MHB Solving for $E(x)$ when X is a Fair Die Toss

[dfraser]

I'm given that: X is the random variable for the number of times a fair die is tossed before a six appears, and asked for E(x).

I know the solution is 5 but am struggling to understand how I'm meant to get there. I understand:

$$E(x) = \sum_{x=0}^{\infty} x\cdot p(x)\ $$
$$= \sum_{x=0}^{\infty} x \cdot (1/6) \cdot (5/6)^x $$
$$= 5/6 \cdot 1/6 \cdot\sum_{x=1}^{\infty} x \cdot (5/6)^{x-1} $$

but am then given that:

$$= \frac{5/6 \cdot 1/6}{{(1-5/6)}^{2}} =5 $$

I recognize the numerator, and know the formula for the infinite sum of a geometric series is $$\frac{a}{1-r}$$ but I don't know how to get the denominator from the prior step or why it is squared. I think this is an easy mistake, but I can't seem to find it. Can anyone help me here?

 

[Evgeny.Makarov]

\[
\sum_{n=1}^\infty nx^{n-1}= \sum_{n=1}^\infty \left(x^n\right)' =\sum_{n=0}^\infty \left(x^n\right)' \overset{(*)}{=}\left(\sum_{n=0}^\infty x^n\right)' =\left(\frac{1}{1-x}\right)' =\frac{1}{(1-x)^2}.
\]
The equality (*) holds for $x$ in the interval of convergence of the right-hand side, i.e., for $|x|<1$.

 

[I like Serena]

I'm given that: X is the random variable for the number of times a fair die is tossed before a six appears, and asked for E(x).

I know the solution is 5 but am struggling to understand how I'm meant to get there. I understand:

$$E(x) = \sum_{x=0}^{\infty} x\cdot p(x)\ $$
$$= \sum_{x=0}^{\infty} x \cdot (1/6) \cdot (5/6)^x $$
$$= 5/6 \cdot 1/6 \cdot\sum_{x=1}^{\infty} x \cdot (5/6)^{x-1} $$

but am then given that:

$$= \frac{5/6 \cdot 1/6}{{(1-5/6)}^{2}} =5 $$

I recognize the numerator, and know the formula for the infinite sum of a geometric series is $$\frac{a}{1-r}$$ but I don't know how to get the denominator from the prior step or why it is squared. I think this is an easy mistake, but I can't seem to find it. Can anyone help me here?

Hi dfraser! Welcome to MHB! :)Let's define:
$$f(y) = \sum_{n=1}^\infty ny^{n-1}$$
Then:
$$\int f(y) = \sum_{n=1}^\infty y^n + C_1 = \sum_{n=0}^\infty y^n + C_2 = \frac 1{1-y} + C_2$$
Taking the derivative, we get:
$$f(y) = \sum_{n=1}^\infty ny^{n-1} = \frac 1{(1-y)^2} \tag{1}$$When we apply $(1)$ to your formula, then:
$$E(x)
= \frac 56 \cdot \frac 16 \cdot\sum_{x=1}^{\infty} x \cdot \left(\frac{5}{6}\right)^{\!{x-1}}
= \frac 56 \cdot \frac 16 \cdot \frac 1 {(1 - \frac 56)^2}$$Edit: for some reason I completely missed Evgeny's response. Ah well.

 

[awkward]

The infinite series approach is fine, but here is another way for the sake of variety: condition on the result of the first throw. Let's say E is the expected number of throws before throwing a 6.

With probability 1/6, you will throw a 6 on the first throw. In this case the number of throws before a 6 is 0.

With probability 5/6, you throw something else (not a 6). You are now in the same state you were in before the first throw, but since you have taken one throw, the total number of throws expected before you get a 6 is E+1.

So $E = (1/6) \; 0 + (5/6) \;(E+1)$. Solve for E.

 

[dfraser]

Thanks everyone, I understand now.