Why isn't this working properly

Question

the output is wrong as an example, when computer brings out rock and the user paper it says it is a tie

here's my code
//Rock paper scissors is a classic two player game. Each player chooses either rock, paper or scissors. The items are compared, and whichever player chooses the more powerful item wins.
// UserInput is defined based on user's decision
var userInput = 'rock';
userInput = userInput.toLowerCase();
//TheComputerChoice give a number between one, two and three.Based on the outcome number, it returns rock, paper or scissors

function getComputerChoice() {
x = Math.floor(Math.random() * 2);
if (x === 0)
return 'rock';
else if (x === 1)
return 'paper';
else if (x === 2)
return 'scissors';
}
// This function determines the winner by doing some simple math

function determineWinner(userChoice,computerChoice) {
if (userChoice === computerChoice)
return 'tie';
else if (userChoice === 'rock' && computerChoice === 'paper')
return 'computer won';
else if (userChoice === 'paper' && computerChoice === 'rock')
return 'user won';
else if ( userChoice === 'scissors' && computerChoice === 'rock')
return 'computer won';

else if ( userChoice === 'rock' && computerChoice === 'scissors')
return 'user won';

else if ( userChoice === 'paper' && computerChoice === 'scissors')
return 'computer won';
else if ( userChoice === 'scissors' && computerChoice === 'paper')
return 'user won';
else if ( userChoice === 'bomb')
return 'user won';
}
// In this function. userChoice and computerChoice were defined in order to use console.logs

function playGame() {
var userChoice = userInput;
var computerChoice = getComputerChoice();

console.log(determineWinner(userChoice,computerChoice));

}
// the results are printed out

console.log('userInput:' + userInput);
console.log('getComputerChoice:' + getComputerChoice());
playGame();

Answer 1

You are calling your getComputerChoice() function twice.
1st time:
console.log('getComputerChoice:' + getComputerChoice());

2nd time within this function
playGame();

Given computer choice is random the 1st result may be different to the 2nd so the choice printed may not be the same as the choice in the game.

Answer 2

You have two mistakes:

+ getComputerChoice() never return 'scissors' until you change:

x = Math.floor(Math.random() * 3); // will now produce integer in [0;3[ (0 included, 3 not included)

+ you log a different computer choice than the one computed for determine winner: put your log (at least the computer choice one) inside your playGame() function, and compute only once the computer choice:

function playGame() {
var userChoice = userInput;
var computerChoice = getComputerChoice();
console.log('getComputerChoice:' + computerChoice);

console.log(determineWinner(userChoice,computerChoice));

}
// the results are printed out

console.log('userInput:' + userInput);
//console.log('getComputerChoice:' + getComputerChoice());
playGame();