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: 
To add one further dimension of insight, I want to point out that the reason the non-normalized density matters is a little subtle. The other answers are right, but it’s not super obvious why. The usual proof (using linearity of expectation) that your two expressions for the variance are equal does not seem to rely on normalization, after all.

In particular, we have

$$E[(X-\mu)^2]=E[X^2-2\mu X+\mu^2]=E[X^2]-2\mu E[X]+E[\mu^2]=E[X^2]-2\mu^2+E[\mu^2]$$

and all this goes through regardless of whether the density is normalized. The normalization matters only for the last step: equating $E[\mu^2]$ with $\mu^2$. These won’t agree unless the density is normalized.
The probability density function integrates to 9 in the interval given, that is why it does not work.