Crackerjack

PBE6
Posers and Puzzles 31 Jan '11 18:23
1. PBE6
Bananarama
31 Jan '11 18:23
I may have asked this one already, but I can't remember so let's pretend it's new!

Imagine that as an avid collector of Crackerjack prizes, you set out to collect them all. Luckily, Crackerjack has scaled back on the variety of prizes available to just 4 types so the task will be less onerous than usual. You decide to order Crackerjack boxes in cases of 12 to take advantage of the savings down at Ludicrous Luke's Low-ball Local Ltd., but before you place your order you feel compelled to calculate the probability of collecting all 4 prizes at a stroke. So, how many cases would you have to order to be:

(a) at least 90% sure of collecting all 4 prizes?
(b) at least 99% sure?
(c) at least 99.9% sure?
2. sonhouse
Fast and Curious
08 Feb '11 03:29
Originally posted by PBE6
I may have asked this one already, but I can't remember so let's pretend it's new!

Imagine that as an avid collector of Crackerjack prizes, you set out to collect them all. Luckily, Crackerjack has scaled back on the variety of prizes available to just 4 types so the task will be less onerous than usual. You decide to order Crackerjack boxes in cases of 12 ...[text shortened]... least 90% sure of collecting all 4 prizes?
(b) at least 99% sure?
(c) at least 99.9% sure?
Looks like 4 factorial, 4X3X2, 24 so two cases to be 99.9% sure of getting them all.
3. 10 Feb '11 20:2012 edits
Assuming all four prizes are equally likely, there are 4^12n total prize distributions where n is the number of cases. Of those 4^12n total prize distributions, there are 3^12n where you never get one prize in particular. Each of these 3^12n combinations can happen in four ways, once for each of the four different prizes being missing. There are 4*(3^12n) prize combinations where you get only three of the four prizes (this introduces some double counting for cases where two prizes are missing, but that should be a very low percentage of the time -- we could add these back with more terms but I doubt it would change the answer much for 12n sufficiently larger than the number of prizes).

Thus the probability of not getting one of the four prizes is 4*(3^12n)/(4^12n) = 4*(3/4)^12n and the probability of not getting any of the prizes is p = 1 - 4*(0.75)^12n.

(a) We want (4*3^12n / 4^12n) < 0.1 so solve the equality:
4*(3/4)^12n = 0.1
log(4) + 12n*log(0.75) = log(0.1)
n = [log(0.1)-log(4)] / [12*log(0.75)]
n = [-1-0.6]/[12*(-.1249)] = 1.6/1.5 = 1.07 so you'd need at least 2 cases.

(b) Same problem with 0.01
n = 2.6/1.5 = 1.73 so you'd need at least 2 cases.

(c) Same problem with 0.001
n = 3.6/1.5 = 2.4 so you'd need 3 cases to pass 99.9% confidence.