Drawing against made Badugi
 

I am trying to get an estimate of what pot odds one would need to draw against a made Badugi.

For example, in a big multi-way pot with A25. We know how many cards that we discarded and that one villian is pat so he has one of the suit that we need and three that we don't.

There are 12 of our suit left. How would we discount the outs that may or may not be good?

Can we really even start with 12 outs if there are several other people drawing one and there is 3/4 of a chance that they have one of our outs?

Re: Drawing against made Badugi
 

[ QUOTE ]
I am trying to get an estimate of what pot odds one would need to draw against a made Badugi.

For example, in a big multi-way pot with A25. We know how many cards that we discarded and that one villian is pat so he has one of the suit that we need and three that we don't.

There are 12 of our suit left. How would we discount the outs that may or may not be good?

Can we really even start with 12 outs if there are several other people drawing one and there is 3/4 of a chance that they have one of our outs?

[/ QUOTE ]

There are 13 cards of the suit you're missing. But it is not the case that your pat opponent necessarily has one of your outs. The A, 2, and 5 don't make you a badugi (so you only have 10 outs to a badugi to start) but your opponent could easily hold one of these and thus not decrease your outs at all. On the other hand, if all four of his cards are 6 or higher then he must hold one of your outs. So the question of how many live outs you have is connected with his hand range in a nontrivial way.

If your opponent has an arbitrary dealt Badugi, his distribution is very different from one made by drawing to a good starter. In the former case it is very heavily skewed toward the high end (as described in last month's 2+2 magazine article on Badugi.)

A while back I tried figuring out what the distribution looked like if you started with three to an 8 badugi or better, and made it in one draw.

The results of this exercise, assuming all combinations were equally likely, are as follows:

K8-K5xx (10 outs): 9.3%
K4-Q5 (9 outs): 10%
Q4-J5 (8 outs): 10%
J4-T5 (7 outs): 10%
T4-95 (6 outs): 10%
94-85 (5 outs): 22.9%
84-75 (4 outs): 14.3%
74-65 (3 outs): 7.1%
6432-542A (2 outs): 5%
532A (1 out): 0.7%
432A (0 outs): 0.7%

Let's say that 3/4 of the time one of your outs is held by the other player, which isn't too bad of an approximation. So there is about 2.3% chance of having 10 outs, 9.5% chance of 9 outs, etc., down to about 1.2% chance of having zero outs.

This gives you a weighted average of 5.2 outs. 5.2 outs in 52-4-4 = 44 unknown cards is 11.8%, or odds of about 7.5 to 1.

Now, if your opponent originally held something stronger than a three-card eight, then the distribution will be worse for you and the odds higher. (But note that the better your opponent's hand is the more likely he is to hold a card from your needed suit that would pair you.)