An other dicey challenge

Probability Level 4

Suppose we have the following two dice:

  • a fair, four-sided die with numbers D 4 = [ 1 , 2 , 2 , 3 ] D_4 = [1,2,2,3]
  • a fair, nine-sided die with numbers D 9 = [ 1 , 3 , 3 , 5 , 5 , 5 , 7 , 7 , 9 ] D_9 = [1,3,3,5,5,5,7,7,9] .

Rolling these two dice and adding their outcomes together results in a sum between 1 and 12. Remarkably, the probability distribution of this sum is the same as that for two regular, six-sided dice. That is, ( D 4 + D 9 ) ∼ ( D 6 + D 6 ) (D_4 + D_9) \sim (D_6 + D_6) .

Now, can you construct another pair of four-sided and nine-sided dice with positive integers different from the above but with the exact same property of ( D 4 + D 9 ) ∼ ( D 6 + D 6 ) ? (D_4 + D_9) \sim (D_6 + D_6)?

If A A is the highest number on the four-sided die and B B is the highest number on the nine-sided die, post the product A ⋅ B A\cdot B .

Inspiration


The answer is 35.

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

This is a follow-up on the solution I wrote for the "Inspiration" problem by Satyen Nabar:

Let D n D_n stand for "a uniformly distributed number in { 0 , 1 , … , n − 1 } \{0,1,\dots,n-1\} ". Then the situation of two dice can be written as 2 + D 6 + D 6 2 + D_6 + D_6 .

Because six is a composite number, we can decompose its outcomes as D 6 = D 2 + 2 D 3 or D 6 = D 3 + 3 D 2 D_6 = D_2 + 2D_3\ \ \ \ \text{or}\ \ \ \ D_6 = D_3 + 3D_2 In other words, random choice of { 0 , 1 , 2 , 3 , 4 , 5 } = random choice of { 0 , 1 } + random choice of { 0 , 2 , 4 } ; random choice of { 0 , 1 , 2 , 3 , 4 , 5 } = random choice of { 0 , 1 , 2 } + random choice of { 0 , 3 } . \text{random choice of}\ \{0,1,2,3,4,5\} = \text{random choice of}\ \{0,1\} + \text{random choice of}\ \{0,2,4\}; \\ \text{random choice of}\ \{0,1,2,3,4,5\} = \text{random choice of}\ \{0,1,2\} + \text{random choice of}\ \{0,3\}.

The example from the problem description is generated by writing 2 + D 6 + D 6 = 2 + ( D 2 + 2 D 3 ) + ( D 2 + 2 D 3 ) = ( 1 + D 2 + D 2 ) + ( 1 + 2 D 3 + 2 D 3 ) . 2 + D_6 + D_6 = 2 + (D_2 + 2D_3) + (D_2 + 2D_3) = (1 + D_2 + D_2) + (1 + 2D_3 + 2D_3). Thus we get the dice D 4 = 1 + { 0 , 1 } + { 0 , 1 } = { 1 , 2 ; 2 , 3 } ; D 9 = 1 + { 0 , 2 , 4 } + { 0 , 2 , 4 } = { 1 , 3 , 5 ; 3 , 5 , 7 ; 5 , 7 , 9 } . D_4 = 1 + \{0,1\} + \{0,1\} = \{1,2;2,3\};\ \ \ \ D_9 = 1 + \{0,2,4\} + \{0,2,4\} = \{1,3,5;3,5,7;5,7,9\}.

Another solution we get by writing 2 + D 6 + D 6 = 2 + ( D 3 + 3 D 2 ) + ( D 3 + 3 D 2 ) = ( 1 + 3 D 2 + 3 D 2 ) + ( 1 + D 3 + D 3 ) . 2 + D_6 + D_6 = 2 + (D_3 + 3D_2) + (D_3 + 3D_2) = (1 + 3D_2 + 3D_2) + (1 + D_3 + D_3). This results in D 4 ′ = 1 + { 0 , 3 } + { 0 , 3 } = { 1 , 4 ; 4 , 7 } ; D 9 ′ = 1 + { 0 , 1 , 2 } + { 0 , 1 , 2 } = { 1 , 2 , 3 ; 2 , 3 , 4 ; 3 , 4 , 5 } . D_4' = 1 + \{0,3\} + \{0,3\} = \{1,4;4,7\};\ \ \ \ D_9' = 1 + \{0,1,2\} + \{0,1,2\} = \{1,2,3;2,3,4;3,4,5\}. The answer is therefore 7 ⋅ 5 = 35 7\cdot 5= \boxed{35} .

