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?
After 9: 71.13%
p99.9: 29
Ran it up to 2 billion. Still holds, but no additional precision! It's somewhere around 71.134%, so it was threatening to round up for a bit.
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.
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.
- 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
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.
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?