It's an optimization that shines its ugly head through. When you use $1 (or any of its friends), no copy is actually made - there's a pointer that still needs to be followed (a pointer into the original string). However, since you are doing another match before querying $hash{a}, the structure in $1 has been overwritten. You have to treat $1 as it were a reference (or maybe alias would be a better term).

The "$1" forces a copy. Or you could swap the assignment:

hash = (
        b => ($b =~ /\d/    ? 1  : 0),
        a => ($b =~ /(\d+)/ ? $1 : 0),
);
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Or not use $1:

hash = (
        a => ($b =~ /(\d+)/)[0] || 0,
        b => ($b =~ /\d/    ? 1  : 0)
);
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It's related to the problem when passing $1 to a subroutine:

$ perl -wE 'sub f {say @_} 3 =~ /(\d)/; f $1'
3
$ perl -wE 'sub f {"1" =~ "1"; say @_} 3 =~ /(\d)/; f $1'
Use of uninitialized value $_[0] in say at -e line 1.
$ perl -wE 'sub f {"1" =~ "1"; say @_} 3 =~ /(\d)/; f "$1"'
3
$
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$1 really is valid only till the next successful match.