# How to derive one percentage from two options

I am very bad at mathematics, so apologies in advance. My confusion comes from the voting system on an online poll. Example video (Picked at random): http://www.funnyordie.com/videos/4d0ac00700/youre-pitiful-weird-al-yankovic-ver-1-from-insane_ian

The percentage is 71%, meaning that some combination of "funny" or "die" votes added up to that. If 10 out of 100 people voted "funny" it would be easy to say that the percentage who thought it was funny is 10%, I am wondering how the number is derived taking both into account.

Hopefully I worded this correctly. I have a feeling it is something simple that I just don't remember. Thanks. I'm not sure what to tag this with, so I'm taking my best guess. Sorry if that's wrong.

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You add up the number of people who voted "funny", and the number of people who voted "die". This gives you a total $X$. You take the number of people who voted overall; this is a total $Y$. To compute the percentage of people who voted $X$ you compute $100X/Y$. –  Arturo Magidin Jun 22 '12 at 2:35
Thank you for the reply Arturo. I'm still not clear though, wouldn't the number of people who voted "funny" and those who voted "die" (which you say is X) be the same as Y? –  Daniel Jun 22 '12 at 2:39
I don't know what all the options are. If $Y$ people voted (for anything, whether "funny", "die", or something else), and $X$ voted for either "funny" or "die", then the percentage of people who voted for either funny or die from among those who voted is $100X/Y$. If everyone who voted voted for either "funny" or "die", then $X=Y$, and the percentage is, of course, 100% (of all the people who voted in the last election, how many people cast a vote? 100%). But if there are other options (including abstensions), then in general you will have $X\leq Y$, with $X\lt Y$ possible. –  Arturo Magidin Jun 22 '12 at 2:41
Thanks for the clarification. For the purposes of this calculation I was wondering about restricting it to only two options. Sorry, forgot to shift+enter. So 100X/Y (Does that mean 100 * (X/Y)?) is good for an unknown amount of options, but what works for just two? –  Daniel Jun 22 '12 at 2:44
First, $(100X)/Y$ gives the same answer as $100*(X/Y)$. Second: if you want to know what percentage $X$ things are out of $Y$, then you compute $100X/Y$. So, to find out what percentage 74 is out of 87, you compute $100(74)/87 \approx 85.06\%$. Doesn't matter how you get the $X$, it just matters how much it is. You can restrict to whatever options you want to compute $X$, the point is that you need to know both how many "good" items you have ($X$), and how many items overall there are ($Y$). –  Arturo Magidin Jun 22 '12 at 3:01