LibraryFool77
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DM me more knowledge. According to fictitious history some guy name Gore got credited, i always preferred lawyers, guns and money.
Werewolves eating Chinese too, fucking scoundrels.
Whose your daddy?
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Werewolves eating Chinese too, fucking scoundrels.
LibraryFool77
Active member
Take it you may be from the northwest?
LibraryFool77
Active member
No comment, been around a lot longer than you think. Was pretty recognized, got sick of the attitudes and burned some bridges, got banned more than once, but always got invited back! Feel free to give me a run for the money on pool knowledge, I enjoy it! I’ll praise you if you can bear me, just like the devil expected going down to Georgia, he lost in both songs, lolIn years? From the guy who's been here all of 16 months?
LibraryFool77
Active member
I also have some online friends here I really respect and enjoy their content, they know who they are.
Is it live or it Memorex
LibraryFool77
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I think I was 6 when Betamax (Sony) and VHS , the rest fought over rights. My Episodes of Miami Vice and TJ Hkooker look great.
LibraryFool77
Active member
I didn’t look forward to Saturday morning Ape grape. I knew Heather was smoking hot when I was 6, that’s how I know if you’re not your born this way. Big fan of Gaga, personally.
can all be done with a pocket calculator.No one is solving that problem on paper.
For the 5-pack version, there are 3168 individual binary operations that must be performed perfectly. A single mistake ruins the calculation. For the 10-pack version, that number jumps to 20,328 individual binary operations, all of which must be perfect. (I'm assuming an algorithm more efficient than the one I wrote, which does over 12x as many calculations as necessary, but was faster to write.) Matrix multiplication is extremely tedious. That means when using pencil and paper, the whole process has to be repeated (at least once) to make sure the final results are identical.
When it comes to statistics, fluency in the R programming language is the equivalent of knowing how to fish. It was designed by statisticians, for statisticians working on statistics problems.
analysis paralysis. no need to think back to univac.
had my first computer timex1000 with 2k of ram. 100 bucks. then an ibm286 for almost 3 grand.
now all they do is surf the internet and watch movies.
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g0ds gReeN
Active member
My dad's a cop, you idiot! He's Detective John Kimble.
Cartune
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If one has a few days to work on it, maybe. As a practical matter? It's not going to happen. It's too slow and far too prone to silent error to do the calculations by hand or with a pocket calculator.
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I figured out where the AI went wrong. It seems to have used this reasoning:
Let p be the chance of running a rack. Suppose you know that by n racks, the chance you have run 10 already is S( n ). S(n+1) can be found by the recursion:
S(n+1) = S( n) +(p^10)*(1-S( n))
which is just the total already plus the chance that this rack is the 10th run in a streak and S(9) = 0.
That recursion gives the AI result. Where does it go wrong? How can the recursion be fixed to give the right result?
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hose your daddy?
First, and foremost, much respect to you for digging deeper to figure out where AI went wrong.
I think you're describing the negative binomial distribution, which "models the number of failures in a sequence of independent and identically distributed Bernoulli trials before a specified/constant/fixed number of successes r occur." It fails on this problem for the same reason the binomial distribution fails as an approach.
When I first encountered this problem independent of classes, while I was an undergrad enrolled in Statistical Theory 1 or 2. It was about making (basketball) five consecutive free throw shots in no more than 100 attempted free throws, given a probability of 30% to make any particular free throw shot. I'm going to revert back to free throws for my explanation here, in the interest of using slightly simpler terminology.
My first inclination was to divide 100 shots into 96 5-shot groups, and treat those 5-shot groups as Bernoulli trials. "[T]he binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own Boolean-valued outcome: success (with probability p) or failure (with probability q = 1 − p)."
The probability of making five shots out of five attempts is p^5 = 0.3^5 = 0.00243. I wanted to subtract the probability that 0 of 96 5-shot groups was successful, which is the complement of at least one 5-shot group being successful, from 1. I applied the binomial distribution's probability mass function: nCr * p^r * q^(n-r), where "nCr" is "n choose r", the number of ways in which r items can be selected from n items, order not important. So, I had:
1 - nCr * p^r * q^(n-r)
= 1 - (96 choose 0) * 0.00243^0 * (1 - 0.00243)^96
= 1 - 1 * 1 * 0.99757^96
= 1 - 0.99757^96
= 1 - 0.79170701
= 0.20829299, an incorrect result.
The reason it doesn't work is the 96 Bernoulli trials are not independent identically distributed (i.i.d.) experiments, violating an assumption of the binomial distribution. Each 5-shot group shares four shots with each adjacent 5-shot group. Failure of one group makes a failure of an adjacent 5-shot group more likely. Suppose your first five shots are success, failure, failure, failure, success. It's impossible for the second 5-shot group to include no failures. The trials are not independent. The negative binomial distribution also requires the Bernoulli trials be independent and identical distributed (i.i.d.).
I asked my Stat Theory professor about the problem, and he confirmed my approach was wrong and that there was a way to calculate the result directly, without a Monte Carlo simulation, but he wouldn't tell me what it was. A couple years later, I learned about Markov chains, and immediately recognized that they could be used to solve the problem. There were six possible states. There was the case in which five consecutive shots had not yet been made and the case where five in a row had been made. The case in which five shots hadn't yet been made could itself be partitioned into five states: the current streak could be exactly 0, 1, 2, 3, xor 4 shots long. I had my six states, and I knew p and q, which was all I needed to model the Markov process.
Besides using Markov chains and Monte Carlo simulations, I'm not aware of any other methods to solve the problem. That doesn't mean there isn't another way, only that I don't know what it is. Every other method I've tried runs up against the i.i.d. assumption.
What about you?
Here's another problem with attempts at mathematical determination of real world phenomena. How was the 30% for any individual shot determined? It has to be based on the real world experience of shooting shots. What was the pool that the average for each individual shot was determined from? What was the distribution of the pool? Was the distribution really averaging every third shot or were there clusters of shots made versus missed. All of that would change the actual probability of making a certain number of shots in a row.
It can be done with recursion, but you have to look farther back than just the previous rack. You have to go back eleven racks for the chance the run has already occurred, and then there is necessarily a miss followed by ten runs, so
S(n+1) = S( n) + (1-S(n-10)) * p^10 * q