If a die is rolled what is the probability of getting an odd number or a number greater than 4

In saying that "5 is counted twice", meaning (presumably) you're removing the duplicate event $(5, 5)$, what you should be doing is saying "The event $(5, 5)$ is part of both of the cases I've considered, so I need to only count it once, so I will remove it once from my calculation."

Instead, what it looks like you've done is removed it from both of your cases, each of which assumes the other case has already counted it.

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?

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