Guess the number of easter eggs to win...

[screw_dog]

At work we are having one of those "Guess the number of easter eggs in the jar to win it" competitions. For those who aren't familiar, there is a plastic jar that is (almost) full of cheap, nasty easter eggs. We all (the 40 odd staff in my area) are allowed one guess as to how many are inside the jar. The person who gets closest will win the entire jar (to share with their section).

As the chocolate is really cheap and nasty, I just want to win it for the bragging rights. So, what is the best way to estimate the number of eggs?

The quantities that I can easily measure are:
Volumes: of the jar (from the outside), of an individual egg.
Mass: of the jar and eggs (together).
Density: of the jar and eggs (together).

I can estimate the mass and density of an egg by purchasing similar eggs. Unfortunately, I have no idea about the mass or density of the jar (without eggs in it).

Any suggestions? If anyone wants any measurements I can get them, was just thinking about formulae at this stage.

[Iamme]

There are 17 eggs in the jar...because Sylvia Browne told me so!







































Have you had a chance to see what size/type jar it is? If so, get a jar like it and start throwing in eggs! If not... if you don't know and it is being kept a secret...then what good will it do you to know a formula? What would you be formulating?

[tkingdoll]

Screw dog, I don't have a decent suggestion for your egg guess but I feel terrible that the only reply you've had so far is from Iamme, so I'm posting anyway.

If you're willing to go to some expense and effort, why not buy some eggs and a jar and make a replica, weigh it, then replace the real jar? You could even add rocks to your new jar to ensure no-one else is even close.

[Amapola]

Tkingdoll has a great idea there..... :D

I am not very mathematical. Probably I would see how many eggs were in the top "row" and try and figure how many "rows" deep the thing was, and use simple arithmetic. (About all I can manage.)

Perhaps my feeble response will inspire a real mathematician to post a good answer.

[Lisa Simpson]

My husband has won many "guess the number" games by counting the number of items on the bottom and multiplying it by the number of rows.

[Soapy Sam]

Isn't tkingdoll's idea the same thing Iamme suggested?

You need the weight of the empty bottle (WE), the weight of the full (WF)bottle, the weight of an egg(WEg), the density of chocolate (D), F ( a fudge factor), your height above sea level (to correct for gravitational variation), your sun sign and the number you first thought of.

The answer is 17.

Blast. Looks like another win for Sylvia.

[tkingdoll]

Isn't tkingdoll's idea the same thing Iamme suggested?



Yep :D

[Zep]

If they will let you weigh it, the weight difference method would be probably the most accurate. Use the scales they weigh parcels with in the office... ;)

Eggs = (weight-full-jar - weight-empty-jar) / weight-one-egg

That's how they count bulk coins in a mint. You would go real close anyway.

[screw_dog]

The weighing method was my first thought too. I have access to the jar, but I haven't been able to find another one to weigh when empty. It seems like very light, but I'm not good at estimating.

I think I'll also estimate by counting the number in each row and multiply by the number of rows. This should give me two estimates which hopefully will be more accurate than just one. Maybe average the two?

I don't actually have access to accurate scales at work, about the only type we have access to is generic spring loaded kitchen scales, good for 100s of grams. I do, however, have an urn, so I'll probably calculate it's weight by measuring its displacement of water in the urn.

[Zep]

...if you lose, put the jar back in the urn and turn it on. ;)

[screw_dog]

Even if I win, I won't want the chocolate. It's that cheap.

[Zep]

YOU DON'T WANT FREE CHOCOLATE! :eek:


Have you checked your pulse lately?

[screw_dog]

You haven't tasted this chocolate. It's like carob, but without the taste.

[T'ai Chi]

At work we are having one of those "Guess the number of easter eggs in the jar to win it" competitions.
...
Any suggestions? If anyone wants any measurements I can get them, was just thinking about formulae at this stage.

What shape is the jar?

And if the easter eggs are all rougly the same sizes, can you get estimates for the measurements of an egg's length, width, and height?

If so, maybe look at doing Estimate of # of eggs = Vol_jar / Vol_egg.

[screw_dog]

The jar is almost cylindrical. It has a broad punt at the bottom and tapers sharply at the top. The measurements of an egg wouldn't be hard to estimate / measure. Just not sure about how densely they are packed. From inspection they are less densely packed than ideal sphere packing, but not sure how much less. I've tried shaking the jar to get them to pack more densely but it seems to make little difference.

If I could bring a camera into work I'd post a picture, but alas...

[T'ai Chi]

The jar is almost cylindrical. It has a broad punt at the bottom and tapers sharply at the top. The measurements of an egg wouldn't be hard to estimate / measure. Just not sure about how densely they are packed. From inspection they are less densely packed than ideal sphere packing, but not sure how much less.


