Why it this probability problem solved in this weird way?

Question

So the task says, we throw a cube until number 1 occurs, what is probability that an even number occured.

My logic of solving was this, let C = an even number occur, so we may get even number in two throws so P(c1) = $\frac{1}{2} * \frac{1}{6}$ (probability that even number appeared * probability that 1 appeared), we can get even number in three throws p(c2) = $\frac{1}{2} *\frac{5}{6} * \frac{1}{6} $ ,we can get even number in 4 throws P(c2) = $\frac{1}{2} * \frac{5}{6}*\frac{5}{6} * \frac{1}{6} $ we did $ \frac{5}{6}$ because when even number occured we don't care what number occur after that as long as it is not one because we are doing it in 4 throws, same for the 5 throw, 6 throws and so on, so P(c) = $\frac{1}{2} * \frac{1}{6}$ + $\frac{1}{2} * \frac{5}{6} * \frac{1}{6} $ + ... P(c) = $\frac{1}{2} * \frac{1}{6}* (1 + {(\frac{5}{6}})^1 + {(\frac{5}{6}})^2$ .... = geometric sequence = $\frac{1}{1-\frac{5}{6}}$ *$\frac{1}{3} * \frac{1}{6} = \frac{1}{2} $

This make perfectly sense to me however their solution is this,https://imgur.com/TEuMI0F , their logic is that they calculated when even number will not appear, and then subtract one by that and get $\frac{3}{4}$, which also makes sense, the problem is my solution and their solution makes sense but I don't know where is the problem in my solution because the results were not the same

Thank you!

Answer 1

So first consider an easier problem, where you assume that you do exactly $n$ throws. This means the last roll is a $1$, none of the first $n-1$ rolls were a $1$, and you want the probability that one of the rolls was even. Then it is equivalent to roll a d5 with sides 2-6 $n-1$ times and ask whether at least one of those rolls is even.

The easiest way to do that while avoiding a double counting error is this "take the complement of the event that none of these rolls were even" trick. The alternative is a big messy inclusion-exclusion problem. Doing that, the probability that an even number showed up would be $1-(2/5)^{n-1}$. (As a sanity check, when $n=1$ this is $0$ as it should be because $n=1$ implies the first throw was a $1$ and $1$ is not even.)

Now we use the easy problem to solve this problem. The probability distribution of the number of throws total is geometric with parameter $1/6$, so the probability of $n$ throws is $(5/6)^{n-1} (1/6)$. We already conditioned on the number of throws earlier, so by the total probability formula, the overall desired probability is

$$\frac{1}{6} \sum_{n=1}^\infty (1-(2/5)^{n-1}) (5/6)^{n-1}=\frac{1}{6} \left ( \sum_{n=0}^\infty (5/6)^n - \sum_{n=0}^\infty (1/3)^n \right ) \\ = \frac{1}{6} \left ( \frac{1}{1-5/6} - \frac{1}{1-1/3} \right ) = \frac{1}{6} \left ( 6 - \frac{3}{2} \right ) = \frac{3}{4}.$$

Answer 2

Here is another alternative. Define $p$ as the probability we are looking for. There are several possible first throw:

1. The first throw is even with probability $\frac{1}{2}$. After this we continue rolling the dice until $1$ appear.
2. The first throw is either $3$ or $5$. The probability is $\frac{1}{3}$. After this we continue rolling the dice until $1$ appear with probability $p$ that we have even number.

$$ \begin{align} p&=\frac{1}{2}+\frac{1}{3}p\\ \\ p&=\frac{3}{4} \end{align} $$