Change the quantifier and I think you'll be good to go. The * is zero or more occurrences.

Are there always three pairings? Maybe something like

(.{3,})((, )|$)

or

([a-z]{3,})((, )|$)

would be better.

The + can be used in place of {3} if there should just be one or more. If there should be exactly three remove the trailing comma. The {} creates a range and with a , it sets a min and a max.

You can change the substitution pattern to:

$1 $1_$2

to get rid of the trailing comma but you'll need to do some right trim to remove the last underscore (in PHP it would be rtrim).

You may try:

(\w+)

Explanation of the above regex:

( - Represents the start of the capturing group

\w - Represents a word character matching from [0-9a-zA-Z_]. If you don't want to include _ then you can provide it manually; something like [0-9A-Za-z]+.

+ - Represents quantifier matching the word characters 1 or more times.

) - Represents the end of the capturing group.

$1 $1_ - For the replacement part you may use the captured group space and captured group along with an _

You can find the demo of the above regex in here.

import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class Main
{
    private static final Pattern pattern = Pattern.compile("(\\w+)");
    public static void main(String[] args) {
        String string = "aaa, bbb, ccc, ddd\n"
     + "some, someone, something, hello";
        
        String subst = "$1 $1_";
        
        Matcher matcher = pattern.matcher(string);
        
        String result = matcher.replaceAll(subst);
        System.out.println(result);
    }
}

You can find the sample run of the above implementation in here.

tl;dr

Skip the regex. Just use string manipulation, in a single statement.

Arrays
        .stream( "aaa, bbb, ccc, ddd".split( ", " ) )  // Parse the text using COMMA with SPACE as the delimiter. 
        .map( ( String s ) -> s + " " + s + "_" )      // Change `aaa` to `aaa aaa_`. 
        .collect( Collectors.joining( ", " ) );        // Join the modified strings together as a single string.

aaa aaa_, bbb bbb_, ccc ccc_, ddd ddd_

String::split

No need for regex.

You can accomplish this goal by splitting the comma-separated words into an array of strings, make a stream of that array, and join them back together again with no extra terminator. And, you can do all than in a one-liner.

Given this input:

String input = "aaa, bbb, ccc, ddd";

…split the string into pieces.

String[] pieces =  input.split( ", " ) ;

Stream

Make a stream of those pieces.

Stream < String > streamOfPieces = Arrays.stream( pieces ) ;

Take each element from the stream and transform each by adding a SPACE, the same string again, and the underscore you want at the end.

Stream < String > streamOfModifiedPieces = streamOfPieces.map( ( String s ) -> s + " " + s + "_" );

Collect results of stream

Terminate the stream by collecting all those transformed elements with a Collector. The Collector implementation we need is provided by a call to Collectors.joining. We pass our desired delimiter to that collector. In our case, the desired delimiter is a COMMA and a SPACE, ,. The collector is smart enough to include the delimiter between the elements yet omit from the end.

String output = streamOfModifiedPieces.collect( Collectors.joining( ", " ) );

output = aaa aaa_, bbb bbb_, ccc ccc_, ddd ddd_

One-liner

Combine all that into the promised one-liner.

String output = Arrays.stream( "aaa, bbb, ccc, ddd".split( ", " ) ).map( ( String s ) -> s + " " + s + "_" ).collect( Collectors.joining(", ") );

…or…

String output =
        Arrays
                .stream( "aaa, bbb, ccc, ddd".split( ", " ) )
                .map( ( String s ) -> s + " " + s + "_" )
                .collect( Collectors.joining( ", " ) );

aaa aaa_, bbb bbb_, ccc ccc_, ddd ddd_