Dice roll - Geometric Distribution Question

Question

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.

Answer 1

This is because you're computing a probability, not an expected value.

There is no need to introduce two variables. Let $X$ the number of times you roll the die before you get a $6$.

You have $P(X=k)=p(1-p)^k$, for $k\geq 0$, with $p=\frac{1}{6}$ (by the way this is a geometric distribution). If you want the expected value, then you have to compute:

$$E(X)=\sum_{k=0}^{\infty}kP(X=k)$$

This is a geometric series so it's easy to find the result.

Answer 2

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.

Answer 3

The expected value is a sort of average value over a large # of trials.

Suppose an elementary event has a Pr = $\dfrac{1}{100}$, i.e. 1 in 100

On an average, how many trials would be needed to get the event once ?

100, isn't it, as the term itself suggests ?

If an elementary event has a probability p, $E[x] = \dfrac{1}{p}$

More complex computations are needed only if you start asking how many trials will be needed to get that event,say, 3 consecutive times !