Two numbers are drawn at random from the integers 1 through 10. What is the expected value of their sum? Does it change if the second draw is done with or without replacement? Click below for the answers.
This puzzle is from Patrick Honner. It's easy to calculate the expected value with replacement. It's just two times the expected value of a random draw from 1...10, so 2 * 5.5, or 11. The interesting part is that when you draw two numbers without replacement the expected sum doesn't change. Why is that?
To find out, take a look at what happens to the expected value of the second draw for each value of the first draw. (Expected value is just the average of the remaining numbers.)
1st draw
E(2nd draw)
$1$
$6$
$2$
$5\frac{8}{9}$
$3$
$5\frac{7}{9}$
$4$
$5\frac{2}{3}$
$5$
$5\frac{5}{9}$
$6$
$5\frac{4}{9}$
$7$
$5\frac{1}{3}$
$8$
$5\frac{2}{9}$
$9$
$5\frac{1}{9}$
$10$
$5$
As the value of the first draw increases, the expected value of the second draw decreases. If you take the sum of each column you get 55. Divide by 10 to get the average and you get 5.5, so when you add them together you arrive back at the solution of 11.
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