Great blog, Tanya! I wish I’d found it years ago.

The minimal probability is indeed 3/32, and — up to switching the labels of the 2 dice —

the unique pair of dice that achieves the minimum is Oscar’s:

{(1,0,0,0,0,1)/2, (2,3,3,3,3,2)/16}.

A proof sketch is below.

Let the distribution (pmf) of the sum of the values on the 2 dice

(translated from 1-6 to 0-5 on each die) be (p[0], …, p[10]), and

let (a[0], …, a[5]) and (b[0], …, b[5]) be the pmf’s for

the 2 individual dice. The 2 key lemmas we need were already (essentially)

given in Tanya’s 13 December 2018 post:

(1) For any m >= 1, if x[i] >= 0 for 1 <= i <= m and sum_{1 <= i <= m} x[i] = C,

then

sum_{1 <= i <= m} x[i]^{2} is uniquely minimized when x[i] = C/m for all i, and

then

sum_{1 <= i = a[0]*b[5] + a[5]*b[0] >= 2*sqrt(a[0]*b[5]*a[5]*b[0]) [by the AM-GM inequality],

so

p[5] >= 2*sqrt( (a[0]*b[0])*(a[5]*b[5]) ) = 2*sqrt(p[0]*p[10]).

Now we relax the conditions of the original problem and seek the minimum of

sum_{0 <= i = 2*sqrt(p[0]*p[10]). We will show that the minimum

for this relaxed problem is 3/32, uniquely achieved when

p = (2,3,3,3,3,4,3,3,3,3,2)/32.

Then we will argue that this unique p is achieved as a convolution of a and b only

for the pair of pmf’s {a,b} that Oscar found.

To solve the relaxed optimization problem, define r[0] := sqrt(p[0]),

r[10] := sqrt(p[10]), and define the slack variable

u >= 0 as u := p[5] – 2*r[0]*r[10].

It will turn out to be more convenient to work instead with the “product” variable

P := r[0]*r[1] and the “quadratic” variable Q := (r[0] + r[1])^{2}, together with u.

Then we can express (p[0]^{2} + p[10]^{2}) + p[5]^2 as a quadratic function of

P, Q, and u with the linear inequality constraints

0 <= Q <= 1, 0 <= P <= Q/4 <= 1/4, 0 <= u= 1, Q + u <= 1.

Furthermore, for any fixed value of p[0] + p[10] + p[5], by Lemma 1 we (uniquely) minimize

the sum of the squares of the 8 remaining p[i]'s by making those 8 p[i]'s all equal.

After a little algebra, we obtain the following transformed minimization problem

(whose global minimum might not be achievable for the original problem, but which will

not lose any solutions to the original problem):

min { (Q^{2} – 4*P*Q + 2*P^{2}) + (2*P + u)^{2} + (1/8)*(1 – Q – u )^{2}) }

s.t. 0 <= Q <= 1, 0 <= P <= Q/4 <= 1/4, 0 <= u <= 1, Q + u 0, b[0] = b[5] > 0, and either

(a[1] = a[4] = 0, b[1] > 0, b[4] > 0) or (b[1] = b[4] = 0, a[1] > 0, a[4] > 0).

We continue by considering p[2] and p[8], obtaining more and more constraints on

the a[i]’s and b[i]’s until all 12 values are nailed down as claimed (up to possible

swapping of the a[] and b[] distributions).

The minimal probability is indeed 3/32, and — up to switching the labels of the 2 dice — the unique pair of dice that achieves the minimum is Oscar’s: {(1,0,0,0,0,1)/2, (2,3,3,3,3,2)/16}. A proof sketch is below.

Let the distribution (pmf) of the sum of the values on the 2 dice (translated from 1-6 to 0-5 on each die) be

(p[0], …, p[10]), and let (a[0], …, a[5]) and (b[0], …, b[5]) be the pmf’s for the 2 individual dice.

The 2 key lemmas we need were already (essentially) given in Tanya’s 13 December 2018 post:

(1) For any m >= 1, if x[i] >= 0 for 1 <= i <= m and sum_{1 <= i <= m} x[i] = C, then

sum_{1 <= i <= m} x[i]^{2} is uniquely minimized when x[i] = C/m for all i, and then

sum_{1 <= i = a[0]*b[5] + a[5]*b[0] >= 2*sqrt(a[0]*b[5]*a[5]*b[0]) [by the AM-GM inequality], so

p[5] >= 2*sqrt( (a[0]*b[0])*(a[5]*b[5]) ) = 2*sqrt(p[0]*p[10]).

