I am having a hard time understanding the concept of a negative binomial distribution. For example the question:
How many times do you expect to roll a six-sided die before landing on the number 6.
The answer to this question is obviously 6 but if I attempt to solve this using a probability function I cannot get the correct answer.
I attempted to answer this question by letting
$X = \text {numbers other than 6 that appear in order to roll a 6}$
$Y = \text{number of rolls required in order to get a 6}$
I know the probability function is defined as $P(X=x) = f(x) = p(1-p)^{x}$
I know that $Y = X + 1$ so $X = Y-1$ then
$P(X=x) = P(X = y-1) = \frac{1}{6}(\frac{5}{6})^{y-1}$ but when I plug in $y=6$ why am I not getting a very high percentage as getting a 6 should be expected on the 6th roll.
Can someone please explain the flaws in my argument.
Best Answer
Another approach is to use the Law of Iterated Expectation, and partitioning on the event of succeeding on the first (next) try.
Let $\bar X$ be the expectation of the count of tries until you encounter a six. If you succeed on the first try, the (conditional) expectation is $1$; this condition has probability $\tfrac 1 6$ of occurring. If you don't then after failing $1$ try you are faced with the exact situation you began with, and the conditional expectation is therefore $1+\bar X$; occurring on probability $\tfrac 5 6$.
Putting this together we find: $$\bar X = \tfrac 1 6 + \tfrac 5 6 (1+\bar X)$$
Solve for $\bar X$ and you are done.