206k views
3 votes
Pick a random number (evenly distributed) between 0 and 1. Continue picking random numbers as long as they keep decreasing; stop picking when you obtain a number that is greater than the previous one you picked. What is the expected number of numbers you pick?

User Rob Darwin
by
7.5k points

2 Answers

2 votes

Answer:

Explanation:

001

User Don Neufeld
by
8.4k points
3 votes

Answer:

e = 2.718

Explanation:

- There're "n!" ways of arranging "n" numbers, supposing that there're n picks, then the first (n−1) picks are in descending order, there are (n−1) ways of choosing the first (n−1) numbers and thus the probability of picking just n numbers is:

(n-1) / n!

- The expected value (E) would be:

E = ∑ n*(n-1)/n!

= ∑ n*(n-1)/n*(n-1)*(n-2)!

= ∑ (n = 2 to infinity) [ 1 / (n-2)! ] = ∑ (n = 0 to infinity) [ 1 / (n)! ]

= e ...... (Maclaurin series approximation)

User Jacob Abraham
by
8.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories