First I wanted to determine the probability of a pat badugi beating a perfect one-card-draw. Here's how I approached it:
KQJT vs 32AX
32A has 9 outs and will draw 3 times. In the FTOP arena, 52-8=44 cards are unknown. So the probability that 32A out draws KQJT is:

and the perfect one-card draw is 62.8% to suck out. Please let me know if this is wrong. I proceed on to a more difficult problem assuming this is correct. I always heard that any pat badugi is a favorite over any one-card draw, and this appears to be a contradiction. Using the same method I determined that a pat JT98 badugi is a 51.1% favorite over 32Ax.

Ill move on to the more difficult problem in a bit, since i have a feeling there's a mistake here. Confirm?

I think if my approach is correct, and we imagine a situation where 32A can draw indefinitely, he could potentially draw 44-9=35 times and still not hit an out. Consequently he should have to draw 36 times for his probability of hitting an out to be 1, but this doesnt seem to be the case. If we sum 5 terms we're already over 1, so I know this has to be wrong.

Is this the birthday problem? Do i need to be using the binomial distribution?

It's wrong. You would have to multiply the 9/43 by the probability that you miss on the first card, and multiply the 9/42 by the probability that you miss on the first 2 cards. It is easier to take 1 minus the probability of missing on all 3 cards:

1 - (35/44 * 34/43 * 33/42)

=~ 50.6%

Still a contradiction, but closer to even.


For the 7 out case, 32Ax draws out about 41.3%, making JT98 about a 58.7% favorite.

1 - (37/44 * 36/43 * 35/42)

=~ 41.3%

^thanks. ill take another crack at the more difficult problem again, now that this is fixed.

Now lets put a pat JT98 against two perfect 1-card-draws, that dont block the others outs, for example Player B has 32A and Player C has 32A. Both draws have 7 outs from 40 unseen cards. I feel like predraw we can consider the events B (player B misses his draw) and C (player C misses his draw) independent, so the event B-intersect-C is just P(B)P(C), or

(33/40 * 32/39 * 31/38)^2 =~0.305

So JT98 is too weak against two 7 out hands that dont block each other.

*Note, if Players B and C are looking for the same suit, (ie B has 32A and C has 42A and they are both looking for the same 7 spades) my JT98 has the same 0.305 equity. This is suspect to me, but interesting if true.

If they block each other once, and both have 6 outs each, JT98 is 0.367 and has enough equity.

thoughts?

They are not independent, though this is close to the exact answer of about 0.3085. When player B draws no spades, it increases the probability that he draws clubs, which decreases the probability that player C draws clubs.

In the following, "spades" refers to the 7 spade outs that player B needs, and "clubs" refers to the 7 club outs that player C needs.

For the exact answer, we must separately compute the probability that player B draws 0,1,2, and 3 clubs, and multiply each of these by the probability that player C draws no clubs given that player B drew that many clubs.

P(B draws no spades AND C draws no clubs) =

P(B draws 0 spades, 0 clubs)*P(C draws 0 clubs | B draws 0 spades, 0 clubs)
+
P(B draws 0 spades, 1 club)*P(C draws 0 clubs | B draws 0 spades, 1 clubs)
+
P(B draws 0 spades, 2 clubs)*P(C draws 0 clubs | B draws 0 spades, 2 clubs)
+
P(B draws 0 spades, 3 clubs)*P(C draws 0 clubs | B draws 0 spades, 3 clubs)


P(B draws no spades AND C draws no clubs) =

(26/40 * 25/39 * 24/38) * (30/37 * 29/36 * 28/35) +
(7/40 * 26/39 * 25/38 * 3) * (31/37 * 30/36 * 29/35) +
(7/40 * 6/39 * 26/38 * 3) * (32/37 * 31/36 * 30/35) +
(7/40 * 6/39 * 5/38) * (33/37 * 32/36 * 31/35)

=~ 0.3085


No, the probability that they both miss is

33/40 * 32/39 * 31/38 * 30/37 * 29/36 * 28/35

=~ 0.289


Try this one, like the first one.

Thanks for your help Bruce. Im hoping to get a better understanding with some of this stuff so I have some foder while I attempt to learn Mathematica.

Quick question on your above conditional probability calculation: since player B might see a club after his first draw, and this affects the probability of player C making his hand, shouldnt we also have some term describing those times when player B is dealt (pre draw) a club? We are assuming 40 unseen cards, but it looks like we've restricted the players first discards to be in the set {,}. amirite?

edit... i guess the 2 would be in this set too, so discards must be all in the 42 cards we havnt explicitly set, all in the 42 cards we havnt explicitly set, and all and not in the set of 7 outs for either player.

That's why I said that "clubs" in this formula refers only to the 7 club outs for player C, and "spades" refers only to the 7 spade outs for player B. Since we are given that each player has 7 outs, these must be in the deck. Player B's first discard could be the K, Q, A, or 2 which are not outs; or the K, Q, 2, or 3 which are not outs; in addition to one of 10 hearts and 10 diamonds. Same for Player C.

It is really up to you whether you want to assume that all 14 outs are in the deck as I did, or to allow each player's fourth card to be a possible out for the other player. In the latter case, you would just change the initial number of cards remaining from 40 to 42.

Another thing to note is that: If you have something like KQJT and you aren't drawing, when he misses the third draw, he is never paying when his A23X misses. So he will be getting more value from his made draw than you get when he misses. Depending on position, this can cost us more or less, and depending whether or not he knows what he's doing and simply calls the earlier streets, but this makes KQJT further behind A23X (not enough to fold on any given street). Though I'm sure you were aware of this, food for thought.