- 19 Aug '09 15:151. Joey loves Crackerjack, but he loves the prizes even more! Consider the case where there are only 4 types of Crackerjack prizes available. If there is always 1 prize per box, and each type of prize is as likely as any other to be found in any given box, how many boxes will Joey have to purchase in order to be at least:

(a) 50% sure that he will end up with at least 1 of each prize type?

(b) 90% sure?

2. Generalize this result for confidence level "c" with "k" prize types. - 20 Aug '09 19:07

Instinct says,*Originally posted by PBE6***1. Joey loves Crackerjack, but he loves the prizes even more! Consider the case where there are only 4 types of Crackerjack prizes available. If there is always 1 prize per box, and each type of prize is as likely as any other to be found in any given box, how many boxes will Joey have to purchase in order to be at least:**

(a) 50% sure that he will end up w ...[text shortened]... ?

(b) 90% sure?

2. Generalize this result for confidence level "c" with "k" prize types.

A) six boxes

B) 10 boxes. - 26 Aug '09 08:41

Yep, definitely harder than it looks. I don't have a general solution (looks to involve lots of combinatorics that I can't be bothered trying to work out!), but with a semi-brute-force approach I get:*Originally posted by PBE6***Close but no cigar! This problem is actually much more complicated than I thought, but I do have answers...**

1.a) 7

1.b) 13

I worked out the following difference relation that helped. If P(n, q) is the probability that if I open n boxes I get q distinct prizes then:

P(1, 1) = 1

P(n + 1, 1) = P(n, 1)/k

and for q > 1

P(1, q) = 0

P(n + 1, q) = P(n, q).q/k + P(n, q - 1).(n - q - 1)/k - 26 Aug '09 18:33

Correct! I spent a lot of my time trying to come up with the number of arrangements possible given the number of boxes "n" and the number of possible prizes "k", but then I realized that I had completely left the probabilities out! After coming up with an ugly but functioning answer, I stumbled upon a much neater one online:*Originally posted by mtthw***1.a) 7**

1.b) 13

http://ranmath.webng.com/problems/crack.htm - 26 Aug '09 20:08 / 3 edits

oops, I did it wrong in my head. Worked out when i wrote it down.*Originally posted by PBE6***Close but no cigar! This problem is actually much more complicated than I thought, but I do have answers...**

4 boxes odds of 1 differnet prizes are 100%

4 boxes odds of 2 differnet prizes are 50%

4 boxes odds of 3 differnet prizes are 25%

**4 boxes odds of 4 differnet prizes are 12.5%**

Therfore, in order to get to 50% we need to figure out how many more % we need.

50- 12.5= 37.5%

We need enough boxes to give us 37.5% more.

So, 37.5/12.5 = 3 boxes.

Therefore, to get 50% we need 4+3 boxes = 7 boxes.

------------------------------------------------------------------------------

To get 90%,

90-12.5=77.5%

To get 77.5 we divide by 12.5 and get 6 boxes. (answer is 6.2 but we can't have .2 of a box..round down!)

Now, we know that 7 boxes equals 50% so 7 boxes plus 6 boxes will give us 90%.

Therefore, 90% = 13 boxes.

- 06 Sep '09 01:41

Your argument can be used to determine how many boxes for 100%, which is impossible.*Originally posted by uzless***oops, I did it wrong in my head. Worked out when i wrote it down.**

4 boxes odds of 1 differnet prizes are 100%

4 boxes odds of 2 differnet prizes are 50%

4 boxes odds of 3 differnet prizes are 25%

[b]4 boxes odds of 4 differnet prizes are 12.5%

Therfore, in order to get to 50% we need to figure out how many more % we need.

50- 12.5= 37.5% ...[text shortened]... boxes equals 50% so 7 boxes plus 6 boxes will give us 90%.

Therefore, 90% = 13 boxes.