Finally, the last possible solution is found as 2 + D 6 + D 6 = 2 + ( D 3 + 3 D 2 ) + ( D 2 + 2 D 3 ) = ( 1 + 3 D 2 + D 2 ) + ( 1 + D 3 + 2 D 3 ) . 2 + D_6 + D_6 = 2 + (D_3 + 3D_2) + (D_2 + 2D_3) = (1 + 3D_2 + D_2) + (1 + D_3 + 2D_3). This results in D 4 ′ ′ = 1 + { 0 , 3 } + { 0 , 1 } = { 1 , 2 ; 4 , 5 } ; D 9 ′ ′ = 1 + { 0 , 1 , 2 } + { 0 , 2 , 4 } = { 1 , 2 , 3 ; 3 , 4 , 5 ; 5 , 6 , 7 } . D_4'' = 1 + \{0,3\} + \{0,1\} = \{1,2;4,5\};\ \ \ \ D_9'' = 1 + \{0,1,2\} + \{0,2,4\} = \{1,2,3;3,4,5;5,6,7\}. The answer is therefore 5 ⋅ 7 = 35 5\cdot 7 = \boxed{35} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I guess my concern in this discussion is how you know that your method finds all the solutions. It finds solutions, yes, and in easy cases finds all the solutions. However, you have already had to extend your method to include the D* forms, and this method is likely to be insufficient as well, as Calvin has pointed out.

Factorising the probability generating functions in term of (irreducible) cyclotomic polynomials is a technique which is assured of finding all solutions.

Mark Hennings - 4 years ago

FYI Thinking of it in terms of the generating functions is a much better way to approach such problems.

Using only these D n D_n greatly restricts the type of dice that you can have.

Calvin Lin Staff - 4 years ago

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Can you give me an example of this restriction? So far I have found all solutions this way, both for six- and twelve-sided dice. Moreover, the decomposition in terms of D D s is easier and more intuitive than factorizing the generating function, and leads to fewer possible permutations.

Arjen Vreugdenhil - 4 years ago

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Find a different pair of dice with 5 and 6 sides, that is equivalent to throwing the pair of dice [1, 2, 3, 4, 5, 6] and [1, 2, 3, 4, 5]

Answer: [1, 2, 2, 3, 3, 4], [1, 3, 4, 5, 7]. It is obvious that this 5 sided dice doesn't arise from any combinations of random choice.

If you consider the generating function, we obtain the answer by:

( x + x 2 + x 3 + x 4 + x 5 + x 6 ) ( x + x 2 + x 3 + x 4 + x 5 ) = [ x ( 1 + x ) ( 1 + x + x 2 ) ( 1 − x + x 2 ) ] [ ( x + x 2 + x 3 + x 4 + x 5 ) ] = [ x ( 1 + x ) ( 1 + x + x 2 ) ] [ ( 1 − x + x 2 ) ( x + x 2 + x 3 + x 4 + x 5 ) ] = [ x + 2 x 2 + 2 x 3 + x 4 ] [ x + x 3 + x 4 + x 5 + x 7 ] ( x + x^2 + x^3 + x^4 + x^5 + x^6 ) ( x+ x^2 + x^3 + x^4 + x^5) \\ = [ x ( 1 + x ) ( 1 + x + x^2 ) ( 1 - x + x^2 ) ] [ ( x + x^2 + x^3 + x^4 + x^ 5 ) ] \\ = [ x ( 1 + x ) ( 1 + x + x^2 ) ] [ ( 1 - x + x^2 ) ( x + x^2 + x^3 + x^4 + x^ 5 ) ] \\ = [ x + 2x^2 + 2x^3 + x^4] [ x + x^3 + x^4 + x^5 + x^7 ]

The main restriction is ensuring that the coefficients are non-negative. That led me to think of ( 1 − x + x 2 ) ( x + x 2 + x 3 + x 4 + x 5 ) ( 1 - x + x^2 ) ( x + x^2 + x^3 + x^4 + x^ 5 ) (which thankfully works).