Now we relax the conditions of the original problem and seek the minimum of

sum_{0 <= i = 2*sqrt(p[0]*p[10]). We will show that the minimum for this relaxed problem is 3/32, uniquely achieved when

p = (2,3,3,3,3,4,3,3,3,3,2)/32.

Then we will argue that this unique p is achieved as a convolution of a and b only for the pair of pmf’s {a,b} that Oscar found.

To solve the relaxed optimization problem, define r[0] := sqrt(p[0]), r[10] := sqrt(p[10]), and define the slack variable u >= 0 as u := p[5] – 2*r[0]*r[10]. It will turn out to be more convenient to work instead with the “product” variable P := r[0]*r[1] and the “quadratic” variable Q := (r[0] + r[1])^{2}, together with u.

Then we can express (p[0]^{2} + p[10]^{2}) + p[5]^2 as a quadratic function of P, Q, and u with the linear inequality constraints

0 <= Q <= 1, 0 <= P <= Q/4 <= 1/4, 0 <= u= 1, Q + u <= 1.

Furthermore, for any fixed value of

p[0] + p[10] + p[5], by Lemma 1 we (uniquely) minimize the sum of the squares of the 8 remaining p[i]'s by making those 8 p[i]'s all equal. After a little algebra, we obtain the following transformed minimization problem (whose global minimum might not be achievable for the original problem, but which will not lose any solutions to the original problem):

min { (Q^{2} – 4*P*Q + 2*P^{2}) + (2*P + u)^{2} + (1/8)*(1 – Q – u )^{2}) }

s.t. 0 <= Q <= 1, 0 <= P <= Q/4 <= 1/4, 0 <= u <= 1, Q + u 0, b[0] = b[5] > 0, and

either (a[1] = a[4] = 0, b[1] > 0, b[4] > 0) or (b[1] = b[4] = 0, a[1] > 0, a[4] > 0).

We continue by considering p[2] and p[8], obtaining more and more constraints on the a[i]’s and b[i]’s until all 12 values are nailed down as claimed (up to possible swapping of the a[] and b[] distributions).

Part 1:

The minimal probability is indeed 3/32, and — up to switching the labels of the 2 dice — the unique pair of dice that achieves the minimum is Oscar’s: {(1,0,0,0,0,1)/2, (2,3,3,3,3,2)/16}. A proof sketch is below.

Let the distribution (pmf) of the sum of the values on the 2 dice (translated from 1-6 to 0-5) be

(p[0], …, p[10]), and let (a[0], …, a[5]) and (b[0], …, b[5]) be the pmf’s for the 2 individual dice.

The 2 key lemmas we need were already (essentially) given in Tanya’s 13 December 2018 post:

(1) For any m >= 1, if x[i] >= 0 for 1 <= i <= m and \sum_{1 <= i <= m} x[i] = C, then

\sum_{1 <= i <= m} x[i]^{2} is uniquely minimized when x[i] = C/m for all i, and then

\sum_{1 <= i = a[0]*b[5] + a[5]*b[0] >= 2*sqrt(a[0]*b[5]*a[5]*b[0]) [by the AM-GM inequality], so

p[5] >= 2*sqrt( (a[0]*b[0])*(a[5]*b[5]) ) = 2*sqrt(p[0]*p[10]).

Now we relax the conditions of the original problem and seek the minimum of \sum_{0 <= i = 2*sqrt(p[0]*p[10]). We will show that the minimum for this relaxed problem is 3/32, uniquely achieved when

p = (2,3,3,3,3,4,3,3,3,3,2)/32.

Then we will argue that this unique p is achieved as a convolution of a and b only for the pair of pmf’s {a,b} that Oscar found.

To solve the relaxed optimization problem, define r[0] := sqrt(p[0]), r[10] := sqrt(p[10]), and define the slack variable u >= 0 as u := p[5] – 2*r[0]*r[10]. It will turn out to be more convenient to work instead with the “product” variable P := r[0]*r[1] and the “quadratic” variable Q := (r[0] + r[1])^{2}, together with u.

