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
tunesmith|3 years ago
But I get it now. If the four prizes are ABCD, then if you calculate your chance of only getting two balls in five purchases, you can do it by calculating your chances to get "A or B" five times in a row. But those include the AAAAA and BBBBB scenarios, which aren't two balls.
Repeat that for AC, AD, BC, BD, and CD.
colmanhumphrey|3 years ago
travisjungroth|3 years ago