Calvin Lin Staff - 4 years ago

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@Calvin Lin – Edit: This is not a valid counterexample.

In fact, there is a simpler counterexample (using the same ideas) - Find a different pair of 6-sided dice, that is equivalent to throwing the standard 6-sided dice.

Answer: [ 1 , 2 , 2 , 3 , 3 , 4 ] [1, 2, 2, 3, 3, 4 ] and [ 1 , 3 , 4 , 5 , 6 , 8 ] [ 1, 3, 4, 5, 6, 8]

The first dice could be written as " 1 + random choice { 0 , 1 , 2 } \{0, 1, 2 \} + random choice { 0 , 1 } \{0, 1 \} " (corresponding to x ( 1 + x ) ( 1 + x + x 2 ) x (1+x)(1+x+x^2) , but the second dice cannot, because it corresponds to ( 1 − x + x 2 ) x ( 1 + x + x 2 + x 3 + x 4 + x 5 ) (1-x+x^2) x ( 1 + x + x^2 + x^3 + x^4 +x^5) . Other than manually verifying, we know that the each factor of ( 1 − x + x 2 ) (1-x+x^2) would require a corresponding factor of ( 1 + x + x 2 ) (1+x+x^2) in order to pair up the "random choice".

Calvin Lin Staff - 4 years ago

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@Calvin Lin – I dealt with this simpler problem here ... The second die can be written as 1 + { 0 , 3 , 6 } + { 0 , 2 } . 1 + \{0,3,6\} + \{0,2\}. I found this solution with my D D symbols because ( 1 − x + x 2 ) ( 1 + x + x 2 + x 3 + x 4 + x 5 ) = ( 1 − x + x 2 ) ( 1 + x + x 2 ) ( 1 + x 3 ) = ( 1 + x 2 + x 4 ) ( 1 + x 3 ) , (1-x+x^2)(1 + x+x^2+x^3+x^4+x^5) \\ = (1 - x + x^2)(1 + x + x^2)(1 + x^3) \\ = (1 + x^2 + x^4)(1 + x^3), which in my notation was 2 D 3 + 3 D 2 2D_3 + 3D_2 .

That is, where the original sum of two dice can be written as 2 + D 6 + D 6 = ( 1 + 3 D 2 + D 3 ) + ( 1 + 2 D 3 + D 2 ) , 2 + D_6 + D_6 = (1 + 3D_2 + D_3) + (1 + 2D_3 + D_2), the new Sicherman dice may be written as 2 + D A + D B = ( 1 + D 2 + D 3 ) + ( 1 + 3 D 2 + 2 D 3 ) . 2 + D_A + D_B = (1 + D_2 + D_3) + (1 + 3D_2 + 2D_3). As you can see, I simply made a swap of D D terms.

Arjen Vreugdenhil - 4 years ago

@Calvin Lin – That is indeed a good example where there is no straightforward decomposition of D 5 D_5 .

