Problem
Suppose that we roll a die until a 6 comes up.
a. What is the probability that we roll the die n times?
b. What is the expected number of times we roll the die?
(Rosen - Discrete Mathematics and Its Appplications, 6th ed., pg. 440 (6.4.12))
Solution
a. If we roll the die n times (assuming n ≥ 1 and the dice is 6-sided and fair), then we must roll n-1 "not 6" rolls followed by 1 "6" roll. The probability of that is:
$ (\frac{5}{6})^{n-1} (\frac{1}{6})^1 $
b. We could roll the die any number of times from 1 to infinite. Consider a random variable R which is the number of rolls to roll a 6. We could determine the expected number in two ways:
i. We could say that R is a geometric RV with a chance of success of $ \frac{1}{6} $, so $ E(R) = \frac{1}{(\frac{1}{6})} = 6 $
ii. We could show i. explicitly, using the formula for expected value:
$ E(R) = \sum_{s \in S} R(s) p(s) $
$ = \sum_{i=1}^\infty (i) ((\frac{5}{6})^{i-1} (\frac{1}{6})^1) $
$ = \frac{1}{6} \sum_{i=1}^\infty (i) ((\frac{5}{6})^{i-1}) $
This summation takes the form $ \sum_{i=1}^\infty (i) (r^{i-1}), \text{ where } |r| < 1 $
$ \sum_{i=1}^\infty (i) (r^{i-1}) = \sum_{i=1}^\infty \frac{d}{dr} (r^i) $
$ = \frac{d}{dr} \sum_{i=1}^\infty (r^i) $
$ = \frac{d}{dr} \frac{1}{1-r} $ (since |r| < 1)
$ = \frac{1}{(1-r)^2} $
Let $ r= \frac{5}{6} \ \ (|r| = \frac{5}{6} < 1.) $. Then,
$ E(R) = \frac{1}{6} (\frac{1}{(1-\frac{5}{6})^2}) $
$ \Rightarrow E(R) = 6 $
Either way, we get the same answer: 6 rolls.
--Thomas34 12:55, 16 October 2008 (UTC)
