# Expected number of die tosses to get something less than 5

Assume you have a die in your hand. Each time you throw it, you look to see what value you would get. If the value is greater than 4, roll the die again. Otherwise you stop.

Let X be the number of times you toss the die before coming to a halt. For instance, if you get $5 → 6 → 1 → stop$, the corresponding value of $X$ will be 3.

a). What's the probability mass function $p_x$?

b). What is the expectation value $E(X)$?

c). Is the value of $(X^2+1)$ odd or even on average? [Just curious]

$P(5,6)=\frac{1}{3}$

$P(1,2,3,4)=\frac{2}{3}$

My solution for a): $$p_x(n)=\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{n-1}where \space n\ge1$$

My solution for b): $$E(X) = \sum_{n=1}^{\infty}n.p_x(n)=\sum_{n=1}^{\infty}n.\left(\frac{2} {3}\right)^{n-1}\left(\frac{1}{3}\right)$$ Simplification then gives me $r=\frac{2}{3}$ to yield $$E(X) = \frac{1}{3}\sum_{n=1}^{\infty}n.r^{n-1}$$

Since

• $n.r^{n-1}$=$\frac{d}{dr}r^{n}$
• along with the fact the sum of a geometric series is given by $\sum_{n=0}^{\infty}r^{n}=\frac{1}{1-r}$
• Also, $\sum_{n=0}^{\infty}r^{n}=1+\sum_{n=1}^{\infty}r^{n}$

Hence, it is possible to write \begin{align} E(X) & = \frac{1}{3}\frac{d}{dr}\left(\sum_{n=1}^{\infty}r^{n}\right)=\frac{1}{3}\frac{d}{dr}\left(-1+\sum_{n=0}^{\infty}\right) \\ & = \frac{1}{3}\frac{d}{dr}\left(-1+\frac{1}{1-r}\right) \\ & = \frac{1}{3}\left[\frac{1}{({1-r})^2}\right] \\ & = \frac{1}{3}\left[\frac{1}{(1-(2/3))^2}\right]\\ & = 3 \end{align}

I'm not sure whether I'm correct for both parts because I am not particularly good at this subject of mathematics.

Could you aid me and confirm that I've done it correctly?

Say that a roll of 1-4 is "good".

The simple solution for (b) is: a single roll produces a 1–4 with probability $\frac23$. Therefore each roll produces $\frac 23$ good results on average.

Since expectations are additive, the expected number of good rolls in $n$ rolls is $\frac 23n$.

How many rolls do we need for the expected number of good rolls to be 1? We want $\frac 23n =1$, so $n=\frac 32$, and this is the answer.

About point a), I would say that $p_x(n)=\left( \frac 23 \right)\left( \frac 13 \right)^{n-1}$, since to stop at the $n$-th toss you need to get 5 or 6 in the first $n-1$ tosses, and 1,2,3,4 in the last.

Thus also the following calculation for $E(X)$ changes: $$E(X)=\sum_{n=1}^{\infty}n\left( \frac 23 \right)\left( \frac 13 \right)^{n-1}=\sum_{n=0}^{\infty}n\left( \frac 23 \right)\left( \frac 13 \right)^{n-1}$$

And now your method works well, since you're dealing with a power series, $\sum_{n=0}^{\infty}r^n$, whose convergence radius is $1$, hence also for its derivative the convergence radius is $1$, then summation and derivation commute for $r=1/3$ since $|1/3|<1$.

You get $$E(X)=\frac 23 \frac d{dr}\left[\frac 1{1-r}\right]_{r=\frac 13} = \frac 32$$

• Appreciate the correction but when I was eliminating the errors, a question popped up in my head. Look at the original post as I have already posted my "curious" question there. @AndreaBianchi Apr 29, 2015 at 12:12