Guess The Number Of M&Ms In A Jar

Holy crap this is so cool. I might pick up that course to take (Solid State Physics) for fun next year or something :)

@atomicrabbit: This is what I came here to comment. Not only that but in the comments of that blog someone linked to this site http://www.madsci.org/posts/archives/feb2002/1013032438.Ch.r.html which says the packing ratio is 59.69% and the density of a single M&M is .45239. Seems odd that each site uses its own [different] calculations and successfully guesses the amount of M&Ms in their own jar.

jensbodal

Ok, someone get the packing fractions of all the popular types of candy used in these contests.

Start with jelly beans and peanut m&m's.

BTW, instead of remembering 0.715 mL and 66.5% packing fraction, just remember this number: 0.93

Simply multiply volume in mL by 0.93 and you'll get your # of plain m&ms.

kftgr

@RosaliaRaphman: NERD ALERT!

@jupiterthunder: That's not LaTeX, it's a symbolic representation of atomicrabbit's equation, in plain ol' ASCII.

LaTeX would look like this:

$$n_{M&M}=\frac{0.665V_{jar}}{V_{M&M}}$$

@dchall8: We had a contest like this in 8th grade algebra, and I won using this method. I only was able to see the jar for about a minute when I guessed but that was enough time to get a rough estimate of the dimensions (in M&M's). My guess was 625, and there were 627.

SteveO3387

this was really interesting! thanks!

Zé Miguel Santos

@RosaliaRaphman: Yup. It was either Freakonomics, or one of Malcolm Gladwell's books, where I first read this. Wisdom of Crowds, maybe? (Which I realize is a third choice). If you have access to the data, aim in the middle.

@Sensai: Indeed. But they were jellybeans, not M&Ms.

sn1per

Yes, that's right.

Volume of Jar: 1150 ml
Volume of 1 M&M: .715 ml

So if the container was filled *completely* (with no space in between M&Ms), the number of M&Ms in the jar is simple division:

1150 / 0.715 = 1608 M&Ms.

According to the paper he found during his research, the packing fraction of plain M&Ms is 66.5%. You use this packing fraction to account for the space in between M&Ms.

1608 x 66.5% ≈ 1070.

The actual number of M&Ms in the jar was 1072. Not bad at all!

UlimaCachu

Heh-heh-heh, I'm so gonna win the next one I see. Not only do I get the M&Ms (and any other prize that comes along with it), I also get any cute girls who think I'm a total genius. :p And thus my quest to rule the world takes another great step!

ZackRaeXavier

Here's my math: The times in my life this would have actually come in handy: 0.

Sexy accent? M&Ms? Hunter S. Thompson tee? Packing ratios? Molecular models?

How much is that puppy in the window?

Very interesting, especially because I just had this exact subject at my college this month (got a C, btw).

Hideki

That was awesome!

Shashank Agarwal

@craigpy: Just weigh a small bag (-bag), and divide by that.

Getting within a bag would no doubt be closer than any other method, especially as you could probably never determine the interior dimensions and curves of a given jar to any sufficient degree of accuracy.

lhmlco

@jupiterthunder:
did you recursive?

craigpy

@lhmlco: To be fair, he does say you could do this, but also says that precision scales aren't always immediately on hand. BTW...if you were to take a weighing approach, then weigh maybe 20 M&Ms and then work out an average M&M weight.

@Parapraxis: I think this makes me Mr Bean.

craigpy

Do those contests ever reveal the volume of the jar?

If not, this method and the last 9 minutes of my life were USELESS.

drdoombot

@.rawr: Mitch Hedberg, of course.

God rest his soul.

Tell you what, you guess how many M&M's I want! If you guessed a handfull, you are correct! [name that stand-up comic]

Man, my solid mechanics professor was a dull tufty haired man who couldn't handle somebody sneezing in class without losing his concentration and ending class early. Wish I had this guy, might have learned something then. =\

russkey

@lhmlco: Did you not watch the video? He said it himself that the easier thing would have been to do that, but because of how small the m&ms are in weight and how much more the actual glass jar weighs, the scale with that much accuracy would cost a lot of cash, which, if you're going to spend that much, you might as well buy m&ms.

Araam Vafai Borhanian

@atomicrabbit:

Candy dish + spoon. No need to touch any of the m&m's.

jupiterthunder

@DixonMaylord:

I just approved all those to illustrate the inherent problem in the comment system.

jupiterthunder

@Dakiall:

So you're saying Gawker comments need to accept LaTeX?

jupiterthunder

@lhmlco:

Mr. Bean, is that you?

Parapraxis

@xvkarbear: The reason this works is because you have to divide the volume of the jar by the volume of the m&ms to get how many m&ms could theoretically fit into the jar assuming they occupy the entire space (i.e. no air, just solid chocolate). Because they don't fit perfectly you need to multiply it by the Packing fraction in order to account for this.

Ferguson1015

@curlymeatball38: no, he said if no one guessed the amount within 5 m&m's, he would give the person who was closest a small pack of m&m's. If someone DOES guess it WITHIN 5 m&m's, they would get the entire jar.

atomicrabbit

I wanted to see him open the bottle and proceed to chug the M&Ms. Needless to say, I was disappointed by the needless inclusions of math. That will not impress my friends and co-workers, it just make them wonder why the hell I spent all that time trying to figure out how many M&Ms were in a jar.

Robotic Bilbo Bagins has no use for fleshy ones

Weigh jar on precision scale. Weigh similar empty jar. Weigh one M&M.

