Dice is tossed until a number greater than $4$ appears 
An unbiased die is tossed until a number greater than $4$ appears.
  What is the probability that an even number of tosses is needed?

I have seen the author's solution in which he has formed a geometric progression and solved the question but I would like to know why the following method is erroneous: 
Since,
$ 1)$ tossing a die an odd number of times 
and
$2)$number greater than 4 appearing are two independent events: 
$P(A\cap B) = P(A)P(B)$
$\implies P(A\cap B) = \dfrac 12 \times \dfrac 26 = \dfrac 1 6$
$\dfrac 12$ because a die can either be tossed an even number of times or odd number of times, there is no other option. 
 A: Would you argue the same if the condition was changed to "until a number greater than $0$ appears"? In that case, the game will always end on the first throw. 
The error you are making is not taking into account that "an even number of tosses is needed" is not only that the condition is fullfilled on that even toss, it has to be not fulfilled on all previous tosses. Since the throw numbering starts at $1$ (and odd number), there is the non-symmetry coming from that your solution does not take into account.
A: I am not sure that the computation in your question makes sense - I don't understand it. But if you want to make a distinction between odd and even outcomes, and also to avoid a geometric progression in favour of a parity arguments, you can do it like this. Let $P_O$ be the probability that a number greater than $4$ first appears on an odd throw, and $P_E$ be the probability for an even throw.
The probability that every throw of an unlimited set gives less than $5$ is zero, so $P_O+P_E=1$
Then think what happens after one throw. If it comes up $5,6$ then you have an odd number of throws already (probability $\frac 13$) and after the first throw has come out $1,2,3,4$ (probability $\frac 23$) you are left with the original situation except with a parity change. So the probability that the first occurrence is on an odd throw is $P_O=\frac 13+\frac 23 P_E$.
Substitute this back to obtain $\frac 13+\frac 23P_E+P_E=1$ which then gives $P_E=\frac 25$

Suppose some event happens with probability $p$ on each throw. We can do similar calculations with
$P_E+P_O=1$
$P_O=p+(1-p)P_E$ so that 
$P_E+p+(1-p)P_E=1$ and 
$P_E=\cfrac {1-p}{2-p}$
In the first version of this answer I mistakenly took $p=\frac 12$ which gave $P_E=\frac 13$
A: Can you make use of this fact?
$$P(\hbox{A roll of 4 or higher}) = \sum_{n=1}^\infty P(\hbox{first
roll of 4 or higher occurs on trial 2n})$$
A: The events that you mention are not independent (see the comment of quasi).
Further note that:$$P(\text{even number needed})=\frac46P(\text{odd number needed})=\frac46(1-P(\text{even number needed}))$$
Leading to: $$P(\text{even number needed})=\frac2{5}$$
So geometric progression can be avoided.
A: The argument has at least two fallacies:
Fallacy #1: Claim that because there are two possibilities for an outcome (odd or even), the probability of each must be $\frac12.$
Fallacy #2: Arbitrarily declare two events to be independent.
There is a third problem, which is that the two "events" are vaguely defined.
In order to work with probabilities of events together, the events have to be defined over some sample space.
What is the sample space in the argument? 
Is it a single throw of the die? In that case $P(A) = 1,$ since a single throw is an odd number of throws.
Or is the sample space the set of all possible sequences of throws that can occur, starting with the first throw of the described procedure and ending with the last throw of the procedure? If so, the procedure says we will stop tossing the die when a number greater than $4$ appears; hence in order for us to end on an odd number of tosses, we must have a number greater than $4.$ You can't have $A$ without $B$, and therefore $P(A \cap B) = P(A).$ The events are not independent.

But let's try applying your approach a little further:
You have computed that $P(A \cap B) = \frac16$ by your method.
What about the $P(A^C \cap B)$ -- that's the probability that the die is tossed an even number of times and comes out a $4.$
You have assumed $P(A) = \frac12,$ therefore $P(A^C) = \frac12,$
and you have assumed $A$ and $B$ are independent, therefore so are $A^C$ and $B$,
so 
$$P(A^C \cap B) = P(A^C)P(B) = \frac12\cdot \frac13 = \frac16.$$
OK, so now we have computed that "odd number of throws and greater than $4$" has (allegedly) probability $\frac16,$
and "even number of throws and greater than $4$" has probability $\frac16.$
Now what? What does this tell  us about the probability that the number of throws to reach a number greater than $4$ will be even?
A: Although a bit redundant (Mark Bennet gave an exhaustive answer for future references if need be), I'm still posting a very simple answer here:
Consider each two throws of the die starting with the first throw. We get the probability of getting a number greater than $4$ to be $1/3$ on the first throw. The probability on the second throw to get the number greater than $4$ is $(1-1/3)\cdot1/3=2/9$ (while not getting it on the first throw, of course). Each following couple of throws (third and fourth, fifth and sixth, etc.) will have the same ratio of probabilities that is 
$$\frac{2}{9}:\frac{1}{3}=\frac{2}{3}$$
The ratio we got is the ratio of the probability of getting a number greater than $4$ on an even throw to the probability of getting it on the odd throw in each two throws (first and second, third and fourth, fifth and sixth, etc.). This ratio is constant for each such couples of throws. Hence the probabilities are $0.4$ and $0.6$. They need to add up to $100\%$ in this situation, and their ratio is $2/3$. 
