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
= $\frac{4}{9}$
