My son recently bought a 20-sided “spin down” die. Here’s a similar one:

It’s really useful in the game “Magic The Gathering” because you use it to keep track of your life points. My other son and I, however, were wondering if it was still a balanced die. We reasoned that all the double-digit numbers were on the same side of the die and that meant that side was lighter, since all the numbers are carved into the die material. We thought that would mean it would affect the roll, possibly giving us slightly more double-digit numbers when rolling since the lighter side might end up on top a little more often.

### Experiment!

So we took a regular 20-sided die (the type where opposing faces add up to 21) and rolled it along with the new die 100 times. Here’s a time series plot of every roll for both along with a simulation with *Mathematica*‘s random number generator:

Hmmm, they all look fairly similar. I thought maybe looking at the histograms of the results would be better:

Note, by the way, that the simulation was re-run between these two images. Here they all still look vaguely similar. But, then, I realized that I wanted to directly compare the single- to double-digits, so I reworked the histogram (remember, we thought there would be more double-digits) . . .

Hmmm, weird. Not at all what was expected, though at first we got excited about the rise in single digits (“maybe it’s not carved but painted” etc), until we looked at the second and third pics.

Really, I think the thing we learned is that 100 rolls isn’t nearly enough. It was fun rolling and recording, though!

### No 12′s?!

If you look above, you’ll see that we didn’t roll any 12′s on the regular die. That had us wondering how rare that is. It promped this tweet from me:

hey statistics tweeps, what's the chance of rolling a 20-sided die 100 times and finding that there is a number that never came up?—

Andy Rundquist (@arundquist) April 08, 2012

which got a lot of responses. I really liked this one from daveinstpaul:

@arundquist Using inclusion-exclusion

p = sum (k=1 to 20) [(-1)^(k-1) * C(20,k) * (20-k)^100] / 20^100;

p ~= 0.113463. First guess was close—

Dave Radcliffe (@daveinstpaul) April 08, 2012

which has some math I wasn’t familiar with. Here it is with better formatting:

Of course, I thought I’d try the old tried-and-true brute force method as well. I “rolled” a 20-sided die 100 times and checked how many different numbers came up. I then repeated that 100,000 times and did a histogram. Here’s the Mathematica command and then the histogram:

check = Table[Length[Union[RandomInteger[{1, 20}, {100}]]], {100000}]; Histogram[check]

The sum of the 17 (12 occurrences), 18 (445), and 19 (10,886) is a fraction of 0.11343 of the total. Very close to Dave’s calculation!

Heya i am for the first time here. I came across this board and I find It truly useful & it helped me out much.

I hope to give one thing again and aid others like you aided me.