5 Fixed typo, n->N

This can answered without any complicated maths.

It can be related to the following: Imagine you have n N marked cards in a pack of m cards and shuffle them randomly. What is the probability that they are all at least distance d apart? Consider dealing the cards out, one by one, from the top of the pack. Every time you deal a marked card from the top of the deck, you then deal d cards from the bottom (or just deal out the remainder if there's less than d of them). Once all the cards are dealt out, they are still completely random. The dealt out cards will have distance at least d between all the marked cards if (and only if) none of the marked cards were originally in the bottom (n-1)d. N-1)d. The probability that the marked cards are all distance d apart is the same as the probability that none are in the bottom (n-1)d.N-1)d.

The points uniformly distributed on a line segment is just the same (considering the limit as m →∞). The probability that they are all at least a distance d apart is the same as the probability that none are in the left section of length (n-1)d. N-1)d. This has probability (1-(n-1)d/L)1-(N-1)d/L)N.

Integrating over 0≤d≤L/(N-1) gives the expected minimum distance of L/(N2-1).

4 more typos

This can answered without any complicated maths.

It can be related to the following: Imagine you have n marked cards in a pack of m cards and shuffle them randomly. What is the probability that they are all at least distance d apart? Consider dealing the cards out, one by one, from the top of the pack. Every time you deal a marked card from the top of the deck, you then deal d cards from the bottom (or just deal out the remainder if there's less than d of them). Once all the cards are dealt out, they are still completely random. The dealt out cards will have distance at least d between all the marked cards if (and only if) none of the marked cards were originally in the bottom (n-1)d. The probability is that the marked cards are all distance d apart is the same as the probability that none are in the bottom (n-1)d.

The points uniformly distributed on a line segment is just the same (considering the limit as n m →∞). The probability that they are all at least a distance d apart is the same as the probability that none are in the left section of length (n-1)d. This has probability (1-(n-1)d/L)N.

Integrating over 0≤d≤L/(N-1) gives the expected minimum distance of L/(N2-1).

3 typo

This can answered without any complicated maths.

It can be related to the following: Imagine you have n marked cards in a pack of m cards and shuffle them randomly. What is the probability that they are all at least distance d apart? Consider dealing the cards out, one by one, from the top of the pack. Every time you deal a marked card from the top of the deck, you then deal d cards from the bottom (or just deal out the remainder if there's less than d of them). Once all the cards are dealt out, they are still completely random. The dealt out cards will have distance at least d between all the marked cards if (and only if) none of the marked cards were originally in the bottom (n-1)d. The probability is that the marked cards are all distance d apart is the same as the probability that none are in the bottom (n-1)d.

The points uniformly distributed on a line segment is just the same (considering the limit as n →∞). The probability that they are all at least a distance d apart is the same as the probability that none are in the left section of length (n-1)d. This has probability (1-(n-1)d/L)N.

Integrating over 0≤d≤1/(N-1) ≤d≤L/(N-1) gives the expected minimum distance of L/(N2-1).

2 typos
1