The Hidden Geometry of Dice
quantquestions
Welcome to What Are the Odds? The show where we answer life’s important questions, like can I outsmart a six-sided die? Today we’re starting small. Two rolls of the dice and one burning question. No magic formulas, just curiosity, some patient counting, and a faint hope that math is on our side. Let’s roll.
Pop quiz: You roll a fair 6-sided die twice. Calculate the probability that the value of the first roll is strictly less than the value of the second roll.
As I always like to say to my niece, “the secret to answering any probability question is to enumerate the outcomes and count the ones we care about.”
\(P(Interesting) = InterestingOutcomes / TotalOutcomes\)
Rolling 1 die once has 6 outcomes: 1 2 3 4 5 6.
Rolling 1 die twice gives us a combination of outcomes, let’s write out a few.
[1 1] [1 2] '... [1 6]
[2 1] [2 2] '...Following this pattern would produce 6 rows of 6 columns, so there must be 36 outcomes. We write down just enough of the pattern to figure out the best way to count it. Now we count how many of those meet the criteria.
[1 1 :no] [1 2 :yes] '...
[2 1 :no] [2 2 :no] [2 3 :yes] '...Logically we should see 5 yeses on the first row, then 4, 3, 2, 1, and 0, which we can ignore. Add them all up and we get 15.
So the answer to the question is 15/36 which reduces to 5/12, dividing top and bottom by the greatest common divisor 3.
You might be thinking that it’s not practical to enumerate everything all the time, I should use the formulas of probability. That’s true, those are marvelous. However, in my experience it is also easy to go wrong reasoning from formulas. It’s harder to go wrong starting with a counting problem, then improving your method of counting. You end up in the same place, but more confident in the answer.
The full enumeration of our simple 2 roll question as a counting problem is just big enough to be too tedious to use only counting.
(let [roll [1 2 3 4 5 6]]
(for [i roll]
(for [j roll]
[i j])))(([1 1] [1 2] [1 3] [1 4] [1 5] [1 6])
([2 1] [2 2] [2 3] [2 4] [2 5] [2 6])
([3 1] [3 2] [3 3] [3 4] [3 5] [3 6])
([4 1] [4 2] [4 3] [4 4] [4 5] [4 6])
([5 1] [5 2] [5 3] [5 4] [5 5] [5 6])
([6 1] [6 2] [6 3] [6 4] [6 5] [6 6]))Identifying the pattern is enough to realize the answer
(+ 5 4 3 2 1)15What a marvelous pattern it is! Predictable, but not flat. Smooth, but not boring. Recursive, and not obvious.
[1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 171 190 210]There’s something special about this sequence. Aren’t those numbers just… pleasing in some way?
This sequence is called the triangular numbers.
.
. .
. . .
. . . .You can find the first 10 or so numbers in your head, and with some paper many more quite quickly.
There is a formula for calculating the nth triangular number: \(T_n = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}\)
The 20th triangular number is (20x21)/2 = (400+20)/2 = 210. Isn’t it wonderful how there’s so many different ways to find the same answer in math?
There is something curious about the formula; it divides by 2 but only produces integers. How can we be certain we will only ever get an integer? So mysterious. n(n+1) is always even! Let’s think about that a bit more, if n is odd, then n+1 is even. If n is even, then n+1 is odd. One of the multiples is always even, meaning that 2 is a factor, so the multiple must always have a factor 2, and be even.
It’s easy now to imagine if we had a 1000 sided dice what the answer would be. But be careful! For a 1000 side die, we want the 999nth triangular number: \((999 \times 1000)/2 = (1000000-1000)/2 = 500000-500 = 499500\) and the total outcomes would be 1000x1000, so the answer would be 0.4995. It’s comforting to see that for a large range, we land closer to 50%.
The point is that once we know what we are counting, it feels more obvious that we used the right formula to count it.
Triangular numbers show up in many situations, my favorite is that they can be used to lay out hexagons. The code that draws Clojure Civitas hexagons is based on the triangular number formula.
Triangular numbers also show up in the number of pairs, handshakes, edges in a complete graph, diagonals sum to triangular numbers, square numbers as sums of consecutive odd numbers, differences of triangulars, acceleration frames, smooth transitions, spacing. Such a beautiful pattern that can be found in so many situations!
Until next time, may your dice be fair and your outcomes interesting.