The rest is obvious, and left as an exercise for the reader...

lhmlco

It turns out that people are, overall, fairly good guessers. While some people may be low, and others high in their predictions, in most instances, another great way of coming up with a guess on how many M&M's would be to take an average of all the guesses. If your sample size of student guesses is large enough, you should come up with an extremely close guess.

I work in a retail store, and we were told by our home office to count the number of gift cards that we had in the store. These gift cards literally cover an entire wall. Just for fun, I had everyone working that day make a guess on the number. The guesses ranged from 800 to 3500 gift cards. And it turns out that it was the average number of all of the guesses that was the most accurate. The average was about 2500, while the actual number was about 2700. Taking into consideration that our sample size was only 5 guesses, I think that is pretty accurate.

Try it out with your experiment!

RosaliaRaphman

@atomicrabbit:
And finally.

Nm&m=(Vjar*0.665)/Vm&m

@madara: The formula is to take the container volume (1150 mil) divided by the volume of an m&m (.715) and that will give you total m&m's if they took up all the volume. Since they don't because of all the gaps you multiply that by .665 (66.5%).

ri59

I bet I could have counted all those M&Ms one by one faster than it took to watch that video.

@atomicrabbit: he said that he was going to give the winner a small package of m&ms.

what a gip.

curlymeatball38

the problem is, if you're at a carnival or something like that, how will you get all of these measurements?

curlymeatball38

Sadly this won't help you win big competitions any more, because it's a pretty good method (my father used a similar technique to win jars full of sweets at church fairs for ages). The big versions of these competitions they have in malls use various different sizes of sweets/toys/whatever so the packing is much denser and more complicated to calculate.

Kludge

@MalvinPolyphemus:Yes I must say, you *are* a terrible explainer. Sorry. Although you were right in the end, the process you took confused the hell outta me.

Saying just "Volume of jar*packing fraction / volume of m&m" would have been sufficient ;) Thanks though

atomicrabbit

@madara: the formula is (in this case): Number of m&ms = (Packing Fraction * Volume of Jar) / Volume of 1 m&m. If you put the numbers in it comes out to be 1070 which is pretty close

Jamie Magee

The packing fraction is the percent of the m&m and the space around it is the actual m&m. So if you have an m&m with a packing fraction of 66%, then the m&m's volume is 66% of the volume an m&m with no space.
therefore:
x*66.5%=0.715 ml
x=1.075
1150/1.075=1069.58 or 1070

so the equation would be Volume of jar*packing fraction / volume of m&m.

I am a terrible explainer and got this by looking at the numbers and figuring it out but hope this helps.

MalvinPolyphemus

@laughingisfree: yep i just figured it out before I refreshed the page. Thanks for posting it though!

Although this site (http://marcos.kirsch.com.mx/2004/08/08/how-many-mms-are-in-the-jar/) says an m&m's packing fraction is 68%

atomicrabbit

1150 ml / .715 ml * .665 = 1,070

In more general terms, if you compacted all the M&M's you could fit 1,608 inside the jar but since you only get 66.5% efficiency you can only stuff 1,070...

ChandlerNaslund

@madara: you divide 1150/.715

Which equals a rounded 1608. Then times 1608 by .665 (the packing fraction) and you get a rounded number of 1069.

I don't know if this is right or not. I never took physics.

xvkarbear

1150/.715 = 1608.39
1608.39*.665 = 1069.5
This is how you get the number...

Chris Sumerlin

@atomicrabbit:

looks to me (1150mL / 0.715mL ) * 0.665 will give you 1070ish

laughingisfree

@atomicrabbit:
1150mL / .715mL = 1608 M&M's * .665 = around 1070 M&M's

Basically, you figure out the number of possible M&M's you could fit in the jar (1608) and then account for the empty space (only about 66.5% of the total space in the jar is taken up by M&Ms)

Bill Groesbeck

@ADiSH: I had blood shooting out of my eyballs at 5 minutes into it.

Or, another way to do it is to use your 8th grade geometry. Hold the jar upside down and count the number of M&Ms across the diameter several times. Get an average. Then count the number of M&Ms from bottom to top of the cylindrical part of the jar. Then use the formula for volume of a cylinder
V=pi x radius(squared) x height
all measured in M&M units. Then estimate the height of the conical part of the container in M&M units and use the formula for volume of a cone to estimate the number of M&Ms in that part. Then add the two volumes for your estimate of the total.

The formula for volume of a cone is left as an exercise for the interested student.

And once again, I have blood shooting from my eyeballs.

Are you guys serious?

1150/.715 = 1608.39
1608.39*.665 = 1069.5

Chris Sumerlin

It's not too tricky.
1150 * .665 / .715 = 1069.something
So it's:
VolumeContainer * PackingFraction / VolumeObject

ZamirAchilles

Are you guys seriously not able to calculate it?

1150/.715 = 1608.39
Then do 1608.39*.665 = 1069.5

DixonMaylord

@ADiSH: What? You sound as though you didn't expect something even remotely mathematical. Sorry to disappoint, but figuring out how much candy is in a jar doesn't ever involve swords or lazers or shouting "Stupify!"

um... he counted each and every m&m?? ... so that means he touched every single m&m? and he's going to give the dirty m&m's to his students?

btw, based on a jar thats 1150 ml and a m&m is 0.715 ml and a packing fraction of 66.5%, how does that equate to 1070? What is the calculation?

atomicrabbit

What's the formula?

jrmbtr

Can some Rainman show me the formula? I can't get it unless I fail to understand his numbers:

1 m&m = .715 mil
container = 1150 mil
Packing Fraction = 66 1/2%

Total number of m&m's = 1072

That was a lot more boring that i expected

I'd be more impressed if someone ate them all.

And what's all this "leeter" talk?