In our home game this afternoon, our resident luckbox was dealt a 5/4 badaci hand...amazingly, someone else drew a perfect ace on the third draw to also make a 5/4! So they chopped.

What are the odds of being dealt a 5/4....I've never seen it before.

Thanks.

My quick guess would be about 1 in 27072

Now we need to have A234 to be a rainbow, and the 5 can be of any suit.

There are 24 ways to arrange A234 into four suits: 4*3*2*1 = 24, and 4 ways to choose the suit for the 5. Giving a total of 96 combinations that give (A234r)5.

The total ways to choose 5 cards from 52 is 52*51*50*49*48/(5*4*3*2*1) = 2,598,960

So the probability is 96/2,598,960 = 1 in 27072.5

And I thought me getting a dealt badugi on Stars (~1 in 11,280) and getting chopped on the river was bad!

Especially bad since we play mid stakes NL...the dealt badugi was milking the table for 50 pre/1/2 draw...then over 1000 goes in on river!

We also play a three card variant, where A23 is the nuts, and you need a three card badugi to chop...this was created since we have 7-8 players usually, and everyone loves these games.

What are the odds of being dealt a three card A23, and then a A23 badugi?

Thanks.

Sigh, where the hell are these types of players at my local casino (St. Louis)? You play a 3-card variant to meet demand? That's pretty sick. The game sounds fun too, at least to me.

A23 any suits: 4^3 = 64 combos
Total combos: 52*51*50/(3*2*1) = 22,100

Probability: 64/22100 = 1 in 345.3

A23 "dugi": 4*3*2 = 24 combos
Total combos: 22100
Probability: 24/22100 = 1 in 920.8

To get this back to back in any given 2 hands, it's (1/345.3)*(1/920.8) =~ 1 in 318k.

But that assumes you only play two hands. If you play 100 hands, the likelihood jumps nearly 100x.