mjhjmtu

% x=abracadabra
% echo $x//a/o
obrocodobro

Hmph...

Is there a way to replace the last occurrence of a pattern using shell substitions (IOW, without rolling out sed, awk, perl, etc.)?

Note: this replaces a trailing occurrence - not quite the same as "the last occurrence"

From the Bash reference manual Section 3.5.3 Shell Parameter Expansion
:

$parameter/pattern/string 

The pattern is expanded to produce a pattern just as in filename expansion. Parameter is expanded and the
longest match of pattern against its value is replaced with string. If
pattern begins with ‘/’, all matches of pattern are replaced with
string. Normally only the first match is replaced. If pattern begins
with ‘#’, it must match at the beginning of the expanded value of
parameter. If pattern begins with ‘%’, it must match at the end of the
expanded value of parameter. If string is null, matches of pattern are
deleted and the / following pattern may be omitted. If the nocasematch
shell option (see the description of shopt in The Shopt Builtin) is
enabled, the match is performed without regard to the case of
alphabetic characters. If parameter is ‘@’ or ‘’, the substitution
operation is applied to each positional parameter in turn, and the
expansion is the resultant list. If parameter is an array variable
subscripted with ‘@’ or ‘’, the substitution operation is applied to
each member of the array in turn, and the expansion is the resultant
list.

So

$ x=abracadabra
$ echo "$x/%a/o"
abracadabro

You can do this with shell substitutions its just a bit tricky. You basically grab everything before the pattern insert your replacement and than grab everything after the pattern. For your example that would look like this:

 x=abracadabra
 echo "$x%a*o$x##*a"

EDIT: Or just do what steeldriver suggested in the comments.