### What's the odds for armour breaks?

It depends on the total number of armour rolls made. He didn't roll 7 sets of 2d6 in isolation. He rolled a whole bunch of armour rolls the whole game and that means there was a sequence of n armour rolls within which there was a sequence which went [no break]-[7 breaks]-[no break], unless the 7 breaks were the first 7 or the last 7 in which case it was [7 breaks]-[no break] or [no break]-[7 breaks].

There are 3 cases there, of which 2 (the start and end) have 1 in n possibilities and the middle has n-[rolls-1] in n possibilities (i.e. they could have happened at any time other than the start or the end and you can fit in that sequence of length n one fewer time than the number or consecutive successful rolls made).

Given the AV break odds of 1/6 and the no-break odds of 5/6 we can say:

Odds of it being the first sequence = (1/6 * 5/6)/n Odds of it being the last sequence = (1/6 * 5/6)/n Odds of it being in the middle of the sequence = (1/6 * 5/6 * 1/6) * (n-[rolls-1])/n

So the total odds are 2* (1/6 * 5/6)/n + (1/6 * 5/6 * 1/6) * (n-[rolls-1])/n

= (10/36)/n + (5/216) * (n-[rolls-1])/n

So if he made 25 successful blocks, and therefore armour rolls, in the game then n = 25 with 7 successful rolls then the total odds are:

= (10/36)/25 + (5/216) * (19/25)

= 0.0111 + 0.231 * 0.76

= 0.0287

=2.87%

So, pretty unlikely but not impossible by any stretch. To answer the question properly, though, we'd need to know the total number of AV rolls made. Onesandskulls.com might help with that.