[bushindo]

I remember long ago McDonald used to run an ad campaign after every major movie. They would make a set of, say, 10 different figurine from a specific movie, and include supposedly a random figurine in every kid's meal. The exciting encouragement after the commercial is always "Collect all 10!".

So, here's the puzzle. Suppose that you are a kid who wants to collect all 10. Assume that every time you go to McDonald, they pick a random figurine from the set of 10 and include it in your meal. How many times, on average, do you expect to whine and drag your parents to McDonalds before you get a complete set of 10?

Edited July 14, 2009 by bushindo

[HoustonHokie]

Well, no, not really.

This may be oversimplifying the calculations, but I think what you're really looking for is the number of trips it takes to get a different toy when you already have 0, 1, 2, 3, ..., 9 of the 10 collectibles. So, if there are 10 toys, and I don't have any yet, then there are 10 new toys available whenever I go. So I expect to get a new toy every 10/10 = 1 trip. After that, there are 9 new toys available, so I'd expect to get a new toy every 10/9 = 1.111 trips. And so on and so forth. Adding it all up, I'd expect a complete set after 29.3 trips.

And there's always the new Batman toys coming out next month, in case I couldn't make my way to McD's every day this month...

[Guest]

i got

n

sum(n/k)

k=1

which is ~29.2897

via trial program:

Average over 50000 iterations = 29.2352 trips

(close enough )

the worst thing about these promotions is that it always seemed like there was ONE particular toy that NO ONE could ever get their hands on.

Edited July 14, 2009 by tpaxatb

[Guest]

The chances of getting all ten toys in ten trips is a whopping 0.0363% Good luck little johnny! If you go to the arches ten times a month in hopes to get the new set each month, it's gonna take you 229.64 years just to finally have collected one complete set. Happy hunting!

[Guest]

i would have to say on average or the central tendency of our data pool would point to an infinite amount of trips since most kids never achived a full set of 10 toys.

[Guest]

FYI...lol

As a past mamager of Micky D's...I can relay this information.

A store is allowed to order 1 time a week.

As these promotions are on going...(and as we ordered),

A store is sent 1 to 2 different toys per week.

And by the end of the "certain" permotions, a store is sent left overs

from the warehouse.

[Guest]

FYI...lol

As a past mamager of Micky D's...I can relay this information.

A store is allowed to order 1 time a week.

As these promotions are on going...(and as we ordered),

A store is sent 1 to 2 different toys per week.

And by the end of the "certain" permotions, a store is sent left overs

from the warehouse.

you beat me to it...lol

[Guest]

i would have to say on average or the central tendency of our data pool would point to an infinite amount of trips since most kids never achived a full set of 10 toys.

This would only be true assuming an infinite number of outcomes (a continuous distribution) -- if there were a continuous distribution, then it is true that a single point would be theoretically impossible (since it would be the equivalent to trying to find the area of a single point under a curve of a continuous probably function). However, in this case, there are a finite number of outcomes (a discrete distribution) -- determining the probability of a single outcome should generally be possible.

[Guest]

in the spirit of the problem

I just thought of this real quickly but i think it works

going backwards you expect to get the last toy in 10 trips, the second to last in 5, the third to last in 10/3, and so on and so on this adds to 29.28968254 trips, so for Americans chances are this would take two days (we love our mcdonalds)

[Guest]

If the parents were smart then It would only take One trip. thiswill work because if you offer the person working there enough money they will dig through the box and give them to you( I know that for a fact)

[Guest]

If the parents were smart then It would only take One trip. thiswill work because if you offer the person working there enough money they will dig through the box and give them to you( I know that for a fact)

the only problem with that is that alot of the collectables have started being put in non see-through bags. so you'd have to bank on being able to feel the difference, assuming there isnt a case of any sort.

[superprismatic]

It is easy to prove that, if you need k more toys, the expected wait to get at least one of them is 10/k. All you need to do is a bit of algebraic manipulation on an infinite convergent power series. So, the exact answer *really* is (10/10)+(10/9)+(10/8)+...+(10/1).

[Guest]

I grew up with a kid named Mastif Koranski. His dad was Bruno "Killer" Koranski, a huge, hairy, professional wrestler. I can still hear Mr. Koranski saying, "Yo-----you there----kid. What'dyous get kid? That right? OK now give it to my kid." Generally only took one trip.