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


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.

/r/bloodbowl Thread