How many jelly beans are in the jar? It is a classic head scratcher. Today I’m going to talk about how I approach these puzzles and share some numbers data because numbers are fun!

## How Many Dice are in the Aquarium?

This was the challenge put together by the local game store. Over 800 people guessed. Do you have a number in mind? It’s okay, I’ll wait.

Ready?

Here is a breakdown of how everyone else guessed.

Click on it for a bigger more readable image.

Here is how I reached that conclusion.

First of all, always try to get more information! I knew this was a yearly challenge so I went back and looked at the past two years.

Here was the previous year picture. There were 2,813 dice in the aquarium then.

That is a huge data point and potential advantage when making a guess. I drew in some lines on the current challenge.

We now know a big triangle has about 2813 dice. We can clearly see two big triangles in the current aquarium, but they have some overlap. The challenge is to figure out how many dice are in the intersection of the two triangles so we can math it up. Triangle 1 + Triangle 2 – Intersection = # of dice in aquarium.

Or

2813 + 2813 – intersection = ???

By drawing a line straight down in the middle of the intersection, we can make two right triangles. That makes the math a bit easier. Then we need to start counting. I counted length and height for both the big triangle and the little intersection triangle. My count of the big triangle was 22 dice wide by 12 tall. We don’t need to worry about the depth because it is all the same depth.

From the previous year, we know that 2813 dice occupied 22×12 triangle. Convert that 22×12 triangle into area (multiply height x base and divide by 2).

22 x 12 / 2 = 132 area. 2813 total dice divided by area (132) = 21.31 dice for each cube of dice volume.

Now count the little triangle. I got 10×4. Each little triangle would have an area of 10×4/2 = 20, but there are two little triangles so it is just like a rectangle.

We now have enough information to make an educated guess.

The intersection is 40 area x 21.31 dice per area unit = 852 dice in the intersection.

2813 + 2813 – 852 = 4774 dice in the aquarium.

There is one more thing that we can do to get a competitive advantage. That is look at everyone else’s guess! By finding the biggest available range of non-guesses we can maximize our odds of winning. If someone guessed 1 and someone else guessed 3, you wouldn’t want to guess 2 and lock yourself to a single number.

Since this challenge was posted publicly to Facebook, it is really easy to grab all of the other guesses and put them into a spreadsheet. I used a free tool called FacePager. It lets you quickly dump all of the posts, comments, etc. from a public page into a spreadsheet. Then you can massage the data as you like. For instance, the simplest way of finding a big gap between guesses is to create a second column next to the original column. Do a simple A2-A1 formula and extend it all the way down the second column. The bigger the number, the bigger the spread in guesses. Look for the biggest spread near the estimate we arrived at above (4774). In my case, someone had already guessed 4779, so I simply did 1 more than that at 4780. That bought be about 20 possible numbers to win. All of that guesstimating improved our odds of winning from 1/800 or 0.125% to about 1/100 or 1%. That doesn’t sound like a lot, but it is nearly a tenfold increase in the likelihood of winning. Not to shabby for a few minutes of work!

So there you have it. Now that you know my secret, I’ll have to off you.

Oh, you’re asking how many dice are actually in the aquarium? I don’t know yet, some unlucky bugger is still counting them!