[/b] - 07 Sep '09 18:46 / 1 edit

not only that, but he said that 77.5% = 6 boxes, and then added the 6 boxes to his previous calculation for 50%... proof by coincidence rather than finding a solution.*Originally posted by AThousandYoung***Your argument can be used to determine how many boxes for 100%, which is impossible.**

essentially, he used a linear approximation algorithm for a non-linear computation; as the percentage approaches 100% the number of boxes diverges further and further from a linear increase. so even though his method is relatively accurate at 50%, it fails for 90% (when implemented properly, his approximation projects an answer of 10.2 boxes, where 13 boxes is correct). - 08 Sep '09 23:31

i think his point was that your method can't possibly be correct because you could use it to find an "answer" for a case that can't possibly be true. by showing your method's failure at the extremal value, he showed a flaw in the logic of your argument that should be re-examined.*Originally posted by uzless***no, as in many other parts of math, we can reject an answer that is impossible.**

yes, the calculation you performed outputted the correct answer in this particular case... but it seems that was largely coincidental, since your method fails to compute accurately for higher percentages (e.g. 90%, 100%, etc.). - 09 Sep '09 15:08 / 7 edits

Look, even Einstein manipulated things to fit his basic design. Discounting an answer that is impossible is just good judgement. This method may not be clean, but it does produce the correct answers.*Originally posted by Aetherael***i think his point was that your method can't possibly be correct because you could use it to find an "answer" for a case that can't possibly be true. by showing your method's failure at the extremal value, he showed a flaw in the logic of your argument that should be re-examined.**

yes, the calculation you performed outputted the correct answer in this ...[text shortened]... ince your method fails to compute accurately for higher percentages (e.g. 90%, 100%, etc.).

This method relies on computational ratios. Ratios have calculations built into them that provide us with short-cuts. If I told you I have 2 apples for every 3 apples you have, we could write that as 2/3. Automatically we know that I have .66 apples for every 1 you have. If someone wanted to know how many apples I had if you had 10 apples, we'd only have to mulitply .66 by 10 to get the answer. The ratio (or percentage if you prefer to think of it that way) calculates the answer for us. In the same sense, we can use ratios to determine how much more of something we need if we know how much we have. In this case we knew how much of a confidence level we had with 4 boxes (12.5% ) so all we had to do was perform a ratio calculation to determine how much more we needed.

The question usually provides some inherent answers too. In this case 50% and 90% were asked for. 50 and 90 give you nice round numbers of boxes. The "correct" answer method provided by mtthw (i think) it was was fine...but use it for say 65% or 85%. Will that answer give you a nice round number of boxes?

The question is providing secret information that you can use to build a ratio calculation. You just have to be prepared to reject an answer that you know cannot be correct...in this case the case for 100% (and 0% )

The question really is giving away the answer. Really, the key here is the boxes. You cannot have parts of a boxes. You can only have whole boxes. So you can either have 4 boxes or 5 boxes or 10 boxes or 11 boxes etc. You cannot have 11.4 boxes. So, really, the question should have been phrased as,

"you have 13 boxes. there are 4 prizes. At what confidence level are you at that you will have all 4 prizes. This way you have no idea what the answer will be...it could be 90%, it could be 91 it could be 84..hell it could be 21.5%, you have no idea until you calculate it.

But, ask how many boxes you need to get to a 90% confidence level and you KNOW that the asnswer is going to be a whole number because you can't have a decimal number for an answer since a decimal number would mean you don't have a whole box. - 09 Sep '09 21:24

what you've failed to recognize though is that computational ratios, in this problem, only provide a LINEAR approximation to the answer which is a NON-LINEAR problem. you say it outputs the correct answers, but in fact your approximation only gave the correct answer for the 50% confidence level (because at that point in the "probability vs. # of boxes" curve a linear approximation is very close to accurate).*Originally posted by uzless***Look, even Einstein manipulated things to fit his basic design. Discounting an answer that is impossible is just good judgement. This method may not be clean, but it does produce the correct answers.**

This method relies on computational ratios. Ratios have calculations built into them that provide us with short-cuts. If I told you I have 2 apples fo ...[text shortened]... decimal number for an answer since a decimal number would mean you don't have a whole box.

when it came time to compute the 90% answer, using a little bit of hazy logic and a misapplication of your computed numbers, you added 6 to your PREVIOUS answer of 50%, rather than to the 4 boxes associated with a 12.5% confidence level. so you achieved the correct answer of 13 boxes, though your method (when computed accurately) should have given you an answer of 10 boxes. it seems to me that this is akin to "fudging the books" when you know what the answer**should**be, but the math comes out wrong.

i don't mean this to come off as antagonistic, or anything of the sort... in fact i like your computational ratio method for the earlier, more linear, part of the curve - it serves as an easy way to get a fairly accurate approximation. but i would ask you to consider some of the arguments put forth here that show your solution is invalid for a wide range of the probabilities that could be asked for, and therefore is not a very sound solution. in fact, try to calculate a 99.99% confidence level using computational ratios and i think even intuition would tell you that your answer is way off base.