Saturday, December 15, 2007

Skittles: taste the bullshit.

Hey! Remember the "guess the number of Skittles/jelly beans/M&M's in the jar!" game? That game rocked—when you won. Basically, the Responsible Adults thought it'd be a great idea to give one lucky kid 2000% of his daily sugar intake in the form of a 32 oz. bag of Skittles they dumped into an attractive-looking jar. Yeah, you remember. Anyway, I've never organized one of those, but I was thinking... how did they figure out that there were 927 Skittles in that jar?

I mean, I assume they didn't literally sit down and count them one by one (Eww! Grubby Fingers on The Sacred Skittles?! Plus, humans make mistakes!). Eh, then again, maybe they did. Gross. I should've called Health and Human Services on their asses.

Now the cleverer Adults would probably have tried to use some sort of weighing technique, but how do you weigh with enough precision to get 927 and not 928 or 938 for that matter?? This is important. It's potentially the difference between going home with 927 Skittles all to yourself, or losing to that pimply faced brat who's got pull with the Adults 'cause he tattles on your ass in exchange for Skittle-bribes, that sonuvabitch.

Presumably, the protocol would go something lie this: weigh one Skittle, then weigh the jar, then weigh the jar with the Skittles, then divide the difference by the Skittle weight. If you wanted to be less rigorous, I suppose you could trust the "32 oz." on the outside of the bag, but that's bound to be off by a couple tenths of an ounce. Pay it no mind, you say? Nay, say I! Those two-tenths of an ounce represent 5.3 precious Skittles!

In all seriousness though (and I'm a very serious person), what scales are we talking about here? If you use your normal letterweight scale, you could probably determine the Skittle's weight to be, say, 0.03 oz. That's only one significant figure! If you divide that by the total weight of the bag, you'd have like a 100-Skittle margin of error!

To do this right, you'd need an all-out analytical balance that can weigh things out on a milligram scale, so the Skittle weight ends up being something like 737.1 mg.* Then if you wanted to be really rigorous, you'd measure 3 or 4 other Skittles so you could get an average and standard error. Your Skittle weight could then be something like 736.8 ± 7 mg. If you divided by the weight of the bag, which you also determined accurately, you'd get a good estimate of the number of Skittles in that bag (complete with confidence intervals!).

But did the Adults take all of these steps to ensure honest accountability? No. They lied to us. They said there were exactly 927 Skittles in that jar. They were wrong.


*Update: Using an analytical scale, I determined the actual mass of one Skittle = 1066 ± 8 mg (N = 10, ± s.e.m.), or 0.0376 ± 0.0003 U.S. ounces. On a scale of 927 Skittles, this translates to a standard error of about 7 Skittles.

7 Comments:

Anonymous Anonymous said...

That was highly entertaining, yet it seems like it could be a waste of your time. You've presented a problem of a standard error of 7 skittles-on a scale of 927 skittles of course-but does this "actual mass of a skittle" bring us any closer to the answer of how many skittles are in that jar?

3/23/2008 11:14 PM  
Anonymous Anonymous said...

If you can just leave the actual mass and volume of a skittle that would be cool.because i dont have one.and no im not going to go by a bag of skittles because the hole point is for me to win them not buy them.i need to know how many skittles are in a 31 oz jar.

12/09/2008 5:24 PM  
Anonymous Anonymous said...

oh in that comment above ^^ i mean a 32 oz jar

12/09/2008 5:25 PM  
Anonymous Anonymous said...

im really glad you wrote this article. im taking a statistics course in high school and i always thought those games with the jar were complete crap, but it wasnt until this year that i really understood that the off feeling i was getting wasn't off at all. this was also very fun to read and very interesting. :]

5/10/2009 5:12 PM  
Anonymous generic cialis said...

Hello, I do not agree with the previous commentator - not so simple

11/24/2010 4:17 AM  
Anonymous Anonymous said...

But the expected error should vary with the square root of the number of skittles, not the raw number. And this changes you main result, because now the expected error over all the skittles is about a quarter-skittle, close enough that the true number can be determined.

10/09/2011 1:11 PM  
Anonymous cialis pill cutter said...

Thank you! I didn't know they picked up on it until I saw your comment.

10/12/2011 2:27 AM  

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