As a first pass, we can approximate the odds of landing on a face by projecting the polyhedron on a sphere, and taking the fraction of the sphere which it covers.

Then the odds of a triangular, square or pentagonal roll overall are $$14.4\%,\ 50.4\%,\ 35.1\%$$ and the odds for an individual triangular, square or pentagonal face are $$0.72\%,\ 1.68\%,\ 2.93\%.$$

This is Simpson's method. It also represents the result of throwing the die high above an adhesive surface, so that the die is well-randomized in the air, and then after touching the surface falls onto the nearest side.

The angles do not have nice exact formulas, but they are easy enough to calculate in Mathematica:

polyhedron = PolyhedronData["SmallRhombicosidodecahedron"]
data       = polyhedron[[1]]
vertices   = data[[1]]
faces      = data[[2,1]]
triangle   = faces[[20]]
square     = faces[[50]]
pentagon   = faces[[62]]
center[x_] := Table[vertices[[x[[i]]]], {i, 1, Length[x]}] // Mean // Simplify
angle[a_, b_, c_] := 2 ArcTan[Abs[a.Cross[b, c]]/
    (Norm[a] Norm[b] Norm[c] + (a.c) Norm[b] + (b.c) Norm[a] + (c.a) Norm[b])]
angleof[face_] := angle[center[face], vertices[[face[[1]]]], vertices[[face[[2]]]]]
    *Length[face]
angles = Map[angleof, {triangle, square, pentagon}] // Simplify
N[{20, 30, 12} angles/(4 Pi)]
N[angles/(4 Pi)]
(angles /. Sqrt[5] -> (u - 10)/4) // Simplify // TeXForm

This gives the following solid angle measures for each triangular, square or pentagonal face: $$6 \arctan \left(\frac{u-4}{24 \sqrt{5u+5}+58 \sqrt{u+1}+9 \sqrt{9 u-3}+4 \sqrt{45 u-15}+\sqrt{3723 u-3903}}\right),\\ 8 \arctan\left(\frac{1}{u+\left(2+\sqrt{5}\right) \sqrt{u+1}}\right),\\ 10 \arctan\left(\frac{1}{\sqrt{10u}+3 \sqrt{u+1}}\right)$$ where $u=10+4\sqrt{5}$.

As a first pass, we can approximate the odds of landing on a face by projecting the polyhedron on a sphere, and taking the fraction of the sphere which it covers.

Then the odds of a triangular, square or pentagonal roll overall are $$14.4\%,\ 50.4\%,\ 35.1\%$$ and the odds for an individual triangular, square or pentagonal face are $$0.72\%,\ 1.68\%,\ 2.93\%.$$

This is Simpson's method. It also represents the result of throwing the die high above an adhesive surface, so that the die is well-randomized in the air, and then after touching the surface falls onto the nearest side.

The angles have exact formulas which are easy enough to calculate in Mathematica:

data = PolyhedronData["SmallRhombicosidodecahedron"];
faces = Map[data[[1, 1]][[#]] &, data[[1, 2, 1]]];
angle[a_, b_, c_] := 2 ArcTan[Abs[a.Cross[b, c]]/
     (Norm[a] Norm[b] Norm[c] + a.b Norm[c] + b.c Norm[a] + c.a Norm[b])];
Map[Length[#] angle[# // Mean // Simplify, #[[1]], #[[2]]] &, faces]
     // FullSimplify // Union

This gives the following solid angle measures for each triangular, square or pentagonal face: $$6 \cot^{-1}\left(2 \sqrt{3} u+\sqrt{124 u-61}\right),\\ 8 \cot^{-1}\left(2u+\sqrt{40 u-21}\right),\\ 10 \cot^{-1}\left(2 \sqrt{5u}+3 \sqrt{2u+1}\right)$$ where $u=5+2\sqrt{5}$.

As a first pass, we can approximate the odds of landing on a face by projecting the polyhedron on a sphere, and taking the fraction of the sphere which it covers.

