Re: 100 vs 6500
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Dont need to know x,y, and z. Unless x is huge relative to z and y you got way the worst of it.
Assumption. The hundred pros dont play any better than the 6500 randoms. We know this is not true. The chance of none of the 100 making the final 9 is equal to 1 minus the chance of the nine seats are all filled by randoms. This number is about 0.05 or one in twenty. For this to be a good bet for you, that 'x' number better be 40 or 50 times bigger than the 'y' number.
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We figured that the average pro from my 100 is 3x more likely to make the final table than the average player from the remainder of the field (assuming a 6600-person field). So the chance of one of my names filling Seat 1 at the final table is equivalent to 300/6800, right? And the chance of this not being the case is 6500/6800?
So if that's not wrong then the chances of there not being a single name from my list at the final table would come to [(6500/6800)(6499/6799)(6498/6798)(6497/6797)(6496/6796)(6495/6795)(6494/6794)(6493/6793)(6492/6792)], or [66.6%].
Am I wrong?
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