aalhour

Martial arts, reading, long walks, lifting weights.

No,

I worked as a manager at a local German company in Germany with a strong Worker's Council team. Performance Management was run through them, they had comments, the process was adapted and managers were onboarded. This is not to say that performance management doesn't suck in general (because once you start having metrics, people start gaming them and you're almost superimposing it on people's way of working), but the Worker's Council didn't block it, because at the end of the day it was neither invasive nor exploitative.

EDIT: typos.

No worries, I think your solution coverges with @Someone's above, if I am not mistaken. You're multiplying the probabilities for all values of j up to i for all values of i.

I was aware that the Random function call gets evaluated in Python (and C) every time the loop iterates, I just couldn't imagine the probability distribution myself, I had assumed that all numbers are uniformally distributed but didn't cater for the sums. You're a legend! Now I see it. I wrote the following code to check out your argument and it checks out indeed :thumbs-up:

  def p(n):
    total = 1
    for i in range(1, n):
      total *= (100-i)/100
    return total


  for i in range(1, 100):
    print(f"p({i}) = {p(i)}")

The first 13 results:

  p(1) = 1
  p(2) = 0.99
  p(3) = 0.9702
  p(4) = 0.9410939999999999
  p(5) = 0.9034502399999998
  p(6) = 0.8582777279999998
  p(7) = 0.8067810643199997
  p(8) = 0.7503063898175998
  p(9) = 0.6902818786321918
  p(10) = 0.6281565095552946
  p(11) = 0.5653408585997651
  p(12) = 0.503153364153791
  p(13) = 0.44277496045533604

p(12) sits at 50%, of course it will be the mean! :D

Yes, that's true. I don't want an average of 50. My initial guess of 50 is wrong. I am trying to understand why the average turned out to be approximately 12.

Not sure that adds up, I translated the right side of the equality into a Python statement and it returns a number I am not sure how to interpret:

  1/99 * sum([(i-1)*i for i in range(2, 100)])   # 3266.666666666667

Both the median and mean are around 12 for 10K runs of the loop.

me too please?

I just changed the DNS on my unrooted Android to AdGuard's, went into the YouTube and I still saw the ads. What am I missing?