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I'm going through Grinstead and Snell's probability. One of the questions in the earlier chapter is: Chose independently 2 numbers $ B $ and $ C $ from the interval $ [0,1] $ with uniform distribution.

What is the probability that: $$ B+C <1/2 $$

This question is easy enough, you can solve it geometrically using the unit square and drawing a line from (0, 1/2) and (1/2,0) and solving for the area of the triangle as:

$ 1/2 * 1/2 * 1/2 $. I've solved it and verified the answer in the solutions manual as 1/8.

My question is, how would you answer this question if it were more than 2 terms? I was trying to estimate how I would answer the question with 3 terms or 4 terms? With three 3 terms I've tried drawing out a cube and tried solving it, but I have no idea how to even begin approach it with 4 terms.

angryavian
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Jackson
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  • Your intuition is correct. The generalization of this problem involves finding the (hyper-)volume of a certain simplex defined by $x_1 + \cdots + x_n \leq \tfrac12$ where each $x_i \geq 0$, produced by cutting off the corner of the $n$-dimensional (hyper-)cube. However, it becomes more interesting if the bounding sum is in the range $[1, n]$ because the region is a more complicated polytope. – Sammy Black Jul 17 '22 at 17:31
  • @SammyBlack Got it. Unfortunately that seems well beyond the scope of my limited math knowledge, I don't even understand a bunch of the term you used in your comment ha. I was hoping maybe there was a relatively simple solution out there, but it seems not. I suppose the authors limited it to 2 dimensions for a reason. Really appreciate you taking the time out to comment on this, thank you! – Jackson Jul 17 '22 at 19:00
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    The distribution of $X_1+...+X_n$ where the $X_k$ are iid. uniform on $(0,1)$ is called the Irwin Hall distribution. – copper.hat Jul 17 '22 at 22:51

1 Answers1

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If $X_1, \ldots, X_n$ are i.i.d. uniform on $[0, 1]$, then the answers here show that $$P(X_1 + \cdots + X_n < 1) = \frac{1}{n!}.$$

You can geometrically show that $P(X_1 + \cdots + X_n < 1/2) = \frac{1}{2^n} P(X_1 + \cdots + X_n < 1),$ so the probability you want is $\frac{1}{2^n n!}$.

angryavian
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  • A curiosity: equivalently, the denominator is given by the double factorial $(2n)!!$ – Sammy Black Jul 17 '22 at 17:33
  • Wow that's much more of a complex solution than I was imagining. Something unbelievably elegant about the final solution though. Do you know if there's a way to approach this problem in a non-geometric way? – Jackson Jul 17 '22 at 17:55
  • @Jackson The answer by user91500 only uses integrals and induction. – angryavian Jul 17 '22 at 18:02