It's the skeeball prize wall of adulthood – a jar filled with candy, coins, or some other tiny trinket. A person at the front of the room compels you to guess the amount in the jar for free, or for a small donation to charity.

You could throw a number out off the top of your head, but why not try to win, and triumph over your colleagues and relatives instead of making a random guess?

Using math and science to inform your estimate could win you some tasty treats, although there may be errors along the way. Here are a couple of easy tricks to send you home with candy you desired so badly that will resort to the use of math in public.

*Top image courtesy of Gyorgy Barna/**Shutterstock.com*

**The container is your best friend**

Is it a jar? Is it cube? You don't have to be able to handle the container in most cases – just seeing a portion of it will work. It's estimating the number of object that you cannot see that is the hard part.

A jar is akin to a cylinder… well, it is a cylinder. The volume of a cylinder can be calculated by multiplying 3.14 times the height times the distance from the center twice. If you remember high school geometry, you'll remember the volume of the cylinder equation is the the image.

The cylinder in the picture is 24 M&Ms (our wonderfully tasty unit of measurement) tall, with a radius of 8 M&Ms, suggesting that around 5000s M&Ms are inside. This is a bit of an overestimation since there is space between M&Ms in a jar and calculating the volume assumes the M&Ms are oriented perfectly - we'll talk about packing density a little later. You can also use the volume tactic with other simple three dimensional objects (although the equations change) like cubes.

A brute force method to approximate the number of objects in a jar (especially if you can see the bottom jar) is to count the bottom layer and then multiply by the height. However, this only works if you can fondle the container for a while.

**Packing density**

Packing density comes into play since there is always *some* space in the jar occupied by air and not candy. Using the maximum packing density, a value for how much of the container is filled with ellipsoids and spherical objects if the jar is packed tightly, you can inform your estimate and work your way into an easy win. It helps if you know the volume of the container, too.

What is the approximate shape of the objects in the container? If it's squares or ellipsoids (M&Ms again!), you're in luck – if the objects are candy hearts, prepare to cry.

The packing of spheres and ellipsoids is well studied thanks to a 16th century need to determine the best way to pack cannonballs . The optimum packing density of spheres of equal size (think chocolate malt balls) is 0.74, suggesting that the spheres would take up 74% of the container if packed tightly under optimum conditions.

Ellipsoids (M&M's!) have a similar packing density value – 0.755. Multiply the volume of the container by the packing density and divide by the volume of the object (grab your iPhone!) and you're a winner. If you are faced with an unusual set of objects, the packing density will change. Cubes of identical size are difficult to analyze, as they don't pack well (bouillon cubes in a jar).

A group of researchers looked at the packing process of oil droplets of random size (akin to a jar filled with spherical candies of various sizes), and determined that the optimum packing density is around 0.7, 70% of the container. If you have a variation of object shapes and sizes, the problem is more difficult - I'd look at making a guess using the geometric method above.

**What if the jar is filled coins?**

If the objects you want are thin and flat, guessing is hellish. Honestly, I'd pick the container up, guesstimate the mass of the container — if the rules allow you to pick up the container.

Divide your estimate of container mass by the individual mass of the object (use your phone again!) and you just calculated the number of objects. The number, however, is only as good as your estimate of the container's mass.

Let's use coins as an example. Pennies weigh 2.5 grams and dimes weigh 2.25 grams, while nickels and quarters come in at 5 and 5.5 grams each.

If it's jumble of coins, you can use an average mass value, taking into account that coin population will probably decrease as a function of value (unless Uncle Scrooge is running the contest). You will probably have a ton of pennies in that jar. I'd suggest 3.5 grams as a general guess for a mixed jar – enough pennies to be in the majority (offset by those lightweight dimes), but high enough for a reasonable amount to still be nickels and quarters.

Let's take a look at a more exact example. Based on a distribution of 65% pennies, 10% dimes, 20% nickels, and 5% quarters, the average coin is worth 3.9 cents and weighs 3.13 grams. A thousand coins with this distribution would come in at 3.13 kilograms, or 6.89 pounds for those of us in the United States.

Does the jar weigh more or less than the dumbbells you lift in the morning? You can use that to gauge how many coins are present. Also, 7 pounds is pretty darn hefty – one would *hope* someone wouldn't put that much metal in a glass jar.

**Get that candy!**

These are not exact analysis techniques, but hopefully they will help you shame your friends and coworkers and walk away with a jar full of calories or cash. If all fails and someone does something stupid like cramming dollar bills into the coin jar, ask a pal to fake a heart attack, and make off with the container in the ensuing chaos.

*Images courtesy of CC and Shutterstock. Sources linked within the article.*

## DISCUSSION

If you have to eyeball a container at an event, estimate the size of the candy and do calculations (even with a phone) then you are probably making so many estimates and assumptions that you might as well guess.