Hey guys, I have a rather complicated (for my tiny brain!) probability question.

I have two compound events that could possibly happen, and I need to work out the probability of EITHER of the two statements occurring. To give some context, it's the probability of stopping a touchdown from being scored in a game of Blood Bowl (a dice-based fantasy American Football game). Basically there are two methods of doing it in the scenario I found myself in, and just for fun (the game has finished now!) I wanted to work out whether I made the right decision, statistically speaking.

I've worked out the basic percentages using a third-party application (and option one, the route I actually took, worked out as having a 56.66% chance of success) but the application isn't advanced enough to deal with the resulting questions I have.

Option two has two parts:

1: A single 5+ die roll to determine whether I pick up the ball or not, and a more complicated series of dice rolls to work out whether my opponent tackles me, the ball doesn't bounce out of bounds, he picks up the ball, then scores. So it's a 33% chance of me succeeding in my part, THEN a 37.04% chance of him succeeding his part. So I figure that's a 62.96% chance of him failing his rolls and thus me succeeding. How do I work out the chance of both parts working out in my favour? Is it simply 33% of 62.96? A 20.78% chance? Or is there more to it than that?

2: Here's where it gets complicated (for me!). If I fail that 33% chance roll (which obviously has a 66% chance of occurring), my opponent still only has a 59.26% chance of barging my player out of the way, picking the ball up and scoring. Thus there being a 40.74% chance of him not doing that (ie, me succeeding). Assuming my maths is correct in the first part of the question, that would make it 66% of 40.74 = 26.88% chance of that happening (me failing the 5+, then him failing his rolls).

So my final question is, how do I then work out the overall chance of ONE OF those options happening. Ie, either the 20.78% chance scenario happens, or the 26.88% chance scenario does? Is it just basic percentage adding? So (20.78+26.88)/2 = 23.83%? Or is it more complicated than that? I don't see why it would be, but... there's lots of things I don't see in mathematics.

Thanks in advance for any help with this. Now that I've written it all out and seen the final tallies, it sounds plausible. But I'm not too confident when it comes to statistics and 'advanced' maths, so I thought I'd ask.