Exactly fucktard (look up the definition of a proof, if you don't believe me idiot). And with regard to averaging out all occurrences, you didn't even understand what I wrote since you are dumber than a bag of rocks. For a ticket inscribed with an A, you would count the number of times that ticket inscribed with an A is redrawn in 100 draws and then solve for K, the total population, using the binomial probability density function above. So now you have a K estimate using Ticket A (call this KsubA). You could then produce a K estimate again using a ticket inscribed with B in the same fashion as A (call this KsubB) and so on for all different inscriptions observed in 100 draws. So, for example, if you only observed tickets inscribed with an A,B,C, and D in those 100 draws, your K estimate would be the weighted average of those 4 K estimates you fucking cretin. So if ticket A was observed 41 times, B 24 times,C observed 20 times, and D observed 15 times your weighted K calculation would be [(41*KsubA) + (24*ksubB) + (20*KsubC)+(15*KsubD)]/100. In my original solution, I even stated quite plainly that it was an estimate you piece of shit. I never, ever said that it was exact. Re-read that original post if you don't believe me fagboy.Originally Posted by OppsIdidItAgain
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