Guys, theres a thin in my shop that if you guess how many dice are in the jar, you win a 100$ buy

Guys, theres a thin in my shop that if you guess how many dice are in the jar, you win a 100$ buy.


Can you help me?

Pic is the jar whit dice

Well, you'd first need to calculate the approximate volume of the jar, and the approximate average volume of a single die. Those look fairly standard, yes? So you could do the latter with a set of your own, but unless they let you take a tape measure to the jar, you'll have to either eyeball its height and diameter, or figure out where it came from and look it up on the net. Both might be better.

Take the dimensions of the jar
calculate how much space each type of die takes up
do the math
?????
Profit

I eyeballe'd at 434, that number seems off?

Unless you factor packing efficiency, you'll get a number that's too high.

just get the same jar and fill it with your dice

>buy them all
>"guess" zero

it looks about 13 dice in height and ~2.5 in radius
so we have
π*(r^2)*h = 3.14 * 5 * 13 = ~200 dice
taking into account that my eyeballing might be way off(add ~100 dice) and that there's space between the dice (-30% of total volume) we get ~210 dice.
then again the d6 is a lot smaller and a lot more compact that a d12 so fuck i dunno man

Well, yes, as well as the dice being irregular, it's far from a perfect cylinder.

Is the the number of dice you came up with? Yeah, I don't think there's 400 dice in there, but I really have no idea how to refine it scientifically. I can't even remember if volume is 2piR piRsquard... maybe ignore the space involving the neck and lid?

Oh, I didn't even think of that method, but I think your counts are off-- I'd make it 11-12 high, say 11 to compensate for the shape, and at least 4 across, more like 5.

Yeah, 12,5 dice tall, 7,5 dice in lenght.

That's one face, the other is the same so, 93.75+93.75.

That's 187.5 in the outhermost layer. If the radius is 2.5 dice. That leaves us whit pilars of 3 dice on the inside.

So, 3*3*3*7.5 gives us 202.5. And 202.5+187.5 its 390. That's not taking in account the lid and the deformation of the bottom.

So... 375?

Hah... 11 x 4 dice - 30% = 386 dice. I think we're getting closer.

I'm an idiot, I've been thinking diameter instead of radius. Yep, 375 is definitely plausible.

If I was tasked with setting up something like this, I'd put a fist sized D20 in the middle.

First it's hypothetical spiders in drow vagina, then marbles in fa/tg/uys colon and now this?

There is a method for this that works very well, even though it doesn't seem like it should. Take all the guesses of everyone who has guessed, the more the better, then average them. Even having read the research on this it sounds like bullshit but some light googling will back me up, crowds of people are very good at this kind of thing in a way individuals are not.

Did you try rolling a spot check first?

A Luck test would be more appropriate

2.5 squared is 6.25, not 5

This guy is actually right. Did a similar thing with a jar of coins in one of my stat classes. Averaged out the number and it was 5 cents off

Given that it's a group accomplishment he'd probably not be able to take the winning pool on his own, so it's moot. If he wants all the spoils he'll have to work it out
Saw that on Duck Quacks Don't Echo.

>I can't even remember if volume is 2piR piRsquard.
you have internet access you retard

I eyeballed half that

Let's get the following assumptions straight first:
>The Radius of the jar is Rjar=100. We can do this, because the actual units are not important, only their relative value is important.
>The jar has 10 layers with a circumference with 13 dice, and 2.5 layers with a circumference with 9 dice, this is to compensate for the bottleneck at the top and bottom of the jar. it might be 14 and 10, but I couldn't be arsed to make the calculations for an even number of dice.
>Packing efficiency is 0.7.
>Dice behave like perfect spheres, this is for simplifying the calculations.
First we have to find out the radii at which the centres of the dice are, because it is not at the edge of Rjar. To do this, we assume that the 13 dice are perfectly alligned in a circle ultimately making a polygon of 13 sides with length 2*Rdice. This polygon has angles of 1980/13 degrees and we can calculate with pythagoras the total width of the figure as a function of Rdice.
This function being:
>Rpolygon13=(cos(76.15)+cos(48.46)+cos(20.77)+0.5)*2*Rdice=4.68*Rdice
Now the radius of the centres of the dice is at:
>4.68*Rdice+Rdice=Rjar=100 -> Rdice=17.61
Now we will calculate the radius of the bottleneck.
>Rbottleneck=(cos(70)+cos(30)+0.5)*2*17.61+17.61=77.77
For the other problem, we have an area between Rjar and Rpolygon which does not fulfill the conditions for a closed pack efficiency of 70%, as the circles cannot come into contact with other circles at the outside to pack.
To find this area, we need to find what part of the dice are encompassed by the radius of the polygons. This can be done with the circle-circle intersection calculator of ambrsoft.
The area you have now found, is the area of the dice that is within the packing efficiency area. Which for Apolygon13=466.062 and for Apolygon9=456.799
The area of the dice is Adice=974.25
Now we can determine the amount of dice in each layer while accounting for packing efficiency.

>Npolygon13=13*(974.25-466.062)/974.25+(PI*82.4148^2)/(PI*17.61^2)*0.7=22.11
>Npolygon9=9*(974.25-456.799)/974.25+(PI*60.16^2)/(PI*17.61^2)*0.7=12.95
We assumed 10 layers with a circumference with 13 dice and 2.5 layers with a circumference with 9 dice, so now it is a simple multiplication:
>Number of dice=10*22.11+2.5*12.95=254

/co/ here.
The answer is 42.

182

Trick question. It's full of d4s stacked neatly inside.

Pelotudo

>d4s stacked neatly inside
Is it even possible to "neatly stack" d4s?

So all the serious attempts so far added up and averaged come out to 326.

Sure it is, but it would be incredibly difficult to do so inside a jar. You'd need to be able to fit your hand, wrist and part of your forearm inside with maneuvering room to spare.

Correction, there was a sixth guess. Average is 308.

I like how you linked that shit math pic baiting for replies to later gloat we can't count, here's [you]

Ask enough people and take the average. It's been proven that with enough people guessing the average will be the number you are looking for.

Looks like six, if I'm not mistaken.

Remember to update us after the contest is over. I wanna taste your tears after you didn't pick my winning 182.

>Ten

>279

This is however assuming that only D20s and D6s are in the jar. If there's another kind present the number goes up.

Not OP, dude. Don't know if he's even still reading, I just chipped in out of casual interest.

Oh yeah, look at the neck, toward the left. They've got freaking mini D6s in there as well.

Tried this once, it's harder than you think. I tried with an MnM jar that I was allowed to measure. I also bought 10 guesses. Spent an hour researching and calculating and still lost.

I might be retarded, though.

2

this

You could just use tweezers or tongs.

what if this is a trick and there's a hollow tube on the inside?

What if the middle is filled with 5mm dice?

534