A good spot to finish up my very own latest outing, and, with a bit of gematria to finish off the Reaper stuff.

Firstly, the queen died exactly 911 days after the WHO declared the pandemic. Secondly, there are 345,600 minutes from then to the coronation (end-day not included), and, alternatively, 7152 hours from then to my birthday, 2 July (end-day included). The Reaper's last post here was post #3456 in the thread, and, the Reaper's (randomly assigned) usernumeral there, at the gematria forum, was 7152. Lastly, the Opps of my new username fit in well with KJ's return, and, next, my going on to mention the OPP. So, I guess that I was right to be wrong. The part about letting go of reality to get to the other side.

Originally Posted by tableplay View Post
For K unknown unique tickets (i.e. no tickets with the same number printed/written on them), let P=1/K where P is the probability of drawing a particular unknown ticket. The probability of drawing that ticket m times (where m is a positive integer greater than 0) in 100 attempts is thus {100!/[m!*(100-m)!]}*[P^m]*[(1-P)^(100-m)] which I will call A. So 100*A is the expected number of times you would observe the drawing of this ticket.
Practically speaking, one would use the ticket with the highest observed frequency and use this number to solve for an estimate of K. However, you could do this for all of the different tickets to get a K for each one and then average the Ks to get a population estimate.

Example: you observe a ticket with the number 1.5 on it being drawn 6 times in 100 drawings (and replacements). M=6 is then used in the equation for A. So for M=6 we have 100*A=100*(1,192,052,400*[(1/K)^6]*{[(K-1)/K)]}^94}=6 . Solve this equation of K (you can use Excel solver or any web based solver if you don't want to do it by hand). After you come up with K for that ticket, you can then come up with a K for other tickets (using their observed frequencies in the 100 draw and replacments) and average those Ks up for a population estimate.

Obviously VCT forum members Kim Lee, Eliot ,Alpax and Drich could come up with better solutions than this since they are much stronger mathematicians than me, but that is my solution. And obviously one such application of this approach would be to estimate the number of cards held in the active bin of a CSM . . .
Special thanks again, Tablepooey, for the bit of junk above. Have you ever studied a math proof to at least know how to begin to fake it?

Originally Posted by accountinquestion View Post
I'm guessing no one will be able to answer this but maybe Tableplay. He seems to be the best at high-level math.
I really thought that after you being you so much, that old Pooey would have told you the same as me. Nothing like getting the "rug" pulled out from underneath, one more time.

Now it's back to the attic for me to get something done. However, it takes less and less time to truly sort things out.