Solution
I received correct solutions from Rob Pratt, Nicholas Meyer, Ken Schilling, Dan Dima, Joseph DeVIncentis, and John Guilford.
Meyer had a particularly slick way of doing it that makes it easy to count the expected number of unseen faces. He wrote:
"The answer is 6(1 − (5/6)n).
Let Xn be the number of distinct numbers that appear in n rolls. Then
Xn = I1 + . . . + In,
where Ik is 1 if a new number appears on the k-th roll, and 0 otherwise.
E(Ik) is simply (5/6)(k −1), because a new number appears on the k-th roll only if it's different from the preceding k − 1. Since expected values are additive, E(Xn) is the sum of an n-term geometric series, with answer given above. Note that as n → ∞, the answer goes to 6, as it obviously should."
Another way of saying the same thing is that the expected number of unseen faces is the sum of the probabilities that i is an unseen face, as i goes from 1 to 6. But each such probability is (5/6)n, so the expected number of unseen faces is 6(5/6)n, which yields the same answer after subtracting from 6.
In order to get the probability that there are exactly k unseen faces, one needs to set up a recursion. The probabilities are:
|
Faces
|
Probability
|
|
1
|
1/7776
|
|
2
|
155/7776
|
|
3
|
25/108
|
|
4
|
325/648
|
|
5
|
25/108
|
|
6
|
5/324
|
One curiosity here is that the probability that there are 3 unseen faces in 6 rolls of a standard die exactly equals the probability of 5 unseen faces.
Thanks again to Julian Beasley for the suggestion.
[Back to Problem 1169]
© Copyright 2013 Stan Wagon. Reproduced with permission.