# Expected number of rolls?

Suppose a fair six-sided die has the following sides: 1, 1, 1, 1, 4, 4. The die is rolled twice. The mixed outcomes [1,4] and [4,1] are considered "successes" while the outcomes [1,1] and [4,4] are considered "failures." What is the expected number of rolls to achieve the first success?

I am having trouble here because the die is rolled twice and am not quite sure how to calculate this expectation.

lets make a table each cell represents a different outcome
roll(2) 1            2
roll (1)        1 (2/3)^2    (2/3)(1/3)
2 (2/3)(1/3)  (1/3)^2


We can see the probability of a successful run is 2*(2/3)(1/3)=4/9.

The phrasing to the question "expected number of....to susses" tell us we are using a geometric distribution. From that we know the expected number of trials to the first susses is just 1/probability. in this example 9/4 times.

think about the probability of rolling a 4 with one and 1 with the other. You have a $\frac{4}{6}$ probability of rolling a 1 and a $\frac{2}{6}$ of rolling a 4 with the other.

Multiply the 2 probabilities and you will see the probability of this is: $\frac{4}{6}$($\frac{2}{6}$)

= $\frac{2}{3}$($\frac{1}{3}$)

= $\frac{2}{9}$

And since it is rolled twice, multiply by 2 to give you

= $\frac{4}{9}$

• You have missed the fact that there are two orders to roll different numbers. If you compute the chance to roll two $4$'s and the chance to roll two $1$'s and add them up, you won't get $1$ Commented Apr 12, 2015 at 5:58
• I forgot about that. Good point Commented Apr 12, 2015 at 13:05