Then we can express (p[0]^{2} + p[10]^{2}) + p[5]^2 as a quadratic function of P, Q, and u with the linear inequality constraints 0 <= Q <= 1, 0 <= P <= Q/4 <= 1/4, 0 <= u= 1, Q + u <= 1. Furthermore, for any fixed value of

p[0] + p[10] + p[5], by Lemma 1 we (uniquely) minimize the sum of the squares of the 8 remaining p[i]'s by making those 8 p[i]'s all equal. After a little algebra, we obtain the following transformed minimization problem (whose global minimum might not be achievable for the original problem, but which will not lose any solutions to the original problem):

min { (Q^{2} – 4*P*Q + 2*P^{2}) + (2*P + u)^{2} + (1/8)*(1 – Q – u )^{2}) }

s.t. 0 <= Q <= 1, 0 <= P <= Q/4 <= 1/4, 0 <= u= 1, Q + u <= 1.

]]>The minimal probability is indeed 3/32, and — up to switching the labels of the 2 dice — the unique pair of dice that achieves the minimum is Oscar’s: {(1,0,0,0,0,1)/2, (2,3,3,3,3,2)/16}. A proof sketch is below.

Let the distribution (pmf) of the sum of the values on the 2 dice (translated from 1-6 to 0-5) be

(p[0], …, p[10]), and let (a[0], …, a[5]) and (b[0], …, b[5]) be the pmf’s for the 2 individual dice.

The 2 key lemmas we need were already (essentially) given in Tanya’s 13 December 2018 post:

(1) For any m >= 1, if x[i] >= 0 for 1 <= i <= m and \sum_{1 <= i <= m} x[i] = C, then

\sum_{1 <= i <= m} x[i]^{2} is uniquely minimized when x[i] = C/m for all i, and then

\sum_{1 <= i = a[0]*b[5] + a[5]*b[0] >= 2*sqrt(a[0]*b[5]*a[5]*b[0]) [by the AM-GM inequality], so

p[5] >= 2*sqrt( (a[0]*b[0])*(a[5]*b[5]) ) = 2*sqrt(p[0]*p[10]).

Now we relax the conditions of the original problem and seek the minimum of \sum_{0 <= i = 2*sqrt(p[0]*p[10]). We will show that the minimum for this relaxed problem is 3/32, uniquely achieved when

p = (2,3,3,3,3,4,3,3,3,3,2)/32.

Then we will argue that this unique p is achieved as a convolution of a and b only for the pair of pmf’s {a,b} that Oscar found.

To solve the relaxed optimization problem, define r[0] := sqrt(p[0]), r[10] := sqrt(p[10]), and define the slack variable u >= 0 as u := p[5] – 2*r[0]*r[10]. It will turn out to be more convenient to work instead with the “product” variable P := r[0]*r[1] and the “quadratic” variable Q := (r[0] + r[1])^{2}, together with u.

Then we can express (p[0]^{2} + p[10]^{2}) + p[5]^2 as a quadratic function of P, Q, and u with the linear inequality constraints 0 <= Q <= 1, 0 <= P <= Q/4 <= 1/4, 0 <= u= 1, Q + u <= 1. Furthermore, for any fixed value of

p[0] + p[10] + p[5], by Lemma 1 we (uniquely) minimize the sum of the squares of the 8 remaining p[i]'s by making those 8 p[i]'s all equal. After a little algebra, we obtain the following transformed minimization problem (whose global minimum might not be achievable for the original problem, but which will not lose any solutions to the original problem):

min { (Q^{2} – 4*P*Q + 2*P^{2}) + (2*P + u)^{2} + (1/8)*(1 – Q – u )^{2}) }

s.t. 0 <= Q <= 1, 0 <= P <= Q/4 <= 1/4, 0 <= u= 1, Q + u 0, b[0] = b[5] > 0, and

either (a[1] = a[4] = 0, b[1] > 0, b[4] > 0) or (b[1] = b[4] = 0, a[1] > 0, a[4] > 0).

We continue by considering p[2] and p[8], obtaining more and more constraints on the a[i]’s and b[i]’s until all 12 values are nailed down as claimed (up to possible swapping of the a[] and b[] distributions).

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