes XsYsZo w.c.p. 33/50. If the first 2 cards were paired XX (e.g. AA), then the 3rd card either matches an existing suit and makes XXsYs w.c.p. 24/50 or is a new suit and makes XsXoYo w.c.p. 24/50. The probabilities that the 4th card makes a 3dugi is as before conditional on these different types of 3 card 2dugis.

Putting it all together, the probability of being dealt a 3 card dug or better (so including the probability of being dealt a pat badugi) is:
(36/51)[(22/50)+(2/50)(22/49)+(4/50)(11/49)+(22/50)(24/49)]+(12/51)[(6/50)(22/49)+(33/50)(24/49)]+(3/51)(24/50)[(22/49)+(11/49)] ~ 59.6%

This was a complicated calculation so I welcome scrutiny and critique.

P.S. If you want the probability of making a 3 card dugi exactly that isn’t a 4 card badugi, then just multiply the probability of your first 3 cards being a tri times the probability that the 4th card pairs or is of the same suit, which is 39/49. This probability is then
(36/51)[(22/50)(39/49)+(2/50)(22/49)+(4/50)(11/49)+(22/50)(24/49)]+(12/51)[(6/50)(22/49)+(33/50)(24/49)]+(3/51)(24/50)[(22/49)+(11/49)] ~ 53.2%

I’ll give some critique but no scrutiny, probability of a tri is 57%

Is 57% strictly a tri or a "tri or better" (tri+dugi)?

Just tri

Did you calculate this figure through simulation?

The reason I gave my analytical calculations is because OP asked specifically how these values were calculated, not what they were.

No exact with help of Buzz and number also number is published in other sources

Yeah I get why you did it, just letting you know there is almost certainly a mistake somewhere, the calcs are not easy

Ignore N choose k language for a moment and let's run through some examples.

Example
If you pick 4 cards from the deck without replacement and we care about order then there are 52 x 51 x 50 x 49 cards. Since we don't care about order we note we have counted the same card (Ah Ad Kd Js as Ad Ah Kd Js, Ah Kd Ad Js etc., same exact hand w/ different order) multiple times.

How many times? This is the number of ways of switching the 4 cards which is 4! = 4 x 3 x 2 x 1. The logic being you have 4 cards to choose from for "position 1", 3 cards for the second position, 2 for the third forcing the last card.

So number of cards is (52 x 51 x 50 x 49)/(4 x 3 x 2 x 1) [or 52!/(50!4!) or 52 choose 4]

Lesson We can pretend we care about order and then divide out number of ways of ordering.

Number of badugis
You could use this to directly get
#number_of_badugis (52 x 36 x 22 x 10)/(4 x 3 x 2 x 1) = 13x12x11x10

However, let's pretend for a moment that we don't care about suit i.e A234 = As 2c 3d 4h = As 2c 3d 4c. Then there are 13x12x11x10/4! distinct badugis.
Since we do care about suit we multiply by number of ways of assigning 4 suits to the 4 cards of different rank (as hand is badugi).

#number_of_badugis_no_suits is 13x12x11x10/4!
#number_of_badugis is (13x12x11x10/4!) x 4! = 13x12x11x10

Lesson The reason this neatly cancels is the ways of ordering 4 cards and ways of assigning 4 cards of different rank a different suit cancel.

This follows logic of my answer to

Method

• Calculate number of 3 cards with different ranks with different suits
• Add remaining 49 cards but split in 4 cases
• For each case we'll work out by what multiple we are repeating the count

Step 1 - number of 3 cards with different ranks with different suits
Let's lock down the 3 suits first. There 4C3 = 4 ways for this

#number_of_3_cards = (4 x (13*3) x (12*2) X (11*1))/3! = 4 x 13 x 12 x 11

Step 2

Fact 1: any 4-card hand with at least a tri can be made by adding a card to 3 cards with different rank and suit.

So if we add all possible cards to the set of 3-card hands we will have at least the set of 4-card hands with a tri at least. There may however be multiple counting as I described.

