After 9 purchases, what's the percentage certainty you'd have all four toys? If you had to pre-purchase all n at once from the outset, what's the value of n to be 99.9% certain you'd get all four?
Got it up to 17 billion and it flipped to 71.135%! I’m away from my computer and can’t remember more digits.
I’m really surprised by this. I knew things converged slowly, but I didn’t appreciate how much and for how long it would move here. I’d be very appreciative if anyone calculated the number.
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
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.
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?
Thanks for doing that. I suspected those answers were very different than the 8.3, which I'm guessing must have been around the 50th percentile. The way the initial question was asked, it seemed to suggest you had a pretty high probability of all four after 8.3, which isn't true.
It hinges on the "expect" in how "much do you expect to pay?".
You can think of it like "expected value", where you're doing a comparison. Those meals are $8 each. If someone says they'll give you $80 for a set, this would be a profitable endeavor. On average, you'd expect to spend $66.67 (8.33*8) to get a collection.
If you think of it like "setting expectations", it's not so good. If a mafia boss told you he was expecting you to get all four toys for his daughter, you're not going to place an order for nine meals and walk in there with 71% confidence. I might show up with 59 of them, which is the p99.9999 value.
