Most Expensive. Collectibles. Ever.

Though many would argue that The Simpsons has overstayed its welcome and is no longer the comedic force it once was, it still has plenty of cultural cachet.  Taking into account the profits from video games, comic books, clothes, and Squishee machines, I’m sure this show continues to keep a number of people very, very rich. Such wealth, of course, comes at the expense of folks like this guy:

One of the more recent trends in the world of Simpsons memorabilia is the advent of the Mini-Figure collections, produced by Kidrobot.  Each series (there have been two so far) consists of around 25 small Simpsons figures, each with his or her own accessories.  The figures cost around \$10 each (\$9.95, to be precise), so an avid collector would need to spend something like \$250 to complete each of the two collections, right?

Well, not quite.  When you buy one of these figures, you have no idea which one you’ll get, because the box containing the figure doesn’t indicate what’s inside.  All you know are the probabilities for each figure, and even those are sometimes missing!  Here’s how things stacked up for the first series of toys (click on the image to embiggen):

Given this information, here’s a natural question: how many of these boxes should you expect to buy if you want to complete the set, and how much will it cost you?

Before we can answer this question, we need to deal with the three toys having unlisted probabilities.  The given probabilities add up to 43/48, so we somehow need to decide on a way to allocate the remaining 5/48 among the three secret toys. Since the other toys all have probabilities of 1/12, 1/24, 1/48, or 1/96, in keeping with these values I’ll assume the three missing probabilities are 1/48, 1/24, and 1/24 (in no particular order). With these values, the total probability is 1, or 100%, as it should be.

Given this list of twenty-four toys and their corresponding probabilities, how can we determine the average number of purchases required to complete the set? The most obvious thing to do is to simply buy a bunch of toys and see what happens. However, to get a good average you’ll have to do this quite a few times, at an expense of thousands of dollars.  Dropping that kind of coin does not seem like a particularly good use of company funds, and so I was forced to find a more cost-effective approach.

That’s where my computer came in handy.  Using the given probabilities, I ran a simulation 100 times to see how many purchases I’d need to make in order to complete the set of 24 figures.  Here’s a list of the number of purchases required for each of the 100 trials:

 147 110 73 132 121 323 141 150 119 310 213 177 185 142 146 142 336 155 124 135 205 147 136 232 206 124 130 138 217 316 92 235 134 142 116 421 124 169 94 173 124 118 155 291 219 255 95 123 141 118 279 778 78 158 136 93 174 201 82 110 162 145 160 168 226 116 148 167 185 104 140 218 289 193 82 128 410 79 361 245 235 171 287 58 121 197 97 176 137 165 272 121 261 167 77 125 201 328 207 117

While it’s theoretically possible to complete the set in just 24 purchases, it’s highly unlikely (bonus question: how unlikely?!).  Among these 100 computer-generated buyers, the first to complete the set needed only 58 purchases, while one unfortunate digital soul needed 778!  The average of this list is 179.06 – let’s call it 180, just to be safe.  If you need to buy an average of 180 of these toys, at a cost of \$9.95 each, that’s \$1,791 you can expect to spend!  Far more than the \$238.80 you’d have to spend if you got to choose your toy.

We can also compare this to what you’d expect to pay if each toy were equally likely to be in the box (i.e. if all the probabilities were 1/24).  This is known as the coupon collector’s problem, and in this case, if you are trying to collect n objects, the expected number of purchases you need to make is equal to

n(1 + 1/2 + 1/3 + … + 1/n)

(you can check the link for more details on this calculation).  In this case, n = 24, and the above expression is approximately 90.62.  In other words, the chosen probabilities nearly double the number of toys you need to buy, compared to the case where all the toys are equally likely!

If you think that’s bad, it gets even worse.  Check out the probabilities for the second series of figures (once again, you can click to embiggen):

This collection has just one additional toy (25 instead of 24), but also has quite a few more hard-to-get toys.  Let’s assume the zombie versions of the Simpson family all have the same probabilities, i.e. zombie Homer and zombie Bart each have a 1 in 100 shot of being inside a given box.  This leaves a 5% probability left over between the two remaining figures with unknown probabilities (Hans Moleman and the afro-sporting mystery toy).  If we assume these two toys are equally likely to be found, then each one has a 1/40 (=2.5%) probability of turning up inside a box. This gives us a complete list of probabilities from which we can do the same analysis as before.

The experimental data hits the wallet even harder this time around:

 306 240 210 182 111 368 445 283 462 192 89 94 238 112 242 186 110 129 176 196 148 340 150 282 107 291 411 278 90 140 101 405 186 220 262 136 336 69 422 189 186 153 173 119 415 196 211 178 195 190 101 309 214 656 182 149 179 183 210 229 381 156 433 129 527 233 222 358 669 178 115 215 128 286 505 112 107 458 204 275 298 555 399 179 230 114 194 193 165 273 288 389 204 130 158 168 316 163 119 306

While the range turns out to be smaller in this set of experimental data (smallest value = 69, largest = 669), the average is over 30% higher.  For this series, you can expect to have to buy 238.94 toys.  Let’s call it 239 – that’s \$2378.05 you’d expect to spend!   Compared to the amount you’d expect to spend if you got to pick your own toy (25 toys for \$248.75), or if each toy were equally likely (around 96 toys for \$955.20), this is not a modest investment.

That’s all well and good.  But the mathematician in me craves something a little more precise.  These experimental averages are nice, but how close are they to the truth? To answer this question, it would be nice to have a general formula: one which would allow you to vary the number of toys, and the probabilities of each toy, to your heart’s delight.  Thankfully, such a formula exists!  If you’d like to learn about it, you can check out this discussion and the references therein (but be warned, beyond here there be calculus).  The general formula is not particularly nice once you start plugging in values for the probabilities, so let me cut to the chase.  The theoretical expected values for the number of purchases you’d need to complete each of these collections are 176.09 for the first series, and 233.35 for the second. So it looks like our experimental averages (179.06 and 238.94) aren’t too far off the mark.

If you look on eBay, you’ll see some of these toys being sold at a significant markup. But when you crunch the numbers, maybe spending up to ten times the retail price isn’t such a bad deal, at least if you’re a collector.

Teachers: interested in exploring some of these ideas with your students? Then check out our new Happy Meal lesson!

### 2 Responses to “Most Expensive. Collectibles. Ever.”

1. [...] Most Expensive. Collectibles. Ever. | Mathalicious [...]

2. [...] post is taken from here. You should check out the original. It has video and [...]