3 ms·
Seeing that it's a uniform distribution, let's start out with assuming our sample mean (the average serial number we find) has the same distribution as the true
by mruts 7y ago
Seeing that it's a uniform distribution, let's start out with assuming our sample mean (the average serial number we find) has the same distribution as the true mean (the actual number of tanks in existence). If this is true, then:
2 x mean
should be an unbiased estimator of the true mean. But because we are probably under sampling the extremes, we could use the Bessel correction:
1/(n-1) x summation_{i=1}^n(sample_i)
I would guess this comes out to a better estimation than what the article says.
Bessel's correction might be a bit of overkill, since it's intended to work with normal distributions. But I still suspect it comes out to a better estimation that what the blog post says.
- comicjk 7y agoI thought the same, and tried it. The mean sample mean after 10 million runs is 500.47, which is very close to the true mean, 500.5. Bessel's correction is not correct here - it's effectively multiplying by n/(n-1), so your estimate would be 536. Bessel's correction is made for estimating true variance using a sample, not for estimating true mean.
- mruts 7y agoYes of course, the mean is the first moment and a Bessel correction would be inappropriate. Now I feel stupid. The mean calculated by sum(p_t x_i) or 1/n sum(x_i) is already the best linear unbiased estimate. Maybe we can't get better than twice the mean?
- nestorD 7y agoWhat if you get into the war at a later point ? Most tank with a small serial number will have been destroyed (tanks tend to have a short life) and, using the mean instead of the maximum, you will get a seriously biaised result. You could adjust for such problems but it seems much easier to use the maximum.