Solving the Paul/Cain Arizona Straw Poll Winner Riddle

 In a memorable movie Kathleen Turner says to her teacher "I happen to know that in the future I will not have the slightest use for algebra, and I speak from experience".

But to solve the riddle of major news organizations (CNN, Reuters, Politico, etc) offering conflicting stories about who won the Arizona Tea Party Straw Poll, Ron Paul or Herman Cain, a little high school algebra does wonders.

Ron Paul did win 49% of one part of an Arizona Tea Party straw poll and 15% of another part of an Arizona Tea Party straw poll.

Herman Cain did win 12% of one part of an Arizona Tea Party straw poll and 22% of another part of an Arizona Tea Party straw poll.

Ron Paul did get 581 straw poll votes 

Herman Cain did get 256 straw poll votes.
About 1600 people did take part in the Arizona Tea Party straw polls.

It is probably true that 1600 people paid up to 125$ to attend an onsite one day event.
It is probably true that 2300 (maybe even 3000) people paid up to 10$ to view this event online.

The other information in the various articles
2300 people took part in an online poll
1600 people took part in a live/onsite poll
are FALSE.

High school algebra is how you sort out truth from sloppy reporting because the number of people taking each version of this poll was never reported correctly by the news media or the straw poll event organizers. Why, is up for someone else to figure out. I just do math.



Paul: 49% X + 15% Y = 581 votes
Cain: 12% X + 22% Y = 256 votes
where X is the number of all votes in the online poll and Y is the number of votes in the onsite poll.

You have two equations, two variables. This is solvable.

Paul: 49% X + 15% Y = 581 votes
Cain: 12% X + 22% Y = 256 votes

49% X + 15% Y = 581
is the same thing as
49% X = 581 - 15% Y
which is the same thing as
X = (581 - 15%Y)/49%

From the Paul equation, we learn that the number of online votes X is (581/.49) votes - 15% Y/49%

We take this X and put it into the Cain equation
12% (581/49% votes - 15% Y /49%
) + 22% Y = 256 votes
which is the same as
581 * 12%/49% votes + 22% Y - 12%*15%/49% * Y = 256 votes
which is the same as
(22% - 12%*15%/49%) Y = (256 - 581 * 12%/49%) Votes
which is the same as
Y = (256 - 581 * 12%/49%)/(22% - 12%*15%/49%) Votes

This means that the number of voters in the onsite poll, Y is 113.7/.183 or 620 votes

Now we take Y and plug it back into the Paul equation
X = (581 - 15% Y)/49%
X = (581 Votes - .15 * 620 Votes)/.49 = 996 votes

And you know what, 996 + 620 = 1616 votes, which is the "about 1600 votes" often quoted.

Now, because they only report percentages with 2 digits, the theoetical accuracy of our math is only good to numbers that are 2 digits (not counting trailing zeroes). So the number of online votes might be anywhere from 995 to 1004 votes and the number of onsite votes might be anywhere from 615 to 624 votes. But consider what happens when it is exactly 996 online voters and 620 onsite voters.

        Total Votes,%   Online Votes,%      Onsite Votes, %
Paul       581  36%          488  49%        93  15%
Cain       256  16%          120  12%       136  22%
Palin      149   9.2%         87   8.7%      62  10%
Pawlenty   143   8.8%         44   4.4%      99  16%

This also lets us know how many votes Romney and Bachman got in the Onsite poll, though not online or total votes.
        Total Votes,%   Online Votes,%      Onsite Votes, %

Romney       ??             ??                40  6.5%
Bachmann     ??             ??                35  5.6%

It would have been nice for the event organizer to report (or the media to demand) who got the
other 257 online votes and 155 onsite votes, but given the mess things were already in, solving the simple case of WHO WON is more important.

Consider it solved.

Now you know why algebra can be useful. ;-)
Jeff