Quick dice rolling maths question

[Checkmate]

A friend and I are playing backgammon, and we both had one piece on the 1 point. We both rolled let's say 20 times (it's about there) and the 1 came in on my friend's 21st roll. What are the odds of him not rolling a 1 until his 21st roll of 2 dice? I'm struggling to manage my brain with the *not* part of this!

[NowWatchThisDrive]

His 21 dice rolls are all independent events, with the probability of his not rolling a 1 on any single event being 5/6

So ((5/6)^20)*(1/6) = about 0.43%

[RizzoTheRat]

Odds of not rolling a 1 on one roll is 5/6.

Odds 2 rolls is 5/6 * 5/6, as you have to not get a 1 on both rolls

Odds on n rolls is (5/6)^n

So on 20 rolls its (5/6)^20 which is 0.026, or 1 in 38


ETA: Beaten by seconds, what's the odds of that? (our different results are because I didn't include rolling the 1 on the 21'st)

Edited by RizzoTheRat on Friday 16th April 21:34

ETA: Oops, just realised it's 2 dice each roll, so (5/6)^40 *11/36 = 0.000208 or 1 in 4810. the 11/36 is the chance of getting one or both of the last pair as a 1

Edited by RizzoTheRat on Friday 16th April 21:46

[NowWatchThisDrive]

NowWatchThisDrive said:

His 21 dice rolls are all independent events, with the probability of his not rolling a 1 on any single event being 5/6

So ((5/6)^20)*(1/6) = about 0.43%

Actually just realised OP said there are 2 dice. So probability of no 1 on any roll of 2 dice becomes (5/6)*(5/6). However as above posted alluded, you're interested in the probability that it takes him at least 21 rolls, not 21 rolls precisely, so including the last multiplication as I did above doesn't make sense.
So...((5/6)*(5/6))^20 = about 0.06%

[RizzoTheRat]

I love that we both realised our mistake and ninja'd each other again

[Checkmate]

That makes sense to me now, thanks guys!
Quite remarkably unlikely then! Explains why it hasn't played out quite like that before...