# Adding Probabilities Together. Or, the probability of EITHER of two things happening.

• Feb 22nd 2013, 08:24 PM
monkeystyxx
Adding Probabilities Together. Or, the probability of EITHER of two things happening.
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. :p

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.