Case 1 - Same suit, not a pair
There are 30 (10 ranks, 3 suits)
The hand will look like
Ax By Cz Dx

The count will be repeated twice:
-- By Cz Dx
Ax By Cz --

Case 2 - Same suit, pair
There are 6 (3 ranks, 2 suits)
The hand will look like
Ax By Cz Ax

There's no repeat of count

Case 3 - Different suit, pair
There are 3 (3 ranks, 1 suit)
The hand will look like
Ax By Cz Aw

The count will be repeated twice:
-- By Cz Aw
Ax By Cz --

Case 4 - Different suit, non-pair
There are 10 (10 ranks, 1 suit)
The hand will look like
Ax By Cz Dw

The count will be repeated 4 times:
-- By Cz Dw
Ax -- Cz Dw
Ax By -- Dw
Ax By Cz --


Final calculation

#card_with_at_least_tri = 4 x 13 x 12 x 11 x (30/2 + 6 + 3/2 + 10/4)
= 4 x 13 x 12 x 11 x 25 = 171600

Odds = 171600 / 270725 = 63.4%

Case 4 is badugi so skip this to get exactly a tri:

#card_with_exactly_tri = 4 x 13 x 12 x 11 x (30/2 + 6 + 3/2)
= 4 x 13 x 12 x 11 x 22.5 = 154440

Odds = 154440 / 270725 = 57.0%

thanks. i'll read all posts over as i hate reading numbers

i attempted this using pen and paper (inspired by zoogenhiem), and not using the n choose blah notations...

here's what i got:



seems like im off by a fraction or so, so i prolly have arithmetic mistakes somewhere or something is wrong with one of my logics

^ah, the top line is wrong as i did 33/50. it's 33/49.

the sum of the total is 57.0468 once i change that!
https://docs.google.com/spreadsheets...it?usp=sharing

Same method as before but non-pair cases split in two

#3_card_hands_7_or_better = 4 x 7 x 6 x 5

4 cases

Case 1 - Same suit, not a pair, D <= 7
There are 12 (4 ranks, 3 suits)
The hand will look like
Ax By Cz Dx

The count will be repeated twice:
-- By Cz Dx
Ax By Cz --

Case 2 - Same suit, not a pair, D > 7
There are 18 (6 ranks, 3 suits)
The hand will look like
Ax By Cz Dx

The count will not be repeated:-
-- By Cz Dx This does not count as D>7

Case 3 - Same suit, pair
There are 6 (3 ranks, 2 suits)
The hand will look like
Ax By Cz Ay

There's no repeat of count

Case 4 - Different suit, pair
There are 3 (3 ranks, 1 suit)
The hand will look like
Ax By Cz Aw

The count will be repeated twice:
-- By Cz Aw
Ax By Cz --

Case 5 -Different suit, non-pair
This leads to badugi - skip

Final calculation


#hands_exactly_tri_7_or_better = 4 x 7 x 6 x 5 x (12/2+18+6+3/2)
= 4 x 7 x 6 x 5 x 31.5 = 26460

Odds = 26460 / 270725 = 9.77%

Similar calcs give:-

Number of unique hands: 270725
Number of badugi hands: 17160
Number of tri hands (any): 154440
Number of tri hands (2 or better): 0
Number of tri hands (3 or better): 900
Number of tri hands (4 or better): 3456
Number of tri hands (5 or better): 8280
Number of tri hands (6 or better): 15840
Number of tri hands (7 or better): 26460
Number of tri hands (8 or better): 40320
Number of tri hands (9 or better): 57456
Number of tri hands (T or better): 77760
Number of tri hands (J or better): 100980
Number of tri hands (Q or better): 126720
Number of tri hands (K or better): 154440

^thanks

i'll use this as a guide and try to work out all the subset of tris myself using whatever method im comfortable with and see if i arrive at the same answers

i literally cant find any source on the internet in 2019 that shows the distribution of starting hands for games such as badeucy, archie, whatever, so i think it is obviously very valuable to be able to solve for these things!

Following on from my logic and simplifying algebra..

#tris_n_or_better = 4 x n x (n-1) x (n-2) x (42-3*(n/2))

odds = 4 x n x (n-1) x (n-2) x (42-3*(n/2)) / 270725

Ditto, but never fear, Nxia is here

I tentatively suggest this figure is wrong, unless I'm misunderstanding what you mean by this

Reverse engineering 12.7% would suggest there's 34440 = 4 x 7 x 6 x 5 x 41 hands that are exactly a tri and the tri is 7 or better. Unless my program is also wrong, with separate logic to my post

i agree with ur answers nxia!

https://docs.google.com/spreadsheets...it?usp=sharing