Checkerboard Expected Value Question A checker is placed in a random square of an ordinary $8\times 8$ checkerboard (with all squares being equally likely). Then, checkers are placed in all squares that are below and to the right of the first checker, including those that are directly below or directly to the right of the first checker. Here is an example where the first checker is shown in red: !Example After this procedure, what is the expected value of the number of checkers on the board?

You have 64 possibilities. If we start in the bottom corner and move left we traverse the possibilities of 1 to 8 checkers. For the next row we go 2, 4, ..., 16 or twice the first row. The third row is three times the first. And so on. Thus, the expected value,

$$\frac{1}{64}\left( (1+2+3+\cdots+8) + 2(1+2+3+\cdots+8) +3(1+2+3+\cdots+8) + \cdots +8(1+2+3+\cdots+8) \right)$$

The sum $1+2+3+\cdots+8=\frac{8\cdot 9}{2} = 36$. Thus, the expected value is,

$$\frac{1}{64}\left( 36 + 2\cdot 36+ 3\cdot 36+ \cdots +8\cdot 36 \right) = \frac{36(1+2+3+\cdots +8)}{64} = \frac{36^2}{64} = \frac{81}{4}$$