[colmanhumphrey]

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]

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?

[travisjungroth]

The reply button is back so I can thank you publicly!