Question

Consider rolling a fair dice. You keep rolling the dice until you see all of the faces (from number 1 to 6) at least once. What is your expected number of rolls?

Answer 1

The expected value of getting any number on the six faces die is 1.

Because when we roll a die then the outcomes is any number as 1, 2, ..., or 6.

Next we need to observe other number than the first observed number.

There are 5 different numbers than the first observed number, so the probability of getting any one of the remaining 5 numbers first time is (1/6)^(k-1)*(5/6)

Which is a geometric distribution with parameter p = 5/6

Therefore the expected value of observing second different number is 1/p = 1/(5/6) = 6/5

Next we need third different number that the first two numbers. The probability of getting the third number first time than the first two number with probability (2/6)^(k-1)*(4/6)

Which is again follows geometric distribution with parameter p = 4/6

Therefore the expected number to get the third different number than the first two is 1/p = (1/(4/6) = 6/4

Similarly the expected number to get the fourth different number than the first three is 1/p = (1/(3/6) = 6/3

The expected number to get the fifth different number than the first four is 1/p = (1/(2/6) = 6/2

The expected number to get the sixth(last) different number than the first five is 1/p = (1/(1/6) = 6/1

Then adding all the expected values(because expectation is a linear function), we get the final answer as below:

1 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 1 + 1.2 + 1.5 + 2 + 3 + 6 = 14.7

Answer: 14.7