In my response to a solution of Mark Hennings to a similar problem, I suggested the introduction of a "pseudo-die" D 3 ⋆ = { 0 , 1 ⋆ , 2 } D^\star_3 = \{0,1^\star,2\} . It combines with other dice by "deleting" and has the property D 3 + D 3 ⋆ = { 0 , 1 , 2 , 1 ⋆ , 2 ⋆ , 3 ⋆ , 2 , 3 , 4 } = { 0 , 2 , 4 } = 2 D 3 ; D 2 + D 3 ⋆ = { 0 , 1 , 1 ⋆ , 2 ⋆ , 2 , 3 } = { 0 , 3 } = 3 D 2 ; D_3 + D^\star_3 = \{0,1,2,1^\star,2^\star,3^\star,2,3,4\} = \{0,2,4\} = 2D_3; \\ D_2 + D^\star_3 = \{0,1,1^\star,2^\star,2,3\} = \{0,3\} = 3D_2; with this concept, we find D 6 = 2 D 3 + D 2 = D 3 + D 3 ⋆ + D 2 . D_6 = 2D_3 + D_2 = D_3 +D^\star_3 + D_2. Then in D 5 + D 6 = D 3 + D 3 ⋆ + D 2 + D 5 D_5 + D_6 = D_3 +D^\star_3 + D_2 + D_5 we can choose to combine the D 3 ⋆ D^\star_3 term with D 5 D_5 instead, and obtain D A = D 3 + D 2 = { 0 , 1 , 2 } + { 0 , 1 } = { 0 , 1 , 1 , 2 , 2 , 3 } , D B = D 5 + D 3 ⋆ = { 0 , 1 , 2 , 3 , 4 } + { 0 , 1 ⋆ , 2 } = { 0 , 1 , 2 , 3 , 4 , 1 ⋆ , 2 ⋆ , 3 ⋆ , 4 ⋆ , 5 ⋆ , 2 , 3 , 4 , 5 , 6 } = { 0 , 2 , 3 , 4 , 6 } . D_A = D_3 + D_2 = \{0,1,2\} + \{0,1\} = \{0,1,1,2,2,3\}, \\ D_B = D_5 + D^\star_3 = \{0,1,2,3,4\} + \{0,1^\star,2\} = \{0,1,2,3,4,1^\star,2^\star,3^\star,4^\star,5^\star,2,3,4,5,6\} \\ \ \ \ \ \ = \{0,2,3,4,6\}. This is the pair you also found.

Arjen Vreugdenhil - 4 years ago

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@Arjen Vreugdenhil – Reflecting on all this, my D D notation is an intuitive approach to the factorization of generating functions. However, it does not go all the way "down" to the prime factors in the ring k [ X ] k[X] , and may therefore miss out on certain solutions. The introduction of the D ⋆ D^\star symbol goes a long ways toward fixing this, but eventually is likely to fail as well.

So, in the end, the method with generating functions is more comprehensive, especially when we start with two dice of different size. It is also less intuitive. By now you have noticed, I think, that I like out-of-the-box intuitive approaches. It was fun developing this one; I will now bow to the power of hard algebra. :)

Arjen Vreugdenhil - 4 years ago

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@Arjen Vreugdenhil – Right. Introducing the pseudo-die is precisely the generating function 1 − x + x 2 1 - x + x^2 .

The follow-up question then would be: Is there a factor of 1 + x + x 2 + … + x n 1 + x + x^2 + \ldots + x^n which do not have coefficients restricted to 0 , + 1 , − 1 0, +1, -1 ?
As it turns out, the answer is yes (which should initially seem surprising). The first counter example at 1 + x + x 2 + … + x 104 1 + x + x^ 2 + \ldots + x^{104} . It is related to 105 = 3 × 5 × 7 105 = 3 \times 5 \times 7 being the product of 3 distinct odd primes (but requires deeper theory to understand why it is relevant).

This is closely linked to cyclotomic polynomials , because 1 − x n 1 − x = ∏ d ∣ n Φ d ( x ) \frac{ 1 - x^n } { 1-x} = \prod_{d \mid n } \Phi_d (x) .

Calvin Lin Staff - 4 years ago

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@Calvin Lin – So up to dice with 100 sides we should be good to go with a few starred D D s. :D

Arjen Vreugdenhil - 4 years ago

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@Arjen Vreugdenhil – Unfortunately no. The other condition that your n D s ∗ nD_s ^* needs, is that the exponents with non-linear coefficients from an arithmetic progression. However, since we can multiply these terms which results in cancellation, there isn't a simple classification of such allowed dice. This is where the understanding of the theory comes to our rescue :)

Checking the list of cyclotomic polynomials, the first instance where that exponent condition doesn't hold is Φ 15 ( x ) = x 8 − x 7 + x 5 − x 4 + x 3 − x + 1 \Phi_{15} (x) = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1 . This polynomial is irreducible, so you cannot hope to break it down into various D s ∗ D_s ^* .

