Javascript if/else statements not working

Question

I am pretty new to Javascript, and it seems like i didnt understand the if else statements correctly.

I have a script which will make the visitor go to 1 of 4 websites, but the 2 last sites in my code does not work.

<script>

setTimeout(function() { 

var r = Math.random();
if(r > 0.49) {
window.location.replace("1.html");
}

else if(r < 0.48) {
window.location.replace("2.html");
}

if (r == 0.48){
    window.location.replace("maybe.html");
}

else if (r == 0.49){
    window.location.replace("4.html");
}

}, 1);
</script>

Is how my code looks like right now. How would it need to look to make it work?

Code seems to be okay but what is possibility of math.random returning 0.48 and 0.49 exactly? it is very rare. — Jenish Rabadiya, Feb 20 '15 at 17:30
@karan http://stackoverflow.com/questions/12051769/why-are-some-javascript-developers-using-settimeout-for-one-millisecond#answer-12051906 — blex, Feb 20 '15 at 17:31
@Karan i set the timeout to 1ms because i wanted to test. Normally there is a loading page on 1.7 seconds before it redirects - however in this case, i wanted it to load instantly:) — Fred Lu, Feb 20 '15 at 18:06
@Teemu As you can see i set the ms to 1, however the page does not redirect sometimes, and it just stucks on the loading screen. I assumed this was because my logic was not correct. — Fred Lu, Feb 20 '15 at 18:07
Please add that to the question itself, it would make the question much more understandable. — Teemu, Feb 20 '15 at 18:11

Jon Surrell · Answer 1 · 2016-06-21T12:13:53.927

Update

I originally said this looked fine, but I just noticed a problem. There is no branch for r > 0.48 && r < 0.49. Values in this range, such as 0.48342... are more likely than hitting 0.48 or 0.49 exactly, and these are completely unaccounted for, which I assume was not your intention. A simple else branch is always a good idea, or you should account for these cases explicitly.

Original

Your logic looks fine to me. Reduce your problem:

function randomizeText() {
  var r = Math.random();
  var result = '';
  if (r > 0.49) {
    result = 'A';
  }
  else if (r < 0.48) {
    result = 'B';
  }
  else if (r == 0.48){
    result = 'C';
  }
  else if (r == 0.49){
    result = 'D';
  }
  document.getElementById('output').innerText = result;
  document.getElementById('random-number').innerText = 'Number was: ' + r;
}

randomizeText();

<button type="button" onclick="randomizeText();">Randomize!</button><br>
<div id="output"></div>
<div id="random-number"></div>

Note that it's going to be very very unlikely that you'll hit either of the last 2 conditions.

cssyphus · Accepted Answer · 2015-02-20T18:22:30.853

You can replace your entire code block with these 2 lines, and they will do what you want:

var r = Math.floor(Math.random() * 4) + 1;
window.location.replace(r+".html");

Explanation:

Your code is actually working. The problem is that the number returned by Math.random() is a random number between 0 and 1 (it might be 0.5544718541204929 ), and will almost NEVER be exactly 0.48 or 0.49, but will almost always be between those two numbers.

A better solution would be:

var r = Math.floor(Math.random() * 4) + 1;

and then test if number is 1, 2, 3 or 4.

Example:

jsFiddle Demo //jsFiddle temporarily not saving fiddles

var r = Math.floor(Math.random() * 4) + 1;

if(r ==1) {
    alert("1.html");
}else if(r==2){
    alert("2.html");
}else if(r==3){
    alert("3.html");
}else{
    alert("4.html");
}

BUT there is no need for the entire IF block. Just do this:

var r = Math.floor(Math.random() * 4) + 1;
window.location.replace(r+".html");
//alert( r + ".html" );

In response to the this question, submitted as a comment: I want it to be page 1 and page 2 has is almost 50/50, and the last 2 is pretty rare

This would give odds of 1% for cases 3 and 4.

var r = Math.floor(Math.random() * 100) + 1; //return number between 1 and 100

if(r <=48) {
    alert("1.html");
}else if(r<=98){
    alert("2.html");
}else if(r==99){
    alert("3.html");
}else{ //r==100
    alert("4.html");
}

If you desire slightly larger odds:

if(r <=40) { //40% chance
    alert("1.html");
}else if(r<=80){ //40% chance
    alert("2.html");
}else if(r<=90){ //10% chance
    alert("3.html");
}else{ //r is between 91 and 100, 10% chance
    alert("4.html");
}

I understand this pretty well. However i want it to be page 1 and page 2 has is almost 50/50, and the last 2 is pretty rare.. So could i not * with 100 and do it the same way as i already did? Just with 48 as an example, instead of 0.48 ? @gibberish — Fred Lu, Feb 20 '15 at 17:57
I accidentally submitted my comment before finishing it. I edited it now :) — Fred Lu, Feb 20 '15 at 18:03
Solved my problem perfectly! Thank you for your help and time!:) — Fred Lu, Feb 20 '15 at 18:30

Javascript if/else statements not working

2 Answers2

Update

Original