Then the odds of a triangular, square or pentagonal roll overall are $$14.4\%,\ 50.4\%,\ 35.1\%$$ and the odds for an individual triangular, square or pentagonal face are $$0.72\%,\ 1.68\%,\ 2.93\%.$$

This is Simpson's method. It also represents the result of throwing the die high in the air above an adhesive surface, so that the die is well-randomized, and then after touching the surface falls quickly onto the nearest side.

The angles do not have nice exact formulas, but they are easy enough to calculate in Mathematica:

polyhedron = PolyhedronData["SmallRhombicosidodecahedron"]
data       = polyhedron[[1]]
vertices   = data[[1]]
faces      = data[[2,1]]
triangle   = faces[[20]]
square     = faces[[50]]
pentagon   = faces[[62]]
center[x_] := Table[vertices[[x[[i]]]], {i, 1, Length[x]}] // Mean // Simplify
angle[a_, b_, c_] := 2 ArcTan[Abs[a.Cross[b, c]]/
    (Norm[a] Norm[b] Norm[c] + (a.c) Norm[b] + (b.c) Norm[a] + (c.a) Norm[b])]
angleof[face_] := angle[center[face], vertices[[face[[1]]]], vertices[[face[[2]]]]]
    *Length[face]
angles = Map[angleof, {triangle, square, pentagon}] // Simplify
N[{20, 30, 12} angles/(4 Pi)]
N[angles/(4 Pi)]
(angles /. Sqrt[5] -> (u - 10)/4) // Simplify // TeXForm

This gives the following angle measures for each triangular, square or pentagonal face: $$6 \arctan \left(\frac{u-4}{24 \sqrt{5u+5}+58 \sqrt{u+1}+9 \sqrt{9 u-3}+4 \sqrt{45 u-15}+\sqrt{3723 u-3903}}\right),\\ 8 \arctan\left(\frac{1}{u+\left(2+\sqrt{5}\right) \sqrt{u+1}}\right),\\ 10 \arctan\left(\frac{1}{\sqrt{10u}+3 \sqrt{u+1}}\right)$$ where $u=10+4\sqrt{5}$.

As a first pass, we can approximate the odds of landing on a face by projecting the polyhedron on a sphere, and taking the fraction of the sphere which it covers.

Then the odds of a triangular, square or pentagonal roll overall are $$14.4\%,\ 50.4\%,\ 35.1\%$$ and the odds for an individual triangular, square or pentagonal face are $$0.72\%,\ 1.68\%,\ 2.93\%.$$

This is Simpson's method. It also represents the result of throwing the die high above an adhesive surface, so that the die is well-randomized in the air, and then after touching the surface falls onto the nearest side.

The angles do not have nice exact formulas, but they are easy enough to calculate in Mathematica:

polyhedron = PolyhedronData["SmallRhombicosidodecahedron"]
data       = polyhedron[[1]]
vertices   = data[[1]]
faces      = data[[2,1]]
triangle   = faces[[20]]
square     = faces[[50]]
pentagon   = faces[[62]]
center[x_] := Table[vertices[[x[[i]]]], {i, 1, Length[x]}] // Mean // Simplify
angle[a_, b_, c_] := 2 ArcTan[Abs[a.Cross[b, c]]/
    (Norm[a] Norm[b] Norm[c] + (a.c) Norm[b] + (b.c) Norm[a] + (c.a) Norm[b])]
angleof[face_] := angle[center[face], vertices[[face[[1]]]], vertices[[face[[2]]]]]
    *Length[face]
angles = Map[angleof, {triangle, square, pentagon}] // Simplify
N[{20, 30, 12} angles/(4 Pi)]
N[angles/(4 Pi)]
(angles /. Sqrt[5] -> (u - 10)/4) // Simplify // TeXForm

This gives the following solid angle measures for each triangular, square or pentagonal face: $$6 \arctan \left(\frac{u-4}{24 \sqrt{5u+5}+58 \sqrt{u+1}+9 \sqrt{9 u-3}+4 \sqrt{45 u-15}+\sqrt{3723 u-3903}}\right),\\ 8 \arctan\left(\frac{1}{u+\left(2+\sqrt{5}\right) \sqrt{u+1}}\right),\\ 10 \arctan\left(\frac{1}{\sqrt{10u}+3 \sqrt{u+1}}\right)$$ where $u=10+4\sqrt{5}$.