Math question for smart guys. Estimating entries.

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 can't really discuss where this is applicable but figured it might be interesting to discuss?

So there is a bowl with an undisclosed number of tickets. These tickets each have a number assigned (but random - not a series). When a ticket is drawn the number is announced on the ticket and the ticket is placed BACK in the bowl. This process is repeated for 100 times. You have results like - 1 number 3 times, 3 numbers 2 times, and 91 numbers that were drawn exactly once.

So how do I go backwards from these numbers to estimate how many tickets are in the bowl?


If I was to figure this out - I would write a sim to try various amounts of tickets - then average results and find the one closest to the sample I took during the drawing. For a reasonable programmer these little toy games are really simple to write. It is pretty much my go to solution anytime I've tried to figure out something like this.

You'll have to better define the range of numerals on the tickets. Beyond that, it has to do with the BP math problem. I explain the latter after tablepooey figures it out. Ha.

P.S. Surely to God, you can ask a more interesting question.

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Originally Posted by OppsIdidItAgain

You'll have to better define the range of numerals on the tickets. Beyond that, it has to do with the BP math problem. I explain the latter after tablepooey figures it out. Ha.

P.S. Surely to God, you can ask a more interesting question.

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BP ?

The range of numerals on the tickets has no bearing on the solution I am seeking. Thats why I said "random - not a series". If you knew the series/range of the numbers then it'd be fairly easy to closely estimate the size of the pool being drawn from. Maybe I should have said a unique random word is on each ticket to clarify.

Bill is that you?

Where "selective reading" got you. I've already mentioned the BP problem, and, the limit on the Merriam-Webster dictionary (of words in common usage, not including chemical names, and the like.) Oh, and who Oops was.

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Originally Posted by OppsIdidItAgain

Where "selective reading" got you. I've already mentioned the BP problem, and, the limit on the Merriam-Webster dictionary (of words in common usage, not including chemical names, and the like.) Oh, and who Oops was.

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So you can't begin to answer my question? lol. figures. I figured out how I'd do it but sims are a cheat.

I'm actually still much more concerned with the mindless stuff like 861 X 168 = 492 X 294 = 144648; 672 X 276 = 384 X 483 = 185472; and, 3852 X 2583 = 1476 X 6741 = 9949716; 6552 X 2556 = 4473 X 3744 = 16746912. I mean, stuff like that "just is", and, so, if you can figure it out, it might mean a lot more. Stuff that I posted here years ago, but recently got around to further working out.


144648 = {(41 + 43 - 2) X 42 X [(-3 + 5) X (-1 + 7^2 - 3^3)]} ---> 41_43__24_42__351_173.

185472 = {[(-3^3 + 7^2 + 1) X (10 + 5 - 3)] X (2 X 42) X (1 + 4 + 3)} ---> 371_153__24_42__41_43.

9949716 = ((41 + 43 - 2) X 42 X {[0^2 + (-001)^03 + 5^02 + 3^2^0] X [+7^3^0 + (-10)^02 + 0]}) ---> 41_43__24_42__153_371.

16746912 = ({-(0^0^3) X -(010^2^0) X (-7) - (3 X 3) + (5^02) X (010^3) X (0^0)} X (2 X 42) X
(1 + 4 + 3)) ---> 173_351__24_42__41_43.


Some similarly thus mindless stuff.

[-(2^0) + Fibonacci-42] = (-1 + 267,914,296) = {[(0 + 03^02) X (0 + 5) X (001 + 10) X (07^02^0 + 300)] X (41 X 43)}.


And some more.

The repeating part of 1/42 = (2^0 / 42) is (2^0 / 42) ---> [({010^2 - [7 X (-3)^2]} X [3 + (-5)^2 - 010]) X 143] --> 173_351__41_43.


All of which sort of fits into a bit of work with something called "self-similar square sum concatenation". I'm trying to develop a generating function for the fine-structure constants, from the math alone. I already have a couple of terms that fit, and, am working on finding a third, at which point there should be a pattern locked in. To do with what many believe to be the "backbone" of math, insofar as sums, and differences, of squares, cubes, and so on. It's ticklish work, mostly farting around in the hope that a few more digits of the numerals sort themselves out, especially as the numerals become larger, and, so, only when time permits.

I've hit a vein to the motherlode of numerals, themselves, in the sense that the above thus numerals form a mathematical basis for all of them. Crazy as it may seem. Ha.

I think the proper solution may involve using stat software to fit a curve to the data or something like that.

What I would try is to use the most stable data point, the % of tickets that never dupe, and work backwards from that.

So if 70% of the (drawn and returned) tickets never duped over 100 trials that would imply (ignoring variance), that the chance of not duping a given ticket in 100 draws is 70%.

.7^(1/100) would be the chance of not duping in 1 trial which is .9964. Then 1 - .9964 gives the chances of duping in 1 trial = .00356, which is 279:1, for 280 total tickets.

This approach ignores all the extra information you have and is still going to be subject to variance (because that 70% figure is subject to variance).

I think the way you use the extra information of double dupes and triple dupes etc to mitigate variance is curve fitting. But idk.

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Originally Posted by smurgerburger

I think the proper solution may involve using stat software to fit a curve to the data or something like that.

What I would try is to use the most stable data point, the % of tickets that never dupe, and work backwards from that.

So if 70% of the (drawn and returned) tickets never duped over 100 trials that would imply (ignoring variance), that the chance of not duping a given ticket in 100 draws is 70%.

.7^(1/100) would be the chance of not duping in 1 trial which is .9964. Then 1 - .9964 gives the chances of duping in 1 trial = .00356, which is 279:1, for 280 total tickets.

This approach ignores all the extra information you have and is still going to be subject to variance (because that 70% figure is subject to variance).

