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| Re: Your analysis is grossly incomplete | |
| Posted By: uberfoop <atkinso2@seattleu.edu> | Date: 1/17/12 7:32 p.m. |
In Response To: Re: Your analysis is grossly incomplete (Hoovaloov) : I think I understand what you're showing, but in one set, you have compared : each number and put them into a specific place to affect the results. : Basically, you are introducing bias in one set, and no bias in another to : prove your point. I simply sorted the data in ascending order. While this : might introduce some bias, it is neutral and affects both sets equally. In a distribution pair where different orderings show different results, what makes you think there's any such thing as a "neutral" ordering? And if there was a way to order it neutrally, wouldn't it be to order them randomly? After all, locking them into ascending order puts a systematic effect on the whole thing; there's no logical reason here that we should compare extrema only with other extrema and means with means and whatnot. Seems like, if anything, your ordering is one of the most biased possible. For instance, I could easily come up with a pair of two distributions which, under your ordering, would have one set always win every time, but which would tend to show a more "fair" layout under random orderings. For instance, suppose you have one "linear" data set, and one data set which is identical but slightly skewed in one direction. Under an ordering like yours... [.1,.2,.3,.4,.5,.6,.7,.8,.9,1;.11,.21,.31,.41,.51,.61,.71,.81,.91,1.01] ...The first group appears to have a 100% win rate. However, randomize it (and note that I didn't do the following intentionally; this just happened when I plugged two sets of random orders off the top of my head): [.5,1,.2,.9,.8,.3,.7,.1,.4,.6;.31,.71,1.01,.11,.41,.91,.21,.61,.81,.51] And now, it's 50%. Which is somewhat off from what would be expected, since the first set seems to have some sort of timing advantage, but it's nonetheless obviously much more reasonable than the first result. That a random ordering would be "less" biased makes sense; one way to measure win % would be to actually have two players fire at each at different pacing rates, which would effectively do the exact same thing as drawing values at random from each table (as opposed to drawing values from the same area of each ordered table, as your approach currently does). Which brings me to my response to your second part: : If this is still unacceptable, how would you suggest arranging the data to
I'm not entirely sure, actually. Off the top of my head, you could do it by taking a random sample from one set, a random sample from another set, and comparing which one wins. Then repeat that thousands of times, keeping track of how many wins each group has. That would emulate the thing I just described above, which is what you're really trying to measure.
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