The simplest answer would be for independent non-mutually exclusive events, given one "roll of the dice," as they say. Given one opportunity, the chance that at least one event would occur would be (1/12)+(1/18)+(1/277), I would think.
It's still not even that simple, because that's not the question that was posed. That's why I asked.
You're trying to boil this hypothetical scenario to a "roll of the dice." The question, as posed, can't be applied to that type of probability.
In a "roll of the dice," for example, let's say there's a single die, and it's weighted. In this case, event 1 happens 1 time in 12, event 2 happens one time in 18, and event 3 happens 1 time in 277. Yes, in that particular case, the probability of either event 1, event 2, or event 3 occurring on a single roll is indeed (1/12)+(1/18)+(1/277).
But you've overlooked one very significant reality in that scenario: when rolling a single die, as in your example, if the result is that event 1 occurs, then necessarily, event 2 does not occur and event 3 does not occur. In this case, yes, it's as simple as adding the three probabilities together.
That's not the question that was posed, and that's why I said there's not enough information given. That's why I asked whether any of the events were exclusive. If there's a chance that event 1 AND event 2 can occur, then the math is no longer as simple as adding the three probabilities; the same is true if event 2 and event 3 can occur simultaneously, or if all three can occur simultaneously.