throwing a dice probability distribution I was asked to find the probability distribution of a random variable, $X$, that counts the number of tries until getting the result $3$  and the result  $2$.
I'm not quiet sure about my answer and I'd like to know if I'm on the 'right path' of solving the question, and if not then is there another way of solving it? Using Geometric distribution maybe?
$P(X=k) = \frac1{6}*\frac1{6}*(1-\frac4{6})^{k-2}*2*(k-1) $
$ \frac1{6}$ = the probability of getting 3 throwing a dice 
$ \frac1{6}$ = the probability of getting 2 throwing a dice 
$1-\frac4{6}$ =the probability of getting anything but  3 or 2 throwing a dice and since there are k tries, then we multiply the possibilities by $k-2$
$2*(k-1)$ = choosing on which try we got 3 (since the last will be 2) or the opposite.
 A: I would try modeling it this way:
First you have to roll one of the two numbers $2$ or $3.$
Let's say this first happens after $X_1$ rolls;
then $X_1$ is $1$ plus a geometric random variable
with parameter $1/3$ (because of the six equally likely outcomes
of each roll, two of them will end this sequence of rolls).
That is, $P(X_1 = 1) = \frac13,$
$P(X_1 = 2) = \frac13\left(\frac23\right),$
$P(X_1 = 3) = \frac13\left(\frac23\right)^2,$ and so forth.
Then, after the first time you roll either $2$ or $3,$
you have to roll the other number.
This takes some number of additional rolls.
If the additional number of rolls required is $X_2,$ then
$X_2$ is $1$ plus a geometric random variable
with parameter $1/6$ (because now on each roll there is only one
outcome that will allow us to end this sequence of rolls).
That is, $P(X_2 = 1) = \frac16,$
$P(X_2 = 2) = \frac16\left(\frac56\right),$
$P(X_2 = 3) = \frac16\left(\frac56\right)^2,$ and so forth.
The total number of rolls is $X = X_1 + X_2.$
Example: Suppose the first roll is $2.$ Then you have already rolled one of the numbers $2$ or $3$ on the first roll, so $X_1 = 1.$
Next you have to roll the "other" number, which in this case is $3.$
Now suppose the next three rolls are $2, 2, 3.$ Since you got your first $3$
on the third roll, $X_2 = 3.$
The entire sequence of rolls was $2, 2, 2, 3,$ 
so it took four rolls to roll both of the numbers,
and indeed $X = X_1 + X_2 = 1 + 3 = 4.$
