Question: **Original Question:** Two fair dice are tossed simultaneously till a combination $(4,3)$ is obtained. Obtain the probability that the number of tosses required is atmost $2$.

**My Reasoning** : The dice are thrown simultaneously. One can't determine or distinguish between the two die. Thus, the combination of $(4,3)$ will **include** the outcomes $(4,3)$ as well as $(3,4)$. With that reasoning, I came up with following solution:

$$x : \text{Number of tosses required to obtain the combination} $$
$$x ~ \text{~ Geometric distribution}\left(p=\frac{2}{36}=\frac{1}{18}\right) $$
$$P(x < 3) = P(x = 1) + P(x = 2) = \left(1-\frac{1}{18}\right)^0\*\left(\frac{1}{18}\right) + \left(1-\frac{1}{18}\right)^1\*\left(\frac{1}{18}\right) = 0.108025$$

But this answer was not accepted. Can someone please point out how is the reasoning wrong?

The answer expected was $0.0547839$ with p = $\frac{1}{36}$. Does order matters?

Answer: 
Your error is simply the fact that $f(x)$ does not define a probability density function, i.e., your $f(x)$ does not satisfy $$\int\_0^{18} f(x)\,\mathrm{d}x = 1.$$
The probability density function integrates to 9 in the interval given, that is why it does not work.