“This was a good fun match that was hard fought, especially with my team's big TV advantage going in.

There was a little disagreement over a +5 with RR roll I did to pick up the ball, so I just wanted to go over the math which we couldn't agree on in match.

I had a 2/6 chance of succeeding on my first roll. If I succeed, no need for further rolls of possibility of failure. So fight off the bat even without a RR I had a 12/36 (or 1/3) chance of succeeding.

Now if I fail on my first roll, I have another go at the same odds. The probability of me failing my first roll is 4/6 and my probability of succeeding on my second roll is 2/6. To calculate the probability we times the numerator and denominators. This comes out at 8/36 chance of failing on my first roll but succeeding on my second roll.

8/36 + 12/36 = 20/36 (aka, 5/9 or more than 50% as I stated).

The chance for failure was 4/6 * 4/6 = 16/36. You can see that 20/36 (my chance of success) + 16/36 (my chance of failure) adds up to 36/36, covering all possibilities.

So it was a risky play, but one where I had a better than 50/50 chance and thought it was worth the risk.

Your 2/6 * 2/6 calculation was wrong because

- If I succeed on the first roll I don't roll again.

- When you calculate the chance of succeeding on the second roll, it's 4/6 * 2/6 not 2/6 * 2/6. I can only roll a second time if I fail on the first roll, the chances of which happening are 4/6.

I'm sorry to go on about it, but it was hard to explain in match and it really helps to be able to gauge probabilities in Bloodbowl.”