Rock, Paper, Scissors in JavaScript

Something to study:

var choices = ["rock", "paper", "scissors"];
var map = {};

choices.forEach(function(choice, i) {
    map[choice] = {};
    map[choice][choice] = "Was a tie"
    map[choice][choices[(i+1)%3]] = choices[(i+1)%3] + " wins"
    map[choice][choices[(i+2)%3]] = choice + " wins"
})

function compare(choice1, choice2) {
    return (map[choice1] || {})[choice2] || "Invalid choice";
}

---

Here's an alternate that will work for expanded sets. The assumption is that there's an odd number of possibilities, and from any given point, of the total number of opposition, reading forward from our given point (and wrapping around when we reach the end) the first half will win over the given point, and the second half will lose.

Or another way to describe it would be that the half of the remaining opponents that precede our given point will lose, and the half that follow will win.

Therefore the proper order in the choices Array is crucial.

var choices = ["rock", "spock", "paper", "lizard", "scissors"];
var map = {};

choices.forEach(function(choice, i) {
    map[choice] = {};
    for (var j = 0, half = (choices.length-1)/2; j < choices.length; j++) {
        var opposition = (i+j)%choices.length
        if (!j)
            map[choice][choice] = "Was a tie"
        else if (j <= half)
            map[choice][choices[opposition]] = choices[opposition] + " wins"
        else
            map[choice][choices[opposition]] = choice + " wins"
    }
})

function compare(choice1, choice2) {
    return (map[choice1] || {})[choice2] || "Invalid choice";
}

I came up with an alternative that should be easy for you to understand and avoids some issues in your code, like excessive repetition and fixed choices. It is thus much more flexible and easier to maintain.

function compare(choice1, choice2) {
    choice1 = choices.indexOf(choice1);
    choice2 = choices.indexOf(choice2);
    if (choice1 == choice2) {
        return "Tie";
    }
    if (choice1 == choices.length - 1 && choice2 == 0) {
        return "Right wins";
    }
    if (choice2 == choices.length - 1 && choice1 == 0) {
        return "Left wins";
    }
    if (choice1 > choice2) {
        return "Left wins";
    } else {
        return "Right wins";
    }
}

choices is var choices = ["rock", "paper", "scissors"];. You can see a demonstration.

---

To generalize the solution to larger lists, this modulo technique can be helpful:

function mod(a, b) {
    c = a % b
    return (c < 0) ? c + b : c
}

Then it's much easier to write the comparison code:

function compare(choice1, choice2) {
    x = choices.indexOf(choice1);
    y = choices.indexOf(choice2);
    if (x == y) {
        return "Tie";
    }
    if (mod((x - y), choices.length) < choices.length / 2) {
        return choice1 + " wins";
    } else {
        return choice2 + " wins";
    }
}

The corresponding jsFiddle.

You were unable to see the issue most likely due to poor indentation of your code. Properly indented the issue is clear:

if (choice1 === "paper") {
    if (choice2 === "rock") {
        return "paper wins";
    } else {
        if (choice2 === "scissors") {
            return "scissors wins";
        }
    }
    if (choice1 === "scissors") {
        if (choice2 === "rock") {
            return "rock wins";
        } else {
            if (choice2 === "paper") {
                return "scissors wins";
            }
        }
    }
}

Your if (choice1 === "scissors") { is within if (choice1 === "paper") {. The code within will never be reached.