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drizzler · Pair Joined: 06 May 2006 Posts: 45

A few days ago I was the victim of a nasty Razz outdraw. This isn't about the bad beat (it happens all the time in Razz) but about calculating the odds of it occurring. Here's the hand:

I'm dealt [72]3 and Mr. Villain is all-in with [T4]K. This was a tournament and he was very short stacked, so rather than bring in and fold, leaving himself with almost no chips, he decided to complete the bet all-in. Hey, it's his money. I called. On 4th street I've got [72]3A and he's got [T4]KQ. Now, here's where I want the odds calculation. All I need is any 9,8,6,5,4 to win outright because the best he can do is T432A. Or, I need to avoid him getting three different cards below a ten other than a 4. Barring those two scenarios, I just need to avoid a miracle. You can see where this is going.

I end up with [72]3AJQ[J] (J732A). Ouch. He gets [T4]KQ5A[2] (T542A) for the win.

According to Mike Caro, the odds of 432A completing an 8-low or better are 0.40:1, or about 71% (ignoring for a moment the other exposed cards.) I'm not an odds expert, but I would think the odds of 732A completing to an 8-low or better are the same. Right? So the combined odds of me not making a 9-low and him drawing three perfect cards have to be enormous. To make it more precise, I should mention that one 3 and one 4 were exposed in other players hands on the table.

Anyone want to walk me through the calculation?

fmny · High Card Joined: 24 Jan 2006 Posts: 8

You win this one 94.9% of the time.

drizzler · Pair Joined: 06 May 2006 Posts: 45

SaylorMarsh · Posted: Sun Jun 18, 2006 1:05 am Post subject:

www.twodimes.net.

For all your poker odds needs.

http://twodimes.net/h/?z=1790996
pokenum -mc 500000 -r 7c 2c 3c ac - td 4d kd qd / 3s 4s
Razz (7-card Stud A-5 Low): 500000 sampled outcomes
cards win %win lose %lose tie %tie EV
Ac 7c 3c 2c 475124 95.02 24876 4.98 0 0.00 0.950
Kd Qd Td 4d 24876 4.98 475124 95.02 0 0.00 0.050

drizzler · Pair Joined: 06 May 2006 Posts: 45

SaylorMarsh · Posted: Mon Jun 19, 2006 1:16 am Post subject:

Weeelllll, let's see. On fourth street there are eight cards out, not counting the other players' folded upcards because you didn't mention them (well, you did mention two, but this is the basic calculation so let's just assume it's a completely heads-up hand). This leaves 44 cards in the deck.

Any one of five different ranks gives you an unbeatable hand. 5 x 4 = 20, but your opponent holds one card you need (the 4) so you have 19 kill outs. Therefore, the chances that you will hit one of the cards you DON'T need on all three of the next streets are approximately 25/44 x 23/42 x 21/40 = 0.163352273 = 16.3%. I say "approximately" because your opponent could (and did) catch cards you need. But it's close enough for government work.

Meanwhile, your opponent needs to hit three consecutive unpaired cards 9 or lower. There are eight ranks of cards which qualify (everything but the four), and since you hold four of them, there are 28 cards which he needs. The chances of hitting that parlay are, again approximately, 28/44 x 24/42 x 18/40 = 0.163636364 = 16.4%.

So the odds of both events happening are approximately 0.163352273 x 0.163636364 = 0.026730372 = 2.6%. This doesn't quite sync up with the official results twodimes.net gives us, but then again, we were taking a lot of probability shortcuts. With multiple streets to go, Razz is a very complicated game, with any one of millions of potential variations. (You can see twodimes.net cuts off their calculations at 500,000 different outcomes.)

But it's not really that big a deal, considering he was shortstacked. He probably didn't take that much of a bite out of you, and his hand was tied since he was so low in chips. And since he was all-in, he had no decisions to make, which means he didn't really make a bad play. If he wasn't all-in by fourth street, then yeah, this is terrible. But bad beats due to sheer un-luck are a part of the game and always will be. I've seen worse and you probably have, too.

drizzler · Pair Joined: 06 May 2006 Posts: 45

Thanks for walking me through the calculation. That's exactly what I was looking for. And you're right, the beat didn't hurt me too badly.

As for having seen a worse beat, hmmm, I'm not so sure. At least not in Razz. I don't think I have ever been in the situation where my opponent needs two specific runners to beat me; ie. 6432A vs. KQ53A and he needs 4,2 to win, which I believe makes 6432A a 99% favorite (or higher with any dead 4s or 2s).

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