How to Win a Jellybean Guessing Contest

March 7, 2007 at 10:40 am (Thought Provoking)

For the uninitiated, a jellybean guessing contest (or jellybean counting contest) is a game held at fairs or parties or anyplace, where someone fills a jar or other large, typically transparent container with a known quantity of jellybeans and gives a prize to whomever guesses closest to the actual number.  Participant estimates often vary by several orders of magnitude but I believe you can increase your odds and minimize the luck involved with a little knowledge (they say a little knowledge is a dangerous thing, but hey, it’s just a jelly bean contest). 

Strategies for Victory

There are a couple of strategies you may use in calculating the number of jellybeans in a jar or other container.  In order of increasing complexity, they are:

1) Guess

It’s just a game. Arm yourself with the knowledge that there are 930 jelly beans in a US gallon (about 245 / litre) and venture a WAG.  I suspect you wouldn’t be reading this if that’s all you were after, loser.  So…

2) Count them

Don’t count each bean, of course, but if you are allowed to lift the container, you can count the number of jelly beans in one row (remember to compensate for tapering at the bottom of the container) and multiply that number times the number of jellybeans the container is tall.  This will yield a very good estimate. 

3) Equate them

This strategy is best if you are “guessing” the number of jellybeans in a known volume.  In fact, it’s almost fool-proof in that situation.  You need to know:

      a) The approximate volume of one jelly bean can be thought of as a small cylinder 2 cm long and 1.5 cm in diameter (Precisely articulated as: Volume of 1 Jelly Bean = h(pi)(d/2)^2 = 2cm x 3 (1.5cm/2)^2 = 3.375 or 27/8 cubic centimeters)

      b) Due to the Jelly Bean spape and irregularities, there is considerable airspace in the container, along with the jelly beans.  It can be assumed that 20% of a given volume is air rather than jellybeans (though for very small or irregularly shaped containers, this figure might be slightly more… never estimate more than 25% air by volume.  Really 20% is the best value to use for n > 100)

So, to get your answer, you will want to determine the number of cubic centimeters in the container volume and multiply that number by can simply use a calculator to divide the volume of the container in cubic centimeters by 2.7 (3.375 * .8 to allow for air space).  Google is a great too for doing all of this.  For example a search for “cubic centimeters per gallon” and Google returns “1 US gallon = 3 785.4118 cubic centimeters”.  You can then use Google (they call it Google Calculator but you use the normal search engine box) to calculate your answer.  5 gallon bucket = 5 x 3785.4 = 18,927.  You just pop ’round the gorn-and-scumbles, and, Jack’s a doughnut, there you are!

What if I’m Stuck?

But what to do in the inevitable scenario of a container that you aren’t allowed to touch and of an unknown volume?  For the cowards and weaklings, there’s always option one above.  For the stout-hearted, you will need to improvise.  The best way is to try to use option 3 with an estimated volume.  Does the container look about the size of any containers of a known volume?  A Gallon of milk?  A two litre bottle?  A 5 gallon bucket?  Containers tend to be an exact volume… 750 ml, 2 litres, etc. If it looks pretty close, it probably is. 

If it isn’t close enough, or if you just want to double check your conclusion, you can attempt to calculate the volume by sight.  As we already mentioned above, for a cylindrical object, volume = (area of the base) * height = π r^2 h.  You will need to estimate both r and h of that equation.  Note that r (radius) is HALF the diameter.  A useful device for helping with length estimates is a sheet of paper.  Various papers like signup sheets or flyers or whatever are often sitting near the container in question.  You can use the known size of the paper (8 1/2″ x 11″) as a comparison for the container.

Calculating the volume of a non-cylindrical contianer is a much more challenging feat.  If the shape is near recatangular, you can obviously just do length x height x width.  If it is nearly spherical, you can use V = 4πr3 / 3.  As the shape gets more interesting, your best best is to attempt to guess at the volume and then estimate.  You’ll still be at an advantage with this preparation, but it isn’t nearly as great as if you knew the volume.  If you are serious about estimating the volume of more odd-shaped containers, see spheroid and ellipsoid at Wikipedia.

The End?

The final dilemma is how to handle the glory, honor and fame that inevitably accompanies jelly bean contest victory.  Should you gloat or fake humble?  Mock the losers or encouragingly pat shoulders?  These questions are really beyond the scope of this treatise, but I wanted to send you off forewarned and prepared for the crazed paparazzi to come. Good luck and good night.