[jcm4ccc]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I'm interested in constructive criticism. It's just not clear that you're understanding the question.

Here's a relevant example:

I have two data points each from two tournaments:

1000 2000 1000 0.3
1000 1000 2000 0.3

1000 2000 1000 0.2
1000 1000 2000 0.2

I want to calculate the expectation of (1000,2000,1000) and (1000,1000,2000). Are you contending that I cannot use all four data points - two for each of the two distributions - to do so? If not, why not?

Edit: make example clearer that distributions are about other stacks, not mine.

eastbay

[/ QUOTE ]

Independence of observations is an assumption of most statistical procedures. Some assumptions can be violated and the results can still be interpreted (for example, the normal distribution assumption). But that one can't.

If I understand your variables, they are:

Variable 1: Chip count of Player 1
Variable 2: Chip count of Player 2
Variable 3: Chip count of Player 3
Variable 4: Your results in the tournament

The problem is that your method may actually obscure your findings. Take a look at your first two observations. In observation #1, player 2 has 2000 chips. In observation #2, player 3 has 2000 chips. And yet the outcome is the same: you ended up in 2nd place. So if we just went by the first two observations, we would have to say that it makes no difference whether player 2 or player 3 has the most chips. But I think that's a mistake in your method, not an actual finding.


[/ QUOTE ]

The issue here is that we are computing several different things from the same pool of data which has some non-independent data. The question is, does the data have to be independent for each quantity that we are computing, or does it have to be independent between quantities as well. Do you see? I think I am going to have to do an experiment to settle it.

eastbay

[/ QUOTE ]

your observations have to be independent. no ifs, ands, or buts, regardless of my poor understanding of ICM. your variables do not have to be independent.