As I wrote in a comment, if you're really going for optimal play, you need to consider the entire remaining game. The game has a finite but considerable number of possible states: There are $7$ different states for each of the $1$s through $6$s slots, $27$ different states for each of the three-of-a-kind, four-of-a-kind and chance slots, three binary states for full house and the two straights, and a count of yahtzees that you can probably safely limit to something like $0$ to $4$, for a total of $7^6\cdot27^3\cdot3^3\cdot6=375141013254$. That's a bit too much to keep in memory, but after two or three plays the number is reduced considerably and it becomes feasible to calculate the value of the remaining game by dynamic programming. That allows you to focus on a single play, since the later stages of the analysis tell you for every given final result of the current play which slot to use it for.
Assuming that you do know which slot to use which final result for, the problem you describe in the question (if I understand it correctly) isn't a problem. Every final result has a certain value, given by the maximum of all game values that would result from using it in the slots it can be used in, and you just need to optimize the expected value of these final result values; the probabilities of the final results add up to $1$, and the probabilities of getting a result that could be used for a particular slot are just a distraction; what counts isn't what slot a result could be used for but only which slot you actually will use it for in optimal play.
[Edit:]
The numbers of states come about as follows: In the slot for the $1$s, you may have entered anything from $0$ to $5$ $1$s, which makes six states, or you may not have entered anything yet, which makes seven. The slots for three-of-a-kind, four-of-a-kind and chance can contain any of the sums from $5\cdot1=5$ to $5\cdot6=30$, which makes $26$ states, or nothing, which makes $27$. The slots for full house and the two straights can each contain either nothing, zero, or the value of that slot, which makes $3$ states. The yahtzee count can in principle go up to the number of plays in the game, $13$, but I've never seen anyone roll more than $3$ yahtzees in a game, and I think it can be safely limited to $4$ or so; that makes five different states for the counts from $0$ to $4$, and one for nothing, which makes six.
When you roll 2 dice there are 36 equally likely options, starting at double one, $(1,1)$, and going all the way to double six, $(6,6)$. Of these, there are 11 equally likely ways to roll a $5$, $(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3),(5,4),(5,6)$, so the odds of rolling a 5 is $\frac{11}{36}$.
Similarly there are 18 equally likely ways to roll an odd number (I'll let you think about which combinations these are), so the odds of rolling an odd number are $\frac{18}{36}=\frac{1}{2}$.
At this point you may think the odds of rolling a $5$ or and odd number should be $\frac{11}{36}+\frac{18}{36}=\frac{29}{36}$, however this is not the case. Note that six of the rolls contain both a $5$ and are odd:
$(2,5), (4,5), (6,5), (5,2), (5,4),(5,6)$
If we just add the two probabilities, we'll count these situations twice, even though they're no more likely to occur than any other combination! In reality there are only $11+18-6=23$ dice rolls which contain a $5$ or are odd (see if you can list them), and so the odds of rolling either a $5$ or an odd combination is $\frac{11}{36}+\frac{18}{36}-\frac{6}{36}=\frac{23}{36}$.
This is a simple application of the "principle of inclusion and exclusion", which you may want to look up.
Best Answer
The probability that you roll 4+ is the probability that you roll a 4, a 5, or a 6. Each of these events has probability $\frac{1}{6}$; so, the probability of rolling 4+ is $\frac{3}{6}=\frac{1}{2}$.
For your other scenario, note that the event that you get a 5+ when you are allowed on re-roll can be broken up in to two disjoint events: event $A$, in which you roll a 5+ on the first try; and event $B$, in which you roll a number 1-4 on your first attempt, and either a 5 or 6 on the second.
Then $P(A)=\frac{2}{6}=\frac{1}{3}$, since this is the event that you roll either a 5 or a 6. For $B$, we have $$ P(B)=\frac{4}{6}\cdot\frac{2}{6}=\frac{2}{9}, $$ since you must roll a 1, 2, 3, or 4 on the first attempt and either a 5 or 6 on the second. So, overall, the probability of getting 5+ when you allow one re-roll is $$ P(A\text{ or }B)=P(A)+P(B)=\frac{1}{3}+\frac{2}{9}=\frac{5}{9}. $$ (Note that we have used here that $A$ and $B$ are disjoint possibilities.)
So, you are more likely to get a 5+ with a re-roll allowed than to get a 4+ with no re-roll.