Any maths / statistics whizzes here?

[jamesbrown]

wasn't me chief - used r for donkeys, since uni days, along with SPSS and sometimes matlab

I'm not a risk analyst, I'm an engineer.

Were you at uni in 2014?



https://www.contractoruk.com/forums/...ml#post2002494

[jamesbrown]

And? Never googled it before, so what?

Keep digging, scootie.

https://www.contractoruk.com/forums/...ml#post2002528

Your cretinism is about to break the internet.

[jamesbrown]

You know full well r is just an extension of s, which is what we used back then - gwd

[NotAllThere]

R I use for simulations - you use what you know.

Show us the code, or it didn't happen.

[sasguru]

Dunno if I understood the prob but here's some R code which gives the average weeks in the order of 1850-ish.

nums <- c(rep(0, 1000))

for (j in 1:1000){
  
  count <- 0
  people <- c(rep(FALSE, 1000))
  while (TRUE) {
    
    count <- count + 1
    
    k <-  sample(1:1000,4,replace=F)
    for (i in 1:4){
      people[k[i]] <- TRUE
    }
    
    if (all(people)) {break}
  }
    nums[j] <- count
  
}

print(mean(nums))

Edit: Actually more like 1860-ish which kind of matches NAT.
Edit2: Assumes sampling with replacement
NATEdit3: Add code tags around it to make it a bit more readable.

[OwlHoot]

I tweaked a program I'm working on, for random walks in asset prices, to test this out.

1000 people
Each week 0.4% (4 people), selected at random, do the thing.

Running 1000 trials...

After 250 weeks, 633 people have done the thing at least once
After 500 weeks, 865 people
After 1000 weeks, 982 people

To reach 1000 people, with a reasonable confidence level, I had to extend the number of weeks to 4000.

I think jamesbrown had the soundest algebraic approach.

The probability, P, of any given person not doing the "thing" after n weeks is given by P = (1 - 1/250)^n

Equating that to, say, 1/250 and taking logs gives n = 1377 (to the nearest integer). But you could choose any (small) value to represent the near certainty of any given person doing "it", and the values of n increase rapidly as P tends to zero.

[OwlHoot]

The solution may be solved using a deterministic approach, you've got everything you need. No need for a probabilistic approach, assuming inputs are not changing with each new week.

Well the OP didn't mention probability, true, but that was inferred as he mentioned thinking "It isn't as simple as 1000 / 0.04".

Also, no mention was made of whether or not this "thing" can be done more than once. So we don't necessarily have all we need.

But, say there were N people in a prison and the "thing" was getting shot by a sadistic guard who chose 0.04 of the inmates at random to shoot each day. Even then there is some ambiguity, because as ladymuck pointed out in so many words (I think? ) we aren't told whether the number shot is 0.04 * N each day or 0.04 of the surviving prison population, which is decreasing daily.

[sasguru]

The solution may be solved using a deterministic approach, you've got everything you need. No need for a probabilistic approach, assuming inputs are not changing with each new week.

Feck off, you fraud.

[Lost It]

There's Lies.

There's damned Lies.

Then there's Statistics.

[Old Greg]

There's Lies.

There's damned Lies.

Then there's Statistics.

Then there's Pooperscooper.