Re: 100 vs 6500
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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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I did the calculations myself and get the same result. Of course I don't know if your method of adding 300 to the total players is correct. But either way it doesn't make a huge difference in the calculation. Just a couple percentage points.
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