Help with combining data sets...

[Foxymoron]

Hi Math heroes!

I have a problem I hope you can help with.
Imagine a competition in which artists were invited to submit work for an exhibition that was to be judged by the public.
Visitors to the gallery could vote in person for their favourite artist, logically the artist with the most votes wins.
Add to this though a second strand of voting via Instagram, where the public could 'like' their favourite work, which would then count as a vote.
As social media can be easily manipulated the gallery owner stipulated that the online votes would count for ten percent of the accumulated votes, and that the competition would be weighted toward 'real world' voting.
The correct percentages of each artists vote quantity can be easily worked out separately in each strand (real world and online), but how does the gallery owner combine the two percentiles to accurately represent the finished result with weighting applied?

I hope that I've explained this clearly enough,
Thanks in advance for any help,
Duncan

 

[lev888]

the online votes would count for ten percent of the accumulated votes

This part is not clear to me. You use "votes" for both real life and Instagram likes, so "accumulated votes" is ambiguous.

 

[Romsek]

\(\displaystyle \text{The way the owner has stipulated things you're not going to be able to decide your}\\
\text{weighting scheme until all the votes, rl and online, are in.}\\~\\
\text{Once this happens let's say you have}
\)

\(\displaystyle \text{$N_r$ real life votes and $N_o$ online votes, and we want to weight online votes so as