I am having trouble figuring out the expected value in situations were the examples are going to infinity.


I have a fair dice with 8 sides. I keep a counter ($k$) of the rounds I play. Each round I increase the counter by 1 and I roll the die. I keep doing this until I roll a 1 or an 8.

So I'll define a random variable $X$ to be the amount of rounds played. I know that the odds of rolling a 1 or a 8 on a 8 sided die is $\frac1 4$ otherwise it is $\frac3 4$. The way I am trying to solve the expected value is by using a geometric series. This is where I am getting stuck I think it should look like this:


$E(X)=1*\frac1 4+2*\frac1 4+3*\frac1 4+...+n*\frac1 4$

I am unsure how to turn this into a sum.

  • 1
    $\begingroup$ You would have to calculate the expected value of the random variable. If it goes to infinity or not doesn't matter, since it is a normal random distribution. The expected value is the sum of the product between the possible results and their probability. E[X] = x1*p1 + [...] + xn*pn $\endgroup$ – StillBuggin Dec 20 '15 at 18:58
  • 2
    $\begingroup$ Check out the geometric distribution en.m.wikipedia.org/wiki/Geometric_distribution $\endgroup$ – David Quinn Dec 20 '15 at 19:05
  • $\begingroup$ The probability $P_k$ that goes with $X_k$ in your sum should be the probability that the random variable $X$ takes on the value $k$. What you’ve got is the probability that you’ll roll a $1$ or $8$, which is not the same thing at all. You should instead have the probability that the $k$th roll is $1$ or $8$ and that none of the preceding rolls were. $\endgroup$ – amd Dec 21 '15 at 0:24

There is an easy trick for this calculation, using the following fact, for a random variable $X$ and a specific value $x$: $$ E(X)=P(X=x) \cdot E(X\mid X=x)+P(X\neq x)\cdot E(X\mid X\neq x) $$ (where $E(X\mid X=x)=x$, of course).

We can let $X$ be the number of throws we do in total (including the last $1$ or $8$), and $x=1$. This gives $$ E(X)=\frac14+\frac34E(X\mid X\neq 1) $$ But $X\mid X\neq 1$ just means "we know we fail on the first throw, and then we keep going as normal", which means that $$ E(X\mid X\neq1)=1+E(X) $$ and this gives $$ E(X)=\frac14 +\frac34(1+E(X)) $$which may be solved as a normal equation.

  • $\begingroup$ The way I am trying to solve this and the way I have seen similar question solved is using a geometric sequence. Since X is the number of rounds played I think my expected value of X would look something like this: $E(X) = X_1 P_1+X_2 P_2 + X_3 P_3 + ... + X_n P_n$ which would turn to $E(X)=1* \frac 1 4 + 2* \frac 1 4 + 3* \frac 1 4 +...+n* \frac 1 4$ $\endgroup$ – Steph Dec 20 '15 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.