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I found a curious connection between probabilities of dice outcomes and number of elements of a hypercube and I can't make sense of it.

I was trying to calculate odds of getting $0,1,..,x$ threes when rolling $x$ (hypothetical) 3-sided dice, i.e. when rolling 5 3-sided dice, what are the odds 2 of them will land on threes etc.

The (AFAICT correct) odds are (for 0,1,2,3,4,5 threes respectively):

  • 1 dice - $2/3$, $1/3$
  • 2 dice - $4/9$, $4/9$, $1/9$
  • 3 dice - $8/27$, $12/27$, $6/27$, $1/27$
  • 4 dice - $16/81$, $32/81$, $24/81$, $8/81$, $1/81$
  • 5 dice - $32/243$, $80/243$, $80/243$, $40/243$, $10/243$, $1/243$

Compare it with number of elements in x-dimentional cube (from wikipedia):

  • 1-cube - 2 vertices, 1 edge
  • 2-cube - 4 vertices, 4 edges, 1 face
  • 3-cube - 8 vertices, 12 edges, 6 faces, 1 cell
  • 4-cube - 16 vertices, 32 edges, 24 faces, 8 cells, 1 4-face
  • 5-cube - 32 vertices, 80 edges, 80 faces, 40 cells, 10 4-faces, 1 5-face

This continues as least to 10 dice/dimensions. Same thing happens with 2-side dice (aka coins) and hyper-tetrahedron.

Why do probabilities of rolling a certain dice face $y$ number of times on $x$ dice would match a number of elements of a $x$-dimentional shape like that?

Grok_Narok
  • 53

3 Answers3

2

Actually the connection is very strong: it's an isomorphism.

The outcome of the roll of $n$ $r$-faced dice is a $n$-tuple, with components value ranging from $1$ to $r$, which also define a $n$-(hyper)cube, with sides $1, \cdots, r$.
Then an outcome = point $(r,r,x,\cdots,x)$, for instance, is a face of dimension $n-2$, etc.

This of course means that you "label" each of the $n$ dice, and in the same way you consider the order in the results.

See also this post. for some further applications.

G Cab
  • 35,272
1

At least I can explain the first three columns.

The events we're considering are the results of throwing $x$ die, which in turn can be written as a tuple $v \in\{0,1,2\}^x$ (here, I will consider your 'three' as a two).

Probability of getting $0$ two's with $x$ die

So our event are tuples in $\{0,1\}^x$, which are exactly the vertices of the $x$-dimensional hypercube.

Probability of getting $1$ two's with $x$ die

Suppose that $v$ lies in this event and its first coordinate is the one that equal $2$, so $v$ can be written as $v = 2\mid\mid u$ where $u\in\{0,1\}^{x-1}$, we can map this tuple to the edge of the $x$-dimensional cube that joins the vertices $0\mid\mid v$ and $1\mid\mid v$. This is a bijective correspondence between our event and the set of edges.

Probability of getting 2 two's with $x$ dies

We proceed the same way, this time the two coordinate that equal $2$ determine the "corners" of the face we map our tuple to.

Daniel
  • 6,999
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You can see this as a "coincidence" explained by:

The number of $m$ faces of an $n$-cube is: $$2^{n-m}\binom{n}{m}$$as explained on wikipedia.

The probability of throwing exactly $m$ times a $3$ by $n$ throws of a $3$-sided die is: $$\binom{n}{m}\left(\frac13\right)^m\left(\frac23\right)^{n-m}$$ because we are dealing evidently with binomial distribution having parameters $n$ and $p=\frac13$.

This equality might inspire to construct some model in which both concepts come together.

That might be done in the other answer(s) to this question :-).

drhab
  • 151,093