Two fair dice tossed simultaneously till combination $(4,3)$ is obtained. Obtain the probability that number of tosses required is atmost $2$. 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?
 A: I agree with your reasoning.  You are not told the dice are distinguishable; even if they were (e.g., one die is red and the other is green), you are also not told which die represents the first outcome in the ordered pair $(4,3)$.
There could be a circumstance by which the dice are indistinguishable and do not correspond to a fixed ordering, but the outcomes $(4,3)$ and $(3,4)$ are distinguishable because of how they are read:  for instance, whichever die lands to the left of the other die corresponds to the first outcome in the pair.  If you cannot visually determine the order from left to right, you read the dice from top to bottom.
But again, this requires an interpretation of the question that is not stated.  It is much more reasonable to infer that the outcomes are indistinguishable, in which case observing one $4$ and one $3$ in a single roll is $1/18$, and observing such an outcome within at most $2$ rolls is simply $$\frac{1}{18} + \frac{17}{18^2}.$$
The only thing working in favor of distinguishable outcomes is the fact the problem wrote $(4,3)$ rather than $\{4,3\}$.  But this notational difference is inadequate:  the problem needs to explicitly state that the dice or their outcomes are distinguishable.  Otherwise, you would on average observe twice as many $(4,3)$ outcomes as you would $(4,4)$, and this is not reflected in the claimed answer of $\frac{71}{1296}$.
A: Remember, when two dice are tossed simultaneously,, there will be $6\times6 =36$ total equiprobable outcomes out of which $(4,3)$ is one and $(3,4)$ is another. (You could imagine that one die is red and the other blue, if it helps you)
Thus P(get desired outcome in at most two tosses)
= $1$- P(not get the desired outcome in two tosses)
$= 1 - (\frac{34}{36})^2$
The book is wrong.
