[benkahuna]

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.

This thread is not about folding aces.

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?