Non-geometric way to calculate expected value of breaks?"A bar is broken at random in two places. Find the average size of the smallest, of the middle-sized, and of the largest pieces."The author gives what seems like a complicated geometric way of calculating the probabilities. He arrives at the solutions 1/9, 5/18, and 11/18. Is there a simpler, non-geometric way of calculating these probabilities?

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Non-geometric way to calculate expected value of breaks?&quot;A bar is broken at random in two places. Find the average size of the smallest, of the middle-sized, and of the largest pieces.&quot;The author gives what seems like a complicated geometric way of calculating the probabilities. He arrives at the solutions 1/9, 5/18, and 11/18. Is there a simpler, non-geometric way of calculating these probabilities?

Damarion Pierce · Accepted Answer

Step 1Here&#039;s a sketch. Pick points   0  ≤  x  ≤  y  ≤  1. Now let&#039;s find the expected value of x given that the interval [0,x] is shortest. That is, we look at the region satisfying                     x                            ≤        y                            x                            ≤        y        −        x                            x                            ≤        1        −        y            You can draw your own pictures. We then get a triangle with   0  ≤  x  ≤  1      /    3,   2  x  ≤  y  ≤  1  −  x. This triangle has area 1/6 (by inspection) or by double-integration, and       ∫    0          1              /            3            ∫          2      x              1      −      x        x    d  y    d  x  =      1    54    , so the expected value of x is                     1        54                    1        6              =            1      9      .Step 2Next case. Suppose the interval [0,x] has middle length. This corresponds to two regions:                    x                            ≤        y                            y        −        x                            ≤        x                            x                            ≤        1        −        y            or                     x                            ≤        y                            x                            ≤        y        −        x                            1        −        y                            ≤        x            Interestingly, these both have area 1/12 and for each of these we get       ∬    x        d    A    =                  5        216            . So the expected value of x in the middle case is                     5        216                    1        12              =            5      18

Lillie Pittman · Accepted Answer

Step 1For the       k          t      h       smallest piece indicated by       S          (      k      )      , the expected length is given by                              E                          (          n                                S                          (              k              )                                )                                    =                  E                          (                      X                          (              k              )                                )                                                  =                  ∑                      i            =            0                                k            −            1                                    1                      n            −            i                              which simplifies to the formula given in the book of 50 challenging problems.Step 2For the question asked,   n  =  3 and calculate the sums for   k  =  1  ,  2  ,  3, which we find to be                              E                          (                      S                          (              1              )                                )                                    =                  1          9                                              E                          (                      S                          (              2              )                                )                                    =                  5          18                                              E                          (                      S                          (              3              )                                )                                    =                  11          18

A bar is broken at random in two places. Find the average size of the smallest, of the middle-sized, and of the largest pieces.

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