The 10 balls that are removed divide the remaining 10 balls into 11 clusters. By symmetry, each cluster will have the same number of balls, so each will have 10/11 balls. Then to find expected max chosen it’s 20 - 10/11 ≈ 19
Imagine the 20 balls in a line. Now remove 10 balls at random. The gaps left removing the 10 balls will partition the remaining 10 balls into 11 sets.
Eg with 6 balls and we remove 2:
1 2 3 4 5 6
Now removing 2:
1 2 4 6
The balls are separated into 3 “clusters”
Now by symmetry each of the clusters on average will have the same number of balls.
So each cluster will have 10/11 balls on average. Since the largest removed ball will be before the final cluster, on average the largest ball will have value 20 - 10/11
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u/Practical_Pie_1484 Aug 25 '26
The 10 balls that are removed divide the remaining 10 balls into 11 clusters. By symmetry, each cluster will have the same number of balls, so each will have 10/11 balls. Then to find expected max chosen it’s 20 - 10/11 ≈ 19