I got 71.13647% on a first pass. So we want P(all 4 in 9) = 1 - P(not all 4), and we can split that out a few ways. To not get all four, we can restrict ourselves to three, so that's (3/4)^9, but there are four ways of doing that, so that's 4 * (3/4)^9. But that counts using singles and pairs too many times. Specifically each version of "three" can be exactly three balls, three ways of one ball, or three ways of exactly two balls ("1 or 2 or 3" = "1&2&3" or "just 1" or "just 2" or "just 3" or "1&2" or "1&3" or "2&3").

- We can then subtract 6 * P(two balls), so 6 * (2/4)^9. Now this counts singles a few times too, in fact it cancels all of them out.

- We then need to add back four singles, so 4 * (1/4)^9

Putting this together gives:

1 - (4 * (3/4)^9 - 6 * (2/4)^9 + 4 (1/4)^9) = 0.7113647