Timeline for What is the expected number of random numbers (generated uniformly) such that their sum of numbers exceeds one?
Current License: CC BY-SA 3.0
6 events
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| Apr 2, 2019 at 0:43 | comment | added | Douglas Zare | The sum is just the tail-sum formula for the expected value of a random whole number. $\sum i P(X=i) = \sum_i \sum_{j=1}^i P(X=i) = \sum_{j=1}^\infty \sum_{i =j}^\infty P(X=i) = \sum_{j=1}^\infty P(X \ge j)$. | |
| Apr 1, 2019 at 10:46 | comment | added | Douglas Zare | @RichardStanley: The $1/0!$ term is the probability that you need to draw the first time, which is $P(X > 0)=1$. The probability that you need to draw the second number is also 1, $P(X > 1)=1/1!$. The probability that you need to draw the third number is $P(X>2)=1/2!$ etc. | |
| Mar 31, 2019 at 16:38 | comment | added | Richard Stanley | How do you get an expectation $\sum_{n=0}^\infty 1/n!$ from the fact that the probability that the sum of the first $n$ numbers is less than 1 is $1/n!$? Where does the term $1/0!$ come from in the sum? | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 12, 2013 at 15:13 | vote | accept | web_ninja | ||
| Sep 12, 2013 at 14:32 | history | answered | Douglas Zare | CC BY-SA 3.0 |