Ok, maybe something like this(?):

If the jar is cylindrical, and a, b, and c are estimates of mean lengths of the axes of an ellipsoidal egg, then

Estimate of # of eggs = K * (Vol_jar / Vol_egg) =

K* (2*pi*radius_jar*height_jar)/[(4/3)*pi*a*b*c)] =

K * (3*radius_jar*height_jar)/(2*a*b*c)

where K is a constant to lessen the estimate due to ellipsoid packing (http://mathworld.wolfram.com/EllipsoidPacking.html), and is about .75, so the above becomes

Estimate of # of eggs ~ (9*radius_jar*height_jar)/(8*a*b*c)

[screw_dog]

Ah! Interesting that an ellipsoid packing constant of about 74% is feasible for random packings (according to that page). Not a bad formulae then.

I think I'll check the result by using it to estimate the weight of the empty jar just to make sure that comes out to a reasonable figure.

Thanks Tai.

[The Central Scrutinizer]

Don't forget they may have hidden a baseball in the middle of the jar to throw the mathmeticians off.

Last fall I went to a small local jazz festival. There was a table there that had a "guess the number of pieces of licorice in the jar" contest. I think the prize was a free (crappy) radio station t-shirt or something trivial. Except there was one problem - there was no licorice in the jar! There were several hundred pieces of red candy in the shape of licorice that people always refer to as "red licorice" in the jar, but no licorice! So I waited until the end, then handed my slip to the girl and told her that I was quite sure I had turned in the only correct answer of the day - zero. She said it would be an hour or so until they went through the answers, and I didn't feel like waiting around, so I left. But I had made my point!

[Art Vandelay]

Calculate the torque to angular acceleration ratio. That will give you the moment of inertia. With that, the mass of the jar, and the density of the eggs, you should be able to figure out the mass of the eggs.

Or, support one third of the bottom of the jar. Measure how much of the weight remains unsupported.

[screw_dog]

Calculate the torque to angular acceleration ratio. That will give you the moment of inertia. With that, the mass of the jar, and the density of the eggs, you should be able to figure out the mass of the eggs.
Good idea. Any suggestions how I can measure the torque and angular acceleration using only items normally found in a kitchen?
Or, support one third of the bottom of the jar. Measure how much of the weight remains unsupported.
I'm not sure how this would help me. Sorry, I'm not physically minded. I am mathematically minded but that doesn't always help in the real world.

[The Mad Hatter]

If you want an easy way of getting the volume (of the surface, at least), and your coworkers don't mind, you could bring a tub of water and measure the displacement.

But I don't know how helpful that'd be with eggs, since there's a lot of empty space in between.

Perhaps you could try estimating the mass of the jar. If it seems like it weighs less than egg, divide the total mass (jar and eggs) by the mass of an egg, and the remainder is the mass of the jar. So if you can estimate if the jar is lighter than one egg, somewhere in between one egg and two eggs, between 3 and 4 eggs, etc., then you can figure the rest out. Of course, this might be very hard, and it'd take a good intuition or sense of mass. But no matter what, the mass of the eggs can only come in certain multiples. Just be glad you're not working with jellybeans.

[tsg]

You could count them. Just a thought.

[Hellbound]

Bring a portable blow-torch in. Heat the jar until all the eggs melt into a gooey mess. Measure the volume this mess takes up (no airspace left to worry about). Using the density of the eggs (measured by purchasing similar eggs), you can easily calculate the number. Zero (there's no eggs, just a melted chucnk of chocolate).

[Phillybee]

A related website:

http://www.cockeyed.com/inside/howmuchinside.html

[Art Vandelay]

Good idea. Any suggestions how I can measure the torque and angular acceleration using only items normally found in a kitchen?If it's a reasonably rigid body (that is, the eggs don't shift much as you tilt it), you can roll it down a ramp, then time how long it goes a measured distance. From that you get the speed. You can calculate how fast it "should" be going based on the height of the ramp, and assuming that the energy isn't going anywhere else (such as friction), the difference between that and its actual speed must be due to angular energy.

If the category "normally found in a kitchen" includes a portable scale and a turntable of some sort, you can push against the jar with a measured torque for a measured time at a measure distance from the center of gravity (which should be in the center, but you should check), and measure the angular velocity which results.

I'm not sure how this would help me. Sorry, I'm not physically minded. I am mathematically minded but that doesn't always help in the real world.Imagine an infinitely dense, but infinitely thin, jar (so that its mass density as a function of radius is a delta dirac function centered at r). Now, let's say that we have a coordinate system so that the jar satisfies the following equation:
max[(x^2+y^2)/r,|2z-h|/h]=1
(This is the equation for a cylinder of radius r, height h, with the origin being the center of the bottom. Hmm, that would make a good puzzle question. "How can you define a cylinder in a single equation?")

Now, all of the cross sections for 0<z<h are the same, so let's consider all of the mass to be concentrated within one of them. Now, suppose we draw a line at x=r/3. Using the fact that r*cos(theta) = x, theta = arccos 1/3.
That gives us 1.23 and -1.23, for a total arc length of 2.46. The whole circle has 2pi, or 6.28. So the amount to the right of the line is 2.46/6.28 = 39%. Another 39% is balanced on the other side, and the total unsupported weight is 22%.

