Basic word guessing game
Don't do too much in one loop


            for(int i = 0; i &lt; word.length(); i ++) {
                if (guess.equals(Character.toString(word.charAt(i)))) {
                    if(!iterated)
                        temp += Character.toString(word.charAt(i));
                    else {
                        holder = Character.toString(temp.charAt(i)).replace("-", guess);
                        temp = temp.substring(0, i) + holder + temp.substring( i + 1, temp.length());
                    }
                } else {
                    if(!iterated) {
                        temp += "-";
                    }
                }
            }



In this, you handle two entirely different purposes.  You change the existing string to display any occurrence of the character.  And on the first iteration, you fill in the string with dummy characters.  This may be slightly more efficient on that first iteration, but it is less efficient on every other iteration.  Consider how this code looks without the !iterated checks.  

            for (int i = 0; i &lt; word.length(); i ++) {
                if (guess.equals(Character.toString(word.charAt(i)))) {
                    holder = Character.toString(temp.charAt(i)).replace("-", guess);
                    temp = temp.substring(0, i) + holder + temp.substring( i + 1, temp.length());
                }
            }


How much harder is it to do the initial fill?  StackOverflow suggests something like 

    String temp;
    if (word.length() &gt; 0) {
        char[] array = new char[word.length()];
        Arrays.fill(array, '-');
        temp = new String(array);
    } else {
        temp = "";
    }


Or you could make a method like they do:  

    String temp = getStringWithLengthAndFilledWithCharacter(word.length(), '-');


Or any other variants there.  

Which code would you rather read?  Those two simple blocks or one large one?  

StringBuilder to build a string

You use repeated calls to substring and string concatenation to build your replacement string.  Behind the scenes, Java probably converts this into a StringBuilder.  Why not embrace that and do it yourself?  

            if (guess.length() != 1) {
                System.out.println("Guesses must each be a single letter.");
                continue;
            }

            char c = guess.charAt(0);
            for (int i = word.indexOf(c); i &gt;= 0; i = word.indexOf(c, i + 1)) {
                temp.setCharAt(i, c);
            }


This reduces the two line solution to one even simpler line in the loop.  

It also eliminates the need to test that the current index is equal to the guessed character.  This reverses the logic, searching for the guess in the string.  

I also added an explanatory message for why multicharacter or empty guesses don't work.  The original just silently goes to the next guess in that instance.  

Iterate with for loops

Your do/while loop skips the first check at the expense of a more complicated loop.  If you change to a for loop, it's easier to read at the expense of one extra comparison.  

Summary

        StringBuilder temp;
        if (word.length() &gt; 0) {
            char[] array = new char[word.length()];
            Arrays.fill(array, '-');
            temp = new StringBuilder(new String(array));
        } else {
            temp = new StringBuilder();
        }

        for (int turn = 0; turn &lt; 10; turn++) {
            System.out.println("Enter your letter guess");
            String guess = keyboard.nextLine();

            if (guess.length() != 1) {
                System.out.println("Guesses must each be a single letter.");
                continue;
            }

            char c = guess.charAt(0);
            for (int i = word.indexOf(c); i &gt;= 0; i = word.indexOf(c, i + 1)) {
                temp.setCharAt(i, c);
            }

            System.out.println(temp.toString());
            if (word.equals(temp.toString())) {
                System.out.println("You guessed correctly!");
                break;
            }
        }


It would be more elegant to break this into more methods, but it will work like this.