Those are some special dice!

We will use a bounding argument to solve this question. Let the four faces of one of the dice be $a$ , $b$ , $c$ and $d$ . Thus,

$\begin{aligned} \dfrac {1}{a} + \dfrac {1}{b} + \dfrac {1}{c} + \dfrac {1}{d} &= 1\\ \iff abc+abd+acd+bcd &= abcd \end{aligned}$

Note that none of the faces contain 1, otherwise the other 3 faces must have a probability of 0 to be rolled, which is impossible since no positive integer, when reciprocated, gives 0. WLOG $a < b < c < d$ . Thus, we have

$abc+abd+acd+bcd < 4bcd$

$abcd < 4bcd$

$a<4$

Now, we will do a case bash.

Case 1: $a=2$

Substituting, we get

$2bc + 2bd + 2cd + bcd = 2bcd$

$2bc + 2bd + 2cd = bcd$

From our WLOG,

$2bc + 2bd + 2cd < 6cd$

$bcd < 6cd$

$b < 6$

We will split this into further cases.

Case 1.1: $b=3$

Substituting, we get

$6c + 6d + 2cd = 3cd$

$cd-6c-6d=0$

$(c-6)(d-6)=36$

From this, we get solutions $(c,d)$ as $(7, 42)$ , $(8, 24)$ , $(9, 18)$ , $(10, 15)$ . Thus, we have solutions $(a,b,c,d)$ as $(2,3,7,42)$ , $(2,3,8,24)$ , $(2,3,9,18)$ and $(2,3,10,15)$ . (Some solutions are omitted in all the cases because they do not satisfy the WLOG assumption.)

Case 1.2: $b=4$

Substituting, we get

$8c + 8d + 2cd = 4cd$

$cd - 4c - 4d = 0$

$(c-4)(d-4)=16$

From this, we get solutions $(c,d)$ as $(5, 20)$ , $(6,12)$ . Thus, we have solutions $(a,b,c,d)$ as $(2,4,5,20)$ and $(2,4,6,12)$ .

Case 1.3: $b=5$

Substituting, we get

$10c + 10d + 2cd = 5cd$

$3cd - 10c - 10d = 0$

There are no solutions in this case.

Case 2: $a=3$

Substituting, we get

$3bc+3bd+3cd+bcd = 3bcd$

$3bc+3bd+3cd=2bcd$

From our WLOG, we have

$3bc + 3bd +3cd < 9cd$

$2bcd < 9cd$

$b \leq 4$

Thus, $b=4$ (From our WLOG assumption).

Substituting, we get

$12c + 12d + 3cd = 8cd$

$5cd -12 -12d = 0$

There are no solutions in this case.

Thus, we have 6 solutions $(2,3,7,42)$ , $(2,3,8,24)$ , $(2,3,9,18)$ , $(2,3,10,15)$ , $(2,4,5,20)$ and $(2,4,6,12)$ .

Thus, $a=6$ , $S=2+3+7+42+2+3+8+24+2+3+9+18+2+3+10+15+2+4+5+20+2+4+6+12=208$ . Therefore, $a+s=6+208=214$ .

The four unit-fractional odds must add up to 1:

$\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ + $\frac{1}{d}$ = 1

None of the denominators (a,b,c,d) can be 1, since there is no way to make the others zero. Also, trying the sum of the largest distinct unit-fractions (one-over-something) and leaving out 1/2 falls short of 1, or:

$\frac{1}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$ + $\frac{1}{6}$ < 1

So any solutions must include 1/2 because any other substitutions in the above line would be with a smaller fraction.

Now we'll look at adding together sets of distinct, progressively smaller, unit-fractions.

Using 1/3 for the second probability (1/b) , we can rule out the third fraction (1/c) being 1/4, 1/5, or 1/6 because the fourth fraction (1/d) would have to be non-positive to still add to one. So there are only 5 possibilities, starting with 1/7 as the third fraction, they are:

$\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{7}$ + $\frac{1}{x}$ = 1

$\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{8}$ + $\frac{1}{x}$ = 1

$\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{x}$ = 1

$\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{10}$ + $\frac{1}{x}$ = 1

$\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{11}$ + $\frac{1}{x}$ = 1

Before reaching:

$\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{12}$ + $\frac{1}{12}$ = 1

which is not distinct. Continuing past 1/12 produces larger values in the fourth fraction (1/d) and would generate repeat results, so the examination of using 1/3 is complete. Five results is easy to try by hand. Four of them produce sets that work.

