In this year's WSOP $50k Poker Players Championship (PPC), there was a hand of Triple-Draw when four-handed at the final table that is one of the worst bad beats ever witnessed on a poker table. Bryce Yockey was eliminated by Josh Arieh in the most peculiar hand of Triple Draw I have ever seen when Yockey was reputed to be around a 99.9% favorite in the hand.

Here is a link to a description of the hand:

https://www.pokercentral.com/article...avorite-loses/

I want to derive my own estimate of the equities in the hand. First, I don't really believe Yockey was a 99.9% favorite. Second, I'd like to try my hand at coming up with a decent estimate.

There are several things we must consider. First, what are the dead cards? The PokerGo broadcast showed Yockey and Arieh's starting hands (of course) and also John Esposito's starting hand (Esposito folded AJ553). However, we don't know Philip Hui's starting hand, though we do know that he folded to Yockey's two-bet fairly quickly. Yockey was so short-stacked relative to the limits that all of his money was going into the pot on this hand, and Arieh definitely had the "pot odds" to continue in the hand.

Second, since this is Triple-Draw, we need some reasonable approximation of Arieh's drawing strategy in the hand in order to derive equity estimates. Yockey was essentially all-in before the first draw and patted throughout the hand with #2 (76432), so there is nothing we need consider regarding betting or folding for either player during the hand.

My first assumption is that Arieh puts Yockey on 97432. Of course, you don't really put an opponent on a single hand, but I assume that Arieh will make his draws to maximize his chance of achieving 97432 or better.

My second assumption is that we will treat Hui's cards as unknowns. He may well have had a 2, 4, or 7 (the key cards that Arieh seeks), but he might just as well not had any of those cards. I think the safest assumption is to simply ignore Hui's cards.

I have identified 10 different paths that give Arieh a better hand than Yockey. Of course, with Yockey holding #2 (76432) the only hand that Arieh can make to beat Yockey is #1 (75432).

Remember, Arieh started with AQ653 (he will clearly discard AQ on the first round). So upon first glance it appears that Arieh (with 653) can never make a wheel (75432). That is, on first glance it appears Arieh is drawing dead!

But as Nick Schulman on the broadcast pointed out, if Arieh catches a 24 giving him 65432, he would discard the 6, and if he caught a 7 he would win with a 75432! Nick proved prescient for that is essentially what Arieh did. He caught a 2 on his first draw, a 4 on his second draw (giving him an unusable 65432 straight), pitched the 6 and caught a 7 giving him a 75432 and breaking Yockey's heart via an unbelievable bad beat!!

Okay, let's walk through all the ways that I think Arieh could win the hand:


Case 1: Catch 2X on first draw (X is A,K,Q,J,T,6,5,3,2), discard X, catch a 4 on the second draw, discard the 6, catch a 7 on the third draw.

Prob = [C(3,1)*C(20,1)+C(3,2)]/C(37,2) * C(3,1)/C(35,1) * C(3,1)/C(34,1)
= 567 / 792,540


Case 2: Catch 4X on first draw (X is A,K,Q,J,T,6,5,4,3), discard X, catch a 2 on the second draw, discard the 6, catch a 7 on the third draw.

Prob = [C(3,1)*C(20,1)+C(3,2)]/C(37,2) * C(3,1)/C(35,1) * C(3,1)/C(34,1)
= 567 / 792,540


Case 3: Catch 24 on first draw, discard the 6, catch a 7 on the second draw

Prob = C(3,1)*C(3,1)/C(37,2) * C(3,1)/C(35,1)
= 27 / 23,310


Case 4: Catch 24 on first draw, discard the 6, catch a brick on the second draw (A,K,Q,J,T,6,5,4,3,2), discard brick, catch a 7 on the third draw.

Prob = C(3,1)*C(3,1)/C(37,2) * C(24,1)/C(35,1) * C(3,1)/C(34,1)
= 648 / 792,540


Case 5: Catch 7X on first draw (X is A,K,Q,J,T,9,7,6,5,3), discard X, catch a 4 on the second draw, discard the 6, catch a 2 on the third draw.

Prob = [C(3,1)*C(24,1)+C(3,2)]/C(37,2) * C(3,1)/C(35,1) * C(3,1)/C(34,1)
= 675 / 792,540


Case 6: Catch 4X on first draw (X is A,K,Q,J,T,6,5,4,3), discard X, catch a 7 on the second draw, discard the 6, catch a 2 on the third draw.

Prob = [C(3,1)*C(20,1)+C(3,2)]/C(37,2) * C(3,1)/C(35,1) * C(3,1)/C(34,1)
= 567 / 792,540


Case 7: Catch 47 on first draw, discard the 6, catch a 2 on the second draw.

Prob = C(3,1)*C(3,1)/C(37,2) * C(3,1)/C(35,1)
= 27 / 23,310


Case 8: Catch 47 on first draw, discard the 6, catch a brick on the second draw (A,K,Q,J,T,9,7,6,5,4,3), discard brick, catch a 2 on the third draw.

Prob = C(3,1)*C(3,1)/C(37,2) * C(28,1)/C(35,1) * C(3,1)/C(34,1)
= 756 / 792,540


Case 9: Catch two bricks on first draw (A,K,Q,J,T,6,5,3), discard both bricks, catch 24 on the second draw, discard the 6, and catch a 7 on the third draw.

Prob = C(20,2)/C(37,2) * C(3,1)*C(3,1)/C(35,2) * C(3,1)/C(33,1)
= 5,130 / 13,076,910


Case 10: Catch two bricks on first draw (A,K,Q,J,T,6,5,3), discard both bricks, catch 47 on the second draw, discard the 6, and catch a 2 on the third draw.

Prob = C(20,2)/C(37,2) * C(3,1)*C(3,1)/C(35,2) * C(3,1)/C(33,1)
= 5,130 / 13,076,910


TOTAL PROB FROM ALL 10 CASES:
= 102,924 / 13,076,910
= 0.79%

Since there is no way Arieh can make the same hand as Yockey (no ties), the remaining probability is, of course, the probability of Yockey winning the hand:

= 100.00% - 0.79%
= 99.21%

Thus, using my assumptions (which I think are reasonable), my itemization of ways Arieh could win the hand (which I think are correct), and the combinatoric calculations of each case (which I hope are correct), I find that Yockey was a 99.21% favorite.

Different assumptions can be made regarding Hui's cards and Arieh's drawing strategy which would yield different equity numbers. But I'd be surprised if they approach the 99.8% or 99.9% figures that I have seen in the media.

As I always say, it is possible that I have made errors in logic or calculations somewhere along the line which I'd be happy to have pointed out.

I don't know, but 99.21% seems low. Like, something tells me it should be higher, but I can't find a mistake in your math.