I think the way you use the extra information of double dupes and triple dupes etc to mitigate variance is curve fitting. But idk.

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Yea, this might be just as good of an estimate as anything. It would definitely be the most straighforward.

My solution was just to sim a ton of them then find the ones with the same results. So if I simmed with 50 then print out results like 71,15,8,6 (for 0, 15 of 1 duplicate, 2 with 3 of the same etc for each trial.. Then just count the exact matches of each simmed pool size with our 1 sample. It would at least could remove variance on the testing side but there will always be the variance in the observed duplicates.

I'd have to think more about your method. It sorta makes sense but I'm not sure how good of an approximation it would be. However, writing a sim to test the 1 size estimated by the above amount would actually be quite easy. Hmm maybe when I have an hour or so free.

The curve fitting approach also seems correct to me but that is over my head. Maybe fresh out of college I'd have some clue..

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 . . .

Oh, boy, tablepooey outdid himself again. It's not about the reappearance of particular tickets, let alone wholly unique ones. And, obviously, a lot of people could solve the thing, let alone many places to ask it. But, the CMS baloney was a nice touch, in the sense that perhaps the drum holds a multiple of the set of tickets, whichever. As I pointed out, earlier, it involves a bit more thinking than a computer simulation to reverse engineer it. (The old n!/r!(n-r)! bit to limit the products won't cut it. Simulation/regression is not particularly useful for reverse engineering stuff.)

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Originally Posted by OppsIdidItAgain

Oh, boy, tablepooey outdid himself again. It's not about the reappearance of particular tickets, let alone wholly unique ones. And, obviously, a lot of people could solve the thing, let alone many places to ask it. But, the CMS baloney was a nice touch, in the sense that perhaps the drum holds a multiple of the set of tickets, whichever. As I pointed out, earlier, it involves a bit more thinking than a computer simulation to reverse engineer it. (The old n!/r!(n-r)! bit to limit the products won't cut it. Simulation/regression is not particularly useful for reverse engineering stuff.)

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Was there ever a time when you said something relevant or useful?

Think back...

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Originally Posted by OppsIdidItAgain

Oh, boy, tablepooey outdid himself again. It's not about the reappearance of particular tickets, let alone wholly unique ones. And, obviously, a lot of people could solve the thing, let alone many places to ask it. But, the CMS baloney was a nice touch, in the sense that perhaps the drum holds a multiple of the set of tickets, whichever. As I pointed out, earlier, it involves a bit more thinking than a computer simulation to reverse engineer it. (The old n!/r!(n-r)! bit to limit the products won't cut it. Simulation/regression is not particularly useful for reverse engineering stuff.)

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What a turd you are Bill.

Thank you for that. I've learned not to answer any of your questions because the result always has to be about you, and, your way of doing it. Go fuck yourself.

Gambling forums should be the last place to ask any math question. Shackleford's shit was a scam, from start to finish. Everything there, including Jacobson, was to suck more people into it. You could find more real and actual mathematicians on Hitler's ass.

Tablepooey and Monet sitting in his tree. Lol lol lol. Ha.


Oh, double PSS - as faker Red used to put it - I just told old V that he couldn't pass a double blind question about the quality of sound from one stereo player to the next, as if it matters given that most of the earlier music was made from shit speakers, etc. He could no more hear it than the musicians could produce it.

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Originally Posted by OppsIdidItAgain

Thank you for that. I've learned not to answer any of your questions because the result always has to be about you, and, your way of doing it. Go fuck yourself.

Gambling forums should be the last place to ask any math question. Shackleford's shit was a scam, from start to finish. Everything there, including Jacobson, was to suck more people into it. You could find more real and actual mathematicians on Hitler's ass.

Tablepooey and Monet sitting in his tree. Lol lol lol. Ha.


Oh, double PSS - as faker Red used to put it - I just told old V that he couldn't pass a double blind question about the quality of sound from one stereo player to the next, as if it matters given that most of the earlier music was made from shit speakers, etc. He could no more hear it than the musicians could produce it.

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I didn't ask you any questions moron. I made a statement that you are a turd.

That was just the closet psychic in me, to post at the same time as you, for you to try to make something of it. Hint, hint: ... you guys sitting in a tree. Ha. But, no need to quote (the other ding-a-ling) when our posts were at the same time.

Obviously, you couldn't fit in anywhere else, so, you cling to sites like this one. Me, I'm all you can eat on any site out there.

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Originally Posted by OppsIdidItAgain

That was just the closet psychic in me, to post at the same time as you, for you to try to make something of it. Hint, hint: ... you guys sitting in a tree. Ha. But, no need to quote (the other ding-a-ling) when our posts were at the same time.

Obviously, you couldn't fit in anywhere else, so, you cling to sites like this one. Me, I'm all you can eat on any site out there.

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https://www.youtube.com/watch?v=npYyM29mkTU

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Originally Posted by accountinquestion

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 can't really discuss where this is applicable but figured it might be interesting to discuss?

So there is a bowl with an undisclosed number of tickets. These tickets each have a number assigned (but random - not a series). When a ticket is drawn the number is announced on the ticket and the ticket is placed BACK in the bowl. This process is repeated for 100 times. You have results like - 1 number 3 times, 3 numbers 2 times, and 91 numbers that were drawn exactly once.

So how do I go backwards from these numbers to estimate how many tickets are in the bowl?


If I was to figure this out - I would write a sim to try various amounts of tickets - then average results and find the one closest to the sample I took during the drawing. For a reasonable programmer these little toy games are really simple to write. It is pretty much my go to solution anytime I've tried to figure out something like this.

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Jack off my big cock into the bowl. Then pick up the load and count the number of tickets attached to it. Repeat the process three times with three fresh loads. Multiple The total number of cum stained tickets by 9