Well?

I will show by induction that for N people, the expected number of records is H_n = 1 + 1/2 + ... + 1/n.

Base case:
1 person, 1 record. H_1 = 1.

Inductive case:
Assume you can expect H_n records with n people. Now add another person at the end. The probability that the new person is the tallest of all n + 1 people is 1/(n+1), so the expected number of records is H_n + 1/(n+1) = H_(n+1).

So we expect H_1000 = 1 + 1/2 + ... + 1/1000, which is approximately 7.485

this topic is also interesting
en.wikipedia.org/wiki/Order_statistic

It does depend on the probaility distribution.

If you have the distribution where P(X=1) = 1 then there is always only one record.

If you instead have the distribution with P(X=1) = 1/2 and P(X=2) = 1/2 then there are slightly more than 1.5 records on average.

I'll admit that my first intuition was binary log 1000, but after thinking about it I realized it's just regular old log with base e 1000

Solution is basically 1 + 1/2 + 1/3 + ... + 1/1000

>Iz true u dun needs to know notin about the distribution
>E-except they r not same height
Can't solve the problem without a specific distribution, maybe you can get a range of values by trying every possibility but nothing more than that

If you look away from people with the same height (as in even extremely minute differences will be noted) then it is enough that there exists a total order, no more info is necessary

What do you mean no more is necessary? Not even your simulation converged to one value so there has to be a range based on all possible distributions of the height, so more is necessary to find an expected value of records rather than its range

Central Limit Theorem, motherfucker. Or some variation/application thereof.

Yeah a normal with unknown parameters because the question is too ambiguous

For a person in spot m there is a 1/m chance he is the tallest in the series 1..m, so you just sum the resulting series.

To the people who think distribution is necessary, assume normal then poisson with whatever parameters and show what the difference would be

>the tallest
That assumes that an unique tallest person exist. If several people have the same height then this doesn't work.

that person you're replying to isn't the one who did the simulation, i am

my gut feeling is that there probably isn't a singular expected value, but that it's good enough to pin a range to the wall

>The probability that the new person is the tallest of all n + 1 people is 1/(n+1)
isn't that only true if heights are uniformly distributed?

A deterministic distribution trivially has only one record.

It's 1 + 1/2 + ... + 1/1000, it's a standard result on record statistics:

books.google.com/books?id=dAS3CgAAQBAJ&lpg=PA229&ots=6qaFPfmTof&dq=order statistics harmonic numbers&pg=PA229#v=onepage&q=order statistics harmonic numbers&f=false

All that is needed is a total order.

that's my point f4m

If you assume that then the whole question becomes pointless. Arguing semantics gets old, it's clearly intended that there are no people equally tall...

that's a question of measurement precision, not statistics. there is no reason to expect there are truly identical heights in this population

DESU you need that for every real number, P(size=0), (maybe you even need that the diwtribution of a person's size has a density with respect to the Lebesgue's measure).

If the sizes are integers almost surely the above reasoning is wrong (because of ties).

I'd guess it's [math]\sum\limits_{i = 1}^{1000} \frac{1}{i} = H_i[/math] which is somewhere between 7 and 8.

If you allow for ties, then the question statement is just wrong about the answer being independent of the distribution. The answer of 1 + 1/2 + ... + 1/1000 is right if there are no ties, but if you imagine a world where everyone is the same height then it obviously won't work there. (You'll get either 1 or 1000 instead depending on whether ties count as new records or not)