Answer by Brendan McKay for Expected summation of dropped intervals?
Each point in $[0,1)$ has probability $p=\prod_{i=1}^\infty (1-2^{-i})$ of being missed by all the intervals, so the expected measure of what is not missed is $1-p$. This is approximately 0.7112119; I...
View ArticleAnswer by Christian Remling for Expected summation of dropped intervals?
This is a linear first order recurrence, so we can certainly solve it. Let me denote by $A_n$ the solution to the homogeneous equation $A_{n+1}=(1-2^{-n-1})A_n$ with the initial value $A_1=1/2$ (for...
View ArticleExpected summation of dropped intervals?
For each $n\in\mathbb{N}$, let $I_n$ be an interval of length $1/2^{n}$. We drop each $I_n$ onto the interval $[0,1]$ uniformly at random (so that there is "wraparound" if need be). What is the...
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