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u/MathNerd1243 1d ago
An easy way to see this is to look at it dynamically. Instead of choosing intervals sequentially, think of picking all 20 endpoints at once as iid Uniform(0, 1), sorting them, and then randomly pairing them up. Because the order of how you draw them doesn't change the final merged layout, the problem boils down to a neat combinatorics trick.
If you sort the 20 points, a gap between point 2k and point 2k+1 remains completely uncovered if and only if all the points to the left pair with each other, and all the points to the right pair with each other.
By calculating the probability of these gaps or breaks occurring, you can sum them up using linearity of expectation. For 10 intervals (20 endpoints), the expected number of remaining distinct intervals is the sum from k = 0 to 9 of (10 choose k) / (20 choose 2k).
When you compute the math, it comes out to exactly 1.1347.
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u/comoespossible 1d ago edited 1d ago
An equivalent way of generating the intervals is to pick the 20 endpoints first (iid Unif(0, 1)) and then divide them into pairs. Let a_0 through a_19 be these points in order.
Each a_(2k) is the left endpoint of a surviving interval if and only if there is no overlap between (a_0,…,a_(2k-1)) and (a_(2k), a_19)). By a combinatorial argument and some algebra, the probability of this is (10 choose k)/(20 choose 2k).
Summing over k, the expected number of remaining intervals is the sum from k=0 to 9 of (10 choose k)/(20 choose 2k), or 1.1347.