If I understand right then you're just looking for a permutation (i.e. the numbers randomised with no repeats) of the numbers 1-10? Maybe try generating a randomised list of those numbers, once, at the start, and then just working your way through those?

This will calculate a random permutation of the numbers in nums:

var nums = [1,2,3,4,5,6,7,8,9,10],
    ranNums = [],
    i = nums.length,
    j = 0;

while (i--) {
    j = Math.floor(Math.random() * (i+1));
    ranNums.push(nums[j]);
    nums.splice(j,1);
}

So, for example, if you were looking for random numbers between 1 - 20 that were also even, then you could use:

nums = [2,4,6,8,10,12,14,16,18,20];

Then just read through ranNums in order to recall the random numbers.

This runs no risk of it taking increasingly longer to find unused numbers, as you were finding in your approach.

EDIT: After reading this and running a test on jsperf, it seems like a much better way of doing this is a Fisher–Yates Shuffle:

function shuffle(array) {
    var i = array.length,
        j = 0,
        temp;

    while (i--) {

        j = Math.floor(Math.random() * (i+1));

        // swap randomly chosen element with current element
        temp = array[i];
        array[i] = array[j];
        array[j] = temp;

    }

    return array;
}

var ranNums = shuffle([1,2,3,4,5,6,7,8,9,10]);

Basically, it's more efficient by avoiding the use of 'expensive' array operations.

BONUS EDIT: Another possibility is using generators (assuming you have support):

function* shuffle(array) {

    var i = array.length;

    while (i--) {
        yield array.splice(Math.floor(Math.random() * (i+1)), 1)[0];
    }

}

Then to use:

var ranNums = shuffle([1,2,3,4,5,6,7,8,9,10]);

ranNums.next().value;    // first random number from array
ranNums.next().value;    // second random number from array
ranNums.next().value;    // etc.

where ranNums.next().value will eventually evaluate to undefined once you've run through all the elements in the shuffled array.

Overall this won't be as efficient as the Fisher–Yates Shuffle because you're still splice-ing an array. But the difference is that you're now doing that work only when you need it rather than doing it all upfront, so depending upon your use case, this might be better.

Generate a range of numbers:

var numbers = [1, 2, 3, 4];

And then shuffle it:

function shuffle(o) {
    for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
    return o;
};

var random = shuffle(numbers);

In practice, your code will be fine. In theory, it is not guaranteed to terminate, but the probability of that happening is literally negligible.

However, what you are doing is picking a random integer between 1 and 3 (both inclusive), with the number not repeating. With only three possibilities, we could draw up a state machine:

states = {
  1: [2, 3],
  2: [1, 3],
  3: [1, 2] };
if (lastRandom === undefined) {
    random = Math.floor(Math.random() * 3) + 1;
}
else {
    random = states[lastRandom][Math.floor(Math.random() * 2)];
}
lastRandom = random;

Unfortunately, that involves generating all the states first, which is impractical for larger ranges. There, we could use this more clever approach:

var min = 1;
var max = 3;
if (lastRandom === undefined) {
    random = Math.floor(Math.random() * (max - min + 1)) + min;
}
else {
    random = Math.floor(Math.random() * (max - min    )) + min;
    if (random >= lastRandom) random += 1;
}
lastRandom = random;

So if we are choosing the first number, everything works as usual. But for any subsequent number, we choose an integer from a set one smaller (to reflect that one number, the lastRandom, can't be chosen). If the random number obtained like this is less than the lastRandom, it can be used as is. Otherwise it has to be incremented by one, so that the lastRandom has been skipped.

Test to compare code:

var nums = [1,2,3,4,5,6,7,8,9,10,11,12];
var gen_nums = [];

function in_array(array, el) {
   for(var i = 0 ; i < array.length; i++) 
       if(array[i] == el) return true;
   return false;
}

function get_rand(array) {
    var rand = array[Math.floor(Math.random()*array.length)];
    if(!in_array(gen_nums, rand)) {
       gen_nums.push(rand); 
       return rand;
    }
    return get_rand(array);
}

for(var i = 0; i < 9; i++) {
    console.log(get_rand(nums));
}

use a variable to check if the number is the same.

something like this: (using LastNumber to store the lastNumber) if it allready is used we gonna try again)

var videoLinks = [
    ....
];

var lastNumber = 0;
var randomNumber = function () {
    var getRandomNumber = Math.floor(Math.random() * 5);  
    if(getRandomNumber != lastNumber){
        var random = videoLinks[getRandomNumber];
        document.getElementById("videoWrapper").innerHTML = random[0];
        lastNumber = getRandomNumber;
    }else{
        randomNumber();
    }
};

randomNumber(); // To call the function on load

There are a number of ways you could achieve this.

Solution A: If the range of numbers isn't large (let's say less than 10), you could just keep track of the numbers you've already generated. Then if you generate a duplicate, discard it and generate another number.

Solution B: Pre-generate the random numbers, store them into an array and then go through the array. You could accomplish this by taking the numbers 1,2,...,n and then shuffle them.

shuffle = function(o) {
    for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
    return o;
};

var randorder = shuffle([0,1,2,3,4,5,6]);
var index = 0;

setInterval(function() {
    $('.foo:nth-of-type('+(randorder[index++])+')').fadeIn(300);
}, 300);

Solution C: Keep track of the numbers available in an array. Randomly pick a number. Remove number from said array.

var randnums = [0,1,2,3,4,5,6];

setInterval(function() {
    var m = Math.floor(Math.random()*randnums.length);
    $('.foo:nth-of-type('+(randnums[m])+')').fadeIn(300);
    randnums = randnums.splice(m,1);
}, 300);

When creating a random number, only use one that isn't yet in your array:

var number = [];
for(i= 0; i <= 5; i++){
    var num;
    while ( number.includes( num = Math.floor(Math.random()*40) ) );
    number[i] = num;
}
console.log(number);

The empty while loop repeatedly selects a number until it finds one that isn't in the number array. Only after that successfully finishes does it then add that number to the array.