It would take a bit of work to find the counterexample because our real dice require non-negative coefficients. One possible candidate is to see that

( x 8 − x 7 + x 5 − x 4 + x 3 − x + 1 ) ( 1 + x ) ( 1 + x 2 ) ( 1 + x 4 ) = x 15 + x 12 + x 10 + x 9 + x 6 + x 5 + x 3 + 1 (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)(1+x)(1+x^2)(1+x^4) \\ =x^{15} + x^{12} + x^{10} + x^9 + x^6 + x^5 + x^3 + 1

and so we can ask about "rolling a 15-sided, and 8-sided dice" is equivalent to which other "15-sided and 8-sided dice"? I've given the 8-sided dice above, and the 15-sided dice would can be written as D 3 ∗ + D 5 D_3^* + D_5 , corresponding to Φ 3 × Φ 5 \Phi_3 \times \Phi_5 . By looking at the factorization, we see that this 8-sided dice cannot be written as some combination of D s ∗ D_s ^ * (assuming I haven't made any logical errors)

Calvin Lin Staff - 4 years ago

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@Calvin Lin – Yes, your example works out. With D A = { 0 , 3 , 5 , 6 , 9 , 10 , 12 , 15 } , D B = D 3 + D 5 D_A = \{0,3,5,6,9,10,12,15\},\ \ \ D_B = D_3 + D_5 we get the same outcomes as D 8 + D 15 D_8 + D_{15} .

The die D A D_A (or the polynomial P ( x ) = x 15 + x 12 + x 10 + x 9 + x 6 + x 5 + x 3 + 1 P(x) = x^{15} + x^{12} + x^{10} + x^9 + x^6 + x^5 + x^3 + 1 ) still has the same remarkable symmetry as all the other ones: subtracting 15 − D A 15 - D_A gives D A D_A again; or x 15 P ( 1 / x ) = P ( x ) x^{15} P(1/x) = P(x) . Why is that? Is it possible to solve this problem with non-symmetric dice?

Also, D A = 3 D 6 ⊕ 5 D 4 ⊖ 15 D 2 D_A = 3D_6 \oplus 5D_4 \ominus 15D_2 , where ⊕ \oplus and ⊖ \ominus are addition and subtraction on bags; that is, P ( x ) = ( x 15 + x 12 + x 9 + x 6 + x 3 + 1 ) + ( x 15 + x 10 + x 5 + 1 ) − ( x 15 + 1 ) . P(x) = (x^{15} + x^{12} + x^9 + x^6 + x^3 + 1) + (x^{15} + x^{10} + x^5 + 1) - (x^{15} + 1). Is that accidental? It would be interesting to try constructing the pair of dice out using this decomposition in terms rather than factors. ("Irreducible" is, after all, only multiplicative...)

