Assume that you have a die with 9 sides that are equality likely to be rolled. The sides are enumerated from 1 to 9. If you throw the die one hundred times, what is the probability that the sum of all the outcomes is an even number?

  • $\begingroup$ Can you brute force the case of two throws? Showing that kind of a simple case would work towards dispelling the thoughts that you just dumped your homework here. This is actually a concern here. I am looking for a reason to undelete my answer, please give me one! $\endgroup$ – Jyrki Lahtonen Dec 7 '17 at 20:14
  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$ – gimusi Jan 24 '18 at 21:48

It’s approximately equal to 50%.


The process is defined by the recurrence system:

$$n_{o+1}=4\cdot n_{o}+5\cdot n_{e}$$

$$n_{e+1}=5\cdot n_{e}+4\cdot n_{o}$$

with: $n_{o,0} = 1$ and $n_{e,0}=1$

$n_{o,i} \equiv$ possibile # of ODD cases after i rolls

$n_{e,i} \equiv$ possible # of EVEN cases after i rolls

In matrix form:

$$n_k=A^k \cdot n_{0}$$


$$A= \begin{bmatrix} 4 &5 \\ 5 &4 \end{bmatrix}$$


for k=2

$$A^2= \begin{bmatrix} 41 &40 \\ 40 &41 \end{bmatrix}$$


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