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View Full Version : HoH, pg 423, chances of qualifying for big real money tourney

benkahuna
09-07-2005, 02:12 AM
There was another thread in which folding aces preflop in a tournament was discussed. It was in particular reference to the problem in Harrington on Hold Em volume 2.

Rather I'm wondering about winning chances based on stack size.

Dan consistently makes the claim that the winning chances in a tournament are related to your skill relative to your opponents and to your stack size. These chances are proportional, blind position notwithstanding. So if skill is equal and stacks 2:1 in your favor, your chance of winning against that opponent is 2:1. He applies the same rules to multistack situations. Also to multiple first place positions (like in a tournament prize package qualifier tourney). Here's the problem:

In a problem in Dan's book with the following chip stacks:

Player1: 4.5k
Player2: 4.5k
Player3: 4.5k
You: 4.5k
Player4: 1k
Player5: 1k

and Harrington (and Robertie) write that you have a 90 percent chance of winning when 4 people qualify. He takes the percentage of your chips relative to the total chip population and multiplies it times 4 to get your 90 percent as you have 22.5 percent of the chips in play.

The problem I have here is that this calculation cannot be correct if you have 25 percent or more of the chips in play. You cannot have a 100 percent or greater chance of winning.

So, is the calculation wrong (even as an estimate) or is there some point where it breaks down and some point where it's extremely accurate?

09-07-2005, 10:48 AM
The formula doesn't account for blinds or ante. Otherwise, with competent play, 25% of the chips, and 4 players qualifying, you would be guaranteed to make it simply by not playing any hands.

benkahuna
09-07-2005, 02:43 PM
No offense, but that seems hopelessly naive.

I've busted in these types of situations when blinds were big when not playing any hands.

That's an aside though, what's wrong with the formula when you have a greater than 100 percent chance of qualifying?

With 100, I'll concede it's possible to have this chance of qualifying theoretically, but I say it's not possible in the real world.

For over 100, it's simply not possible to have that chance of winning one of the 4 spots, so in what way is the formula wrong?

pzhon
09-07-2005, 02:58 PM
A useful model to use for this type of question is one that is usually called the Independent Chip Model in the 1TT forum. One way of looking at it is that the first place is determined proportionately. Then, that player's chips are removed, and the next place is determined proportionately, etc. Another way of looking at it is that the individual chips are shuffled. Players are ranked by their highest chips.

The independent chip model says you have the following probabilities of placing in each position:

1st: 22.5%
2nd: 21.97%
3rd: 21.07%
4th: 19.22%
5th: 11.62%
6th: 3.63%

The chance to finish out of the top 4 is estimated to be 15.24%.

No one claims the ICM is exactly accurate. However, it is more consistent than simple heuristics, and it is probably much more accurate in this case.

benkahuna
09-08-2005, 02:16 AM
Thanks. That makes sense. I like the fact that you can't violate reality by having an over 100 percent chance of finishing in the top 4. Who came up with the ICM? How have people found out about it?

SumZero
09-08-2005, 04:19 AM
[ QUOTE ]
Thanks. That makes sense. I like the fact that you can't violate reality by having an over 100 percent chance of finishing in the top 4. Who came up with the ICM? How have people found out about it?

[/ QUOTE ]

Read STT forum. ICM is critically important in bubble play as there are a lot of situations that are +chip EV that are -\$ EV because of the payout table and the fact that chips have declining marginal value. Knowing how to push these edges is a key skill that is very important in late stages of tournaments.