Arjen Vreugdenhil - 4 years ago

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@Arjen Vreugdenhil – I notice, for instance, that ( 2 D 3 + 3 D 3 ) ⊕ ( 4 D 3 + D 3 ) ⊖ ( 2 D 5 + D 3 ) = 5 D 3 (2D_3 + 3D_3) \oplus (4D_3 + D_3) \ominus (2D_5 + D_3) = 5D_3 : { 0 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 10 } ⊕ { 0 , 1 , 2 , 4 , 5 , 6 , 8 , 9 , 10 } ⊖ { 0 , 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 6 , 6 , 7 , 8 , 8 , 9 , 10 } = { 0 , 5 , 10 } . \small \{0,2,3,4,5,6,7,8,10\} \oplus \{0,1,2,4,5,6,8,9,10\} \ominus \{0,1,2,2,3,3,4,4,5,6,6,7,8,8,9,10\} \\ \small = \{0,5,10\}. Writing also 2 D 3 = D 3 ⋆ + D 3 2D_3 = D^\star_3 + D_3 , we get D 15 = 5 D 3 + D 5 = ( ( D 3 ⋆ + 3 D 3 ) ⊕ 4 D 3 ⊖ 2 D 5 ) + D 3 + D 5 , D_{15} = 5D_3 + D_5 = \left((D^\star_3 + 3D_3) \oplus 4D_3 \ominus 2D_5\right) + D_3 + D_5, so that the combination of a 15-sided die and an 8-sided die factors as D 15 + D 8 = ( ( D 3 ⋆ + 3 D 3 ) ⊕ 4 D 3 ⊖ 2 D 5 ) + D 3 + D 5 + 4 D 2 + 2 D 2 + D 2 . D_{15} + D_8 = \left((D^\star_3 + 3D_3) \oplus 4D_3 \ominus 2D_5\right) + D_3 + D_5 + 4D_2 + 2D_2 + D_2. Separating out D 3 + D 5 D_3 + D_5 for a new die, we get D A = ( ( D 3 ⋆ + 3 D 3 ) ⊕ 4 D 3 ⊖ 2 D 5 ) + 4 D 2 + 2 D 2 + D 2 . D_A = \left((D^\star_3 + 3D_3) \oplus 4D_3 \ominus 2D_5\right) + 4D_2 + 2D_2 + D_2. This can be further reduced to D A = ( 2 D 4 + 3 D 2 + 3 D 3 ) ⊕ ( 4 D 2 + D 12 ) ⊖ ( 2 D 4 + D 10 ) , D_A = (2D_4 + 3D_2 + 3D_3) \oplus (4D_2 + D_{12}) \ominus (2D_4 + D_{10}), which indeed gives { 0 , 2 , 3 , 3 , 4 , 5 , 5 , 6 , 6 , 6 , 7 , 7 , 8 , 8 , 9 , 9 , 9 , 10 , 10 , 11 , 12 , 12 , 13 , 15 } ⊕ { 0 , 1 , 2 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 , 7 , 8 , 8 , 9 , 9 , 10 , 10 , 11 , 11 , 12 , 13 , 14 , 15 } ⊖ { 0 , 1 , 2 , 2 , 3 , 3 , 4 , 4 , 4 , 5 , 5 , 5 , 6 , 6 , 6 , 6 , 7 , 7 , 7 , 7 , 8 , 8 , 8 , 8 , 9 , 9 , 9 , 9 , 10 , 10 , 10 , 11 , 11 , 11 , 12 , 12 , 13 , 13 , 14 , 15 } = { 0 , 3 , 5 , 6 , 9 , 10 , 12 , 15 } = D A . \small\{0,2,3,3,4,5,5,6,6,6,7,7,8,8,9,9,9,10,10,11,12,12,13,15\} \\ \small \oplus \{0,1,2,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,13,14,15\} \\ \small \ominus \{0,1,2,2,3,3,4,4,4,5,5,5,6,6,6,6,7,7,7,7,8,8,8,8, \\ \small \ \ \ \ \ 9,9,9,9,10,10,10,11,11,11,12,12,13,13,14,15\} \\ \small = \{0,3,5,6,9,10,12,15\} = D_A. Earlier I noted that also D A = 3 D 6 ⊕ 5 D 4 ⊖ 15 D 2 D_A = 3D_6 \oplus 5D_4 \ominus 15D_2 , which is, of course, much more elegant. I don't know if my D D -calculus can easily be modified to make such elegant reductions.

Arjen Vreugdenhil - 4 years ago

@Arjen Vreugdenhil – Referring to the cyclotomic polynomial theory, we can conclude that all of these dice that are a breakdown of "standard dice [ 1 , 2 , … , n ] [1, 2, \ldots, n ] ", will have the symmetry property. The roots of 1 − x n 1 − x \frac{ 1 - x^n } { 1-x} are the roots of unity, and since Φ n \Phi_n is a polynomial with real (in fact, integer) coefficients, so if ω ∣ Φ n \omega \mid \Phi_n , then ω ˉ = 1 ω ∣ Φ n \bar{\omega } = \frac{1}{ \omega} \mid \Phi _n . As such, x m ϕ n ( 1 x ) = ϕ n ( x ) x^m \phi_n ( \frac{1}{x} ) = \phi_n (x) as desired.

Thus, it is not possible to solve this problem with non-symmetric dice. Isn't that an interesting application of roots of unity and complex conjugates ?

It is not immediately obvious to me why / how. What your claim amounts to is:

A polynomial with (non-negative) integer coefficients of degree n n , whose roots are roots of unity (not necessary the nth roots of unity), can be written as linear combinations of 1 − x d ( k + 1 ) 1 − x d \frac{1-x^{d(k+1)}}{1-x^d} where d k = n dk = n .

This feels really surprising, esp since the example has 16th roots of unity, which are "magically" accounted for. I will think further about this.

Calvin Lin Staff - 4 years ago

@Mark Hennings FYI In case you're interested in this discussion.

Calvin Lin Staff - 4 years ago

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