~~rand($') is the same as int(rand($'))

I may be missing something very obvious, but could someone step through this one step at a time? The way I'm reading it, something like the below happens, and it doesn't make much sense to me.

Input: 2d4+10
Code: s/d/q(+~~rand($')+d)x$`/e;

2d4+10
2(q(+~~rand($')+d)x$`)4+10
2(q(+int(rand($'))+d)x$`)4+10
2(q(+int(rand(4+10))+d)x2)4+10
2('+int(rand(4+10))+d'x2)4+10
2('+int(rand(4+10))+d''+int(rand(4+10))+d')4+10
=?????
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2d4+10

2(q(+~~rand($')+d)x$`)4+10

2(q(+int(rand($'))+d)x$`)4+10

2(q(+int(rand('4+10'))+d)x2)4+10

2('+int(rand('4+10'))+d'x2)4+10

2('+int(rand('4+10'))+d'.'+int(rand('4+10'))+d')4+10

2+int(rand(4))+d+int(rand(4))+d4+10

# or looking at each term

2             # since rand return 0..N-1 instead on 1..N, we add one f
+or every die

+int(rand(4)) # int in 0..3

+d            # 0

+int(rand(4)) # int in 0..3

+d4           # also 0, the 'd' are here to help ignore the 4

+10           # desired extra

# then eval that :)
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