Question about Tony Guerrera article

[helter skelter]

I've been trying to learn about ICM and have been reading threads referred to in FAQ's, their associated links, etc. In one of them, I came across this. Since I am interested in MTT's I read the article by Tony Guerrera referred to in this post:



[ QUOTE ]
Ultimately, the proper way to do tournament analysis is to take into account everyone's pushing and calling distributions as a function of stack size and to calculate explicitly the optimal moves to make as a function of those distributions and the current relative stack sizes.

However, once the average stack is less than 10BB, there's not too much of a skill edge that you can exploit, especially at the higher buy-in STTs.

In general, the amount of risk you should shy away from in tournaments is grossly over estimated. I know this is an STT forum, but here's a link to an article I wrote about this topic, as it applies to early round play in MTTs:

http://www.pokerhelper.com/when-to-r...tournament.php

As people gravitate more towards playing an optimal endgame in STTs, it's going to become tough to have any sort of significant edge in the endgame, meaning that your edge actually needs to come from outplaying your opponents early on and getting yourself to the endgame crapshoot with a chip advantage more consistently than your opposition.

[/ QUOTE ]


(This is the link to the thread, for anyone interested.)


http://forumserver.twoplustwo.com/sh...age=0&vc=1


Anyway, I understood the article (mostly) until I got to this point:


"P(2nd) equals the probability of doubling up to half the chips in the tournament minus the probability of winning the tournament. P(3rd) is the probability of doubling up to a third of the chips in the tournament minus the probabilities of finishing in second and first. In general, the probability of finishing in ith place is the probability of doubling up to a stack of minus the sums of the probabilities of finishing in a higher place."



I can conceptualize P(2nd). When you get to 2nd, its only 1 double up to 1st, but I don't understand how P(3rd) would be the probabilty of getting 1/3 the chips.

It's hard to conceptualize, but the only way I can conceptualize P(3rd) would be taking half the chips from the tourney and putting them in 1 stack, the other 2 players each having 1/4 chips and then one of the quarter stacks doubling through the other 1/4 chips. So the player knocked out would have only reached 1/4 chips.

Preceding that, you'd have 1/2 chips in one stack, 1/4 in one stack and two 1/8 chip stacks, etc.

I'm sure I must be missing something, but is it even possible to double to 1/3 the chips? If you doubled up again, then you'd have 2/3 the chips.


He goes on to say:


"In general, the probability of finishing in ith place is the probability of doubling up to a stack of N/i minus the sums of the probabilities of finishing in a higher place"


Am I right in the way I am looking at it? Is he right in the way he is looking at it? Or are we both wrong?