Now, suppose the jar is completely solid (no eggs at all, just glass all the way through). We need to figure out how much is to the right of that line. If you add on the triangle whose sides are the radii and the chord, the total of that will be 39%. The triangle has base x, height 2y, and y is .943, so the area is 31.4%. That leaves 7.76% to the right of the line, or 84.5% unsupported.

So you have two extremes. Let J be weight of the jar, E be the weight of the eggs. If the jar is infinitely thin, the total unsupported weight is .22*J+.845*E. If the eggs are infinitely thin, it will be .845(J+E). If you know what J+E is, and you know what the unsupported weight is, you can figure out where you are between those two extremes.

The top and bottom of the jar makes it a bit more complicated, but not much.

[screw_dog]

Mr. Vandelay, you sir are a genius. I like your idea very much and will try it out. Watch this space!

[Humphreys]

Buy the same jar, and the same eggs, and fill yours up until it looks the same.

Or torture the guy until he tells you.

[screw_dog]

Nah, it's the bosses daughter who's in charge. Torture is out of the question. Also, I want to win on smarts, not by cheating. Less gloating if I cheat.

[Humphreys]

Seduce her.

[screw_dog]

Not possible. See this thread (http://forums.randi.org/showthread.php?t=53496). Although, I would count that as an intellectual victory, particularly as she's a lesbian.

[Psi Baba]

There was a table there that had a "guess the number of pieces of licorice in the jar" contest. I think the prize was a free (crappy) radio station t-shirt or something trivial. Except there was one problem - there was no licorice in the jar! There were several hundred pieces of red candy in the shape of licorice that people always refer to as "red licorice" in the jar, but no licorice! So I waited until the end, then handed my slip to the girl and told her that I was quite sure I had turned in the only correct answer of the day - zero. She said it would be an hour or so until they went through the answers, and I didn't feel like waiting around, so I left. But I had made my point!
:thanks :clap: :clap:
Let's here it for Licorice solidarity! :Banane07: <--Licorice. NOT licorice --> :Banane24:

Ontopic: Of course it throws everything to hell if they've stuck a pingpong ball in the middle of the jar. Screw_dog, my suggestion if you win is to sell the chocolate-like substance to a cow orker for about half of its retail value. You both win.

[Iamme]

If they will let you weigh it, the weight difference method would be probably the most accurate. Use the scales they weigh parcels with in the office... ;)

Eggs = (weight-full-jar - weight-empty-jar) / weight-one-egg

That's how they count bulk coins in a mint. You would go real close anyway.

Did I tell you guys how I once entered a national contest run by Bumble Bee Foods, to figure out how many cans of their product completely filled 4 compartments of a speed boat and the winner would get all the cans of fish products in the boat + the boat + the motor + the trailer? I spent many hours, by hand, at the kitchen table working this out...only not winning. I called up the judging company in new York to ask them why I didn't win. :) Seriously. (There were one or two...I forget correct entries!)

Do you know that you can subscribe to contest newsletters where you can actually try to figure out answers to stuff, rather than relying on luck, as you do in sweepstakes?

[Archangel]

The quantities that I can easily measure are:
Volumes: of the jar (from the outside), of an individual egg.
Mass: of the jar and eggs (together).
Density: of the jar and eggs (together).



My suggestion would be to get the volume of the jar, and then one night before you go home turn the heat up as high as you can get it.

That should (with any luck) melt the easter eggs into one solid mass, giving you a good idea of how many were in there by the now accurate volume produced (assuming I havent stuffed something up in my thinking).

Of course this may be considered cheating :D

**** Edited to add****
D'oh that'll teach me not to read a thread in full before posting, Huntsman beat me to it (although to be fair his idea requires more tools than mine :P )

[Hellbound]

D'oh that'll teach me not to read a thread in full before posting, Huntsman beat me to it (although to be fair his idea requires more tools than mine :P )

Just a blow torch. And the heat might not get high enough to melt em well, the blow torch is more sure :)

Although, you could just stick it in an oven or put it on a hot plate...but really, if you have a blow torch, who's going to argue with you?

:D

[Jorghnassen]

You haven't tasted this chocolate. It's like carob, but without the taste.

Ah yes. Some evil candy people have been producing some sort of very cheap chocolate-like substance (in appearance only) that they sell for Easter in the shape of rabbits, eggs, chickens, etc... That stuff is disgusting, possibly poisonous and definitely both blasphemous and heretical. That's why I buy Easter chocolate made by Trappist monks. Now there's a guarantee of quality and divine taste (I do believe it's considered an indulgence for gluttony too).

[Soapy Sam]

Do you know that you can subscribe to contest newsletters where you can actually try to figure out answers to stuff, rather than relying on luck, as you do in sweepstakes?

Doing promotional competitions methodically was quite a big hobby in the UK for years. There was even at least one magazine. The law requires that details of any judging system must be available in the public domain. The judges must also be named.

I know one man who made a serious hobby of this- even studying the personalities of judges. He won a number of major prizes, including one car and several holidays.

His interest declined after his son was born and I get the impression such competitions are less common these days anyway.

[T'ai Chi]

Did any of our methods help?

[T'ai Chi]

So what was the answer? :)

[T'ai Chi]

Hope he comes back. :)