[tunesmith]

This took me way too long to understand. :)

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]

exactly! It's a version of https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_pr...

[tunesmith]

Don't know if you're still reading this thread, but one thing I'm stumped on is that if I plug in 8.333 instead of 9, I get a probability of around 65%.

If I plug in 7, I still get a probability of above 50%, which seems to conflict with the 8.333 answer calculated upstream.

Wouldn't a positive EV imply that a 50% probability is the break-even?