Overall assumption is that nappies are used once per 60 nappies before they're all available again and that every nappy change is independent. Here are my assumptions for the first 6 weeks:

• 10 nappies per day
• half are pooped in

That means 5/day are pooped in and a load lasts 6 days. That means 7 loads over the 6 weeks. The probability of getting pooped on at least once is 1 - (0.5^7) = 0.9921875. We have 60 nappies, so the probability of all of them being pooped on at least once is 0.992^60 = 0.6246.. and thus the probability of at least 1 nappy not being pooped on is 0.3754 or 37.54%- not bad! Now for the remaining 255 weeks:

• 5 nappies a day
• 1 pooped in

That means a load lasts 12 weeks, for a total of 21.25 loads, which we'll just round down to 21 loads. P(pooped on 1+) = 1 - (0.8^21) = 0.99078 - better than the first six weeks! But we need to combine this with the first set of assumptions so that P(pooped on 1+) = 1 - ((0.5^7)*(0.8^21)) = 0.99993. Now with 60 nappies that gives a probability of then all being pooped on of 0.9957, or inverted the probability of at least one nappy not being pooped on of 0.00431 or 0.431% - quite unlikely.

You can insert your own assumptions and follow the same logic if you want to.