Java - Unexpected exiting of if else statement, possible improper use of scanner comparison? [duplicate]

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• How do I compare strings in Java? 23 answers

I am new to Java. I am working on this bit of code which is being used to create a menu. This is just an example for school, not a real project. I looked for resources online and found some discussion about using string1.equals(string2) when comparing string items. I also have used the method shown below for our other projects to date without any issue.

I did remove the scanner line of code and hardcoded the variable age of 21 and a "S" or a "T" for the variable choice to see if the proper output would print, and it did! I am thinking that there is an issue with my scanner input. Can anyone assist me in figuring it out?

package chapter3.pkg5a;

/** * * @author */

import java.util.Scanner;

public class Chapter35a {

public static void main(String[] args) {
    Scanner stdin = new Scanner(System.in);
    String choice;
    int age = 21;
    choice = stdin.nextLine();
    if (choice == "S" || choice == "T" || choice == "B" && age <= 21) {
    if (choice == "S")
        System.out.println("vegetable juice");
    else if (choice == "T")
        System.out.println("cranberry juice");
    else
        System.out.println("soda");         
}
    else if (choice == "S" || choice == "T" || choice == "B" && age > 21) {
    if (choice == "S")
        System.out.println("cabernet");
    else if (choice == "T")
        System.out.println("chardonnay");
    else
        System.out.println("IPA");  
}   
    else
System.out.println("invalid menu selection");
}

}