Working similarly through 1/4 for the second probability, there are three possibilities with unit-fractions smaller than 1/4, they are:

$\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{5}$ + $\frac{1}{x}$ = 1

$\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{6}$ + $\frac{1}{x}$ = 1

$\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{7}$ + $\frac{1}{x}$ = 1

Before reaching:

$\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{8}$ = 1

Which again is not distinct, and indicates -- as with 1/3-- time to stop. Finding x above produces two more viable results. We happen to have all the results, but how do we know that?

With 1/5 as the second probability, 1/6 in the third fraction requires 13/15 for the last fraction. Using 1/7 for the third fraction requires a number larger than it to add up to 1, so we can stop examining 1/5:

$\frac{1}{2}$ + $\frac{1}{5}$ + $\frac{1}{6}$ + $\frac{13}{15}$ = 1

$\frac{1}{2}$ + $\frac{1}{5}$ + $\frac{1}{7}$ + $\frac{11}{70}$ = 1

Moving on to the second fraction as 1/6, it immediately requires a fourth fraction larger than the largest available third fraction (1/7), so we can stop the entire process of progressively examining sums of smaller fractions:

$\frac{1}{2}$ + $\frac{1}{6}$ + $\frac{1}{7}$ + $\frac{4}{21}$ = 1

There are no smaller unit-fractions (1/b) that can still add with other smaller unit-fractions to reach 1.

Sometimes crunching it out can be gratifying, especially once it is clear there is an end in sight. In total, this required 12 by-hand calculations of adding fractions.

These six sets make a = 6 and the sums of the denominators found through this process was 208 for a total of 214.

Here's a python script that finds all the ways to form a target number as a sum of integer reciprocals. e.g. calling ReciprocalSum(1,4) outputs

[[1/2, 1/3, 1/7, 1/42], [1/2, 1/3, 1/8, 1/24], [1/2, 1/3, 1/9, 1/18], [1/2, 1/3, 1/10, 1/15], [1/2, 1/4, 1/5, 1/20], [1/2, 1/4, 1/6, 1/12]].

I ran my code in a sage notebook to get exact fractions.

def ReciprocalSum(target, p, n_min=None, partial_sol=None, sols=None):
#returns all ways to express target as the sum of "p" integer reciprocal pieces with increasing denominator >= n_min

#default values
if n_min is None:
    n_min = 1
if partial_sol is None:
    partial_sol = []
if sols is None:
    sols = []

#base cases
if target < 0:
    return False
if p == 0 and target == 0:
    return [partial_sol]
elif p==0 or target == 0:
    return False       
n_max = (target/p)^(-1)
if n_min > n_max:
    return False

for n in range(n_min, int(n_max + 2)):
    new_sols = ReciprocalSum(target - 1/n, p-1, n+1, partial_sol + [1/n])
    if new_sols:
        sols += new_sols    

return sols

There is a more logical way of doing this. We need four unique positive integers where $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 1$ . There are so few of these, I'll find them all. In order to do that, I'll find cases where $\frac{1}{n} = \frac{1}{a} + \frac{1}{b}$ and build up to the four-case.

First I'll note that $a + b = c$ and $0 < b \leq a$ implies the larger one, $a$ , must be in this range $\frac{c}{2} \leq a < c$ . This is intuitive, but here's a proof anyway: $b \leq a$ implies $a + b \leq a + a$ but this is $c \leq 2a$ . $0 < b$ implies $-b < 0$ so $a = a + b - b < c + 0$ .

Now I will try to find $\frac{1}{c} = \frac{1}{a} + \frac{1}{b}$ . Without loss of generality, $a$ will be the larger of the two. I need only search $\frac{1}{2c} \leq \frac{1}{a} < \frac{1}{c}$ , so I will start $a = c+1$ and count up to $a = 2c$ , writing down the cases where there exists a $b$ such that $\frac{1}{c} = \frac{1}{a} + \frac{1}{b}$ .

I'll stop here. You'll see in a minute why I can stop at 1/6 and skip 1/5.

Now I can make a tree of all of the ways to get to 1.

1 = 1/2 + 1/2 = 1/2 + 1/4 + 1/4 = 1/2 + 1/4 + (two ways to make up 1/4 using unique fractions)

1 = 1/2 + 1/2 = 1/2 + 1/3 + 1/6 = 1/2 + 1/6 + (one way to make up 1/3 using unique fractions)

Oops, we already counted that in the previous case.

1 = 1/2 + 1/2 = 1/2 + 1/3 + 1/6 = 1/2 + 1/3 + (four ways to make up 1/6 using unique fractions)

Those are all of the ways to get to 1 = 1/2 + 1/2 = 1/2 + 1/a + 1/b + 1/c. If the dice had more than 4 sides, then I would need to further split the fractions. 1/5 and 1/24 might have many decompositions. But I don't need to do that now, because it would result in more than 4 terms.

Hence the dice faces are