“Expected number of viewings” is perfectly well-defined. Let’s say that we have a very large number of movie-watchers, each with a DVD player with a busted remote, and they all watch the movie again and again until they see all of the endings. Some of them will only watch it 3 times, some of them will watch it 4 times, a very small proportion of them will watch it a gazillion times, and so on. Add up all of those viewings, and divide by the number of viewers, and you’ll have the expected number of viewings.
But couldn’t it be interpreted that way?
Suppose, to simplify the problem, there are only two endings, each with a 0.50 probability of showing up each time the movie is played. (The three-ending case is more complicated, and is left as an exercise for the interested reader. 
) Suppose you keep watching until you’ve seen both endings. What’s the expected number (in the long-term-average sense) of times you’d have to watch?
If my reasoning is correct*, there’s a 1/2 probability you’d have to watch exactly two times, a 1/4 probability you’d watch exactly 3 times, a 1/8 probability you’d watch exactly 4 times, a 1/16 probability you’d watch exactly 5 times, etc.
So the expected value of the number of times would be
2*(1/2) + 3*(1/4) + 4*(1/8) + 5*(1/16) + …
= 3**.
*Here’s my reasoning on the probabilities:
For your first two viewings, the endings you see are either AA, AB, BA, or BB, giving you a 2 out of 4 (= 1/2) probability of seeing both.
The probability that you’ll need exactly 3 viewings is the probability that the first two weren’t enough (1/2) times the probability that the third time you see the viewing you haven’t seen yet.
The probability that you’ll need exactly 4 viewings is the probability that the first three weren’t enough (1 – 1/2 – 1/4 = 1/4) times the probability that the fourth time you see the one you haven’t seen.
And so on.
**I got the value of 3 by brute-force calculation (in Excel). Does anyone know of an easy way to prove that the sum of this infinite series is 3?
You are describing a different problem than me. Maybe that’s part of the issue.
I was describing how to calculate the odds of getting at least one head for a number of flips given in advance. The odds of getting at least one head in two flips is indeed 75% (that is, 25% of the outcomes have no heads).
You are describing the person who keeps flipping until he gets the first head then stops. In that case, the answer is more complicated, similar to **Indistinguishable’s ** answer.
Now allow me to move from argumentative mode to “I don’t get it” mode. 
There’s still something I don’t get about this solution. Let’s go back and look at **Indistinguishable’s ** answer of 5.5. I understand what this means mathematically but not in practical terms. How does this answer the OP question, " the number of showings most likely needed to see all 3 endings"? Does that mean if I start viewing until I see all three endings then stop, that my most likely stopping point will be 5.5 viewings? If so I see where we’re going but I didn’t understand that was what the OP was asking. As I said a second ago, I thought the OP was looking for the a priori number of viewings needed to see all three endings.
Yes. Let’s assume, for the time being, that the series does converge, to some value K. Then we have K = 2/2 + 3/4 + 4/8 + 5/16 + … = 1/2 + 1/2 + 1/4 + 2/4 + 1/8 + 3/8 + 1/16 + 4/16 + … = (1/2 + 1/4 + 1/8 + 1/16 + …) + (1/2 + 2/4 + 3/8 + 4/16 …) = 1 + (1/2 + 2/4 + 3/8 + 4/16 …) = (1 + 1/2) + 1/2 * (2/2 + 3/4 + 4/8 + 5/16 + …) = 1 + 1/2 + 1/2*K. Solving for K, we get 3.
Cool!
It doesn’t.
The “most likely” necessary number of viewings is 3 or 4, with equal probability. There is a 22.22% chance the third viewing will be the clincher, a 22.22% chance that the fourth viewing will be the clincher, and a probability which slowly diminishes to zero for larger numbers.
The average number of necessary viewings, over a large number of trials, will be 5.5.
For completeness’s sake, as for how to show that the series actually does converge, one can use the ratio test (the limiting ratio of successive terms will be 1/2; as this is less than 1, the series must converge).