More Dice Roll

We will consider three instances of utilising the concept of expected values for estimating the number of trials for a given outcome.

Getting a 6

How many times must a 6-sided fair die be rolled for a 6?

There is a 1 in 6 chance that the expected number of rolls is 1, i.e., get a six in the first roll. There is a 5/6 chance that the roll results in another number. In that case, the game will need at least another toss, and the game will repeat. Let E be the expected number of rolls until a 6, using the expected value formula:
E = (1/6)x1 + (5/6) x (E+1)
E = 1 + 5E/6
E = 6

Getting a 66

How many times must a 6-sided fair die be rolled for a 6 followed by a 6?

Let E₆₆ be the required value. We have seen in the earlier section that the expected number of rolls for a six is 6. Once that happens, there is a (1/6) chance that the next roll will give a six and a (5/6) chance that the game will start again.
E₆₆ = 6 + (1/6)x1 + (5/6) x (E₆₆+1)
E₆₆ = 6 + 1 + 5E₆₆/6
E₆₆ = 42

Getting a 65

Here, we estimate the expected number of rolls for a six followed by a number other than six. It is more complicated than the earlier case. After getting the first 6, there are three possibilities:

1. A 5 and game over.
2. A 6 and have another chance for a 65 (665).
3. A number other than 5 or 6 and restart. 

Let E₆₅ be the required value, the expected number of rolls for 65 from the start. We will need to define another term, E6, which is the expected number of rolls for 65 from 6.

E₆ = (1/6)x(E₆+1) + (4/6)x(E₆₅+1) + (1/6)x1
E₆₅ = (1/6)x(E₆+1) + (5/6)x(E₆₅+1)

E₆₅ = 36