If you are dealt 5 cards , what is the probability of having a badugi in your hand?

0%

By having badugi , i meant if your hand contains a 4 card badugi . Ex: Ac ks Qh Jd 2s is considered having a badugi.

Not sure if right and maybe more complex than I think?

There are 52C5 combos total so 2,598,960 distinct starting hands

There's 13 x 12 x 11 x 10 combos of badugis.
There's then 47 cards left in the deck, so we take the product of the above.

13x12x11x10x47 = 806,520.

806,520/2,598,960 = 31.03%

I'm not good enough at combinatorics to say for sure, but I worry the above method doesn't factor in the fact that it doesn't have to be the first 4 cards that are a badugi. Then again, if that were the case, I think it'd give a lot lower answer than 31% so maybe you're correct. My main concern is if you do this calculation for having a 3 card badugi or better in a 4 card hand, you get

13x12x11x48/270,725 = 30.4%

which seems too close to the answer you got and not what the 3c probability is (which is >57%)

No, there's 48.
52-4 = 48

Even so, the method
13*12*11*10*48 does not work as you are double counting...
take 2 four card badugis
(1) 3s Tc 5d Ah
(2) Tc 5d Ah 2s
Then (1)+2s = (2)+3s =
3s Tc 5d Ah 2s

If 5th card is a pair then we don't double count if it is we do.

13*12*11*10*(12 + 36/2) = 13*12*11*10*30 = 514800

Answer is 514800 / 2598960 = 19.8%

badugi is 52x36x22x10/4! which is 13x12x11x10.
tri or better follows a completely different logic, which i don't know how to do.

i do know the comparison ur attempting with 13x12x11x48 is an underestimation for tris+, because that calculation (the pattern u followed) is going to assume there are only 3 suits in a deck when there are 4.

thanks! however i don't get why the non-pairs are double counted.

the way i see it,

(1) xx 2h 3c 4d 5d
(2) As xx 3c 4d 5d
(3) As 2h xx 4d 5d
(4) As 2h 3c xx 5d
(5) As 2h 3c 4d xx

(1)+As == (2)+2h == (3)+3c == (4)+4d == (5)+5d

i don't know what to make of this... i guess i don't get why this results in exactly double counting for the 36 non-pairs for the 5th card.

is there something obvious i'm overlooking?

Sorry I should have been clearer.

In your example (1), (2) & (3) are not badugis.

Goal: we want to create all 5-card unique hands with a 4 card badugi in them so that we can calculate

#five_card_hands_w_badugi/#five_card_hands

Fact 1: any 5-card hand with a badugi can be made by adding a card to 4-card badugi.

So if we add all possible cards to the set of 4-card hands we will have at least the set of 5-card hands with a badugi. There may however be multiple counting as I described.

Fact 2: when we add a card to a badugi this will pair one and only one suit. Which leads to 2 cases...

Case 1: card added pairs the badugi
The 5 cards will look like:
Ax By Cz Dw Ay
There is one and only one 4 card badugi in this hand.

Case 2: card added does not pair the badugi
The 5 cards will look like:
Ax By Cz Dw Ex

There is exactly 2 badugis:
-- By Cz Dw Ex
Ax By Cz Dw --

12 cards pair the badugi (four cards, three suits)
The remaining 36 cards do not but exactly double counts the 5-card hands they generate.

as soon as u said this, i realize where i went wrong. thanks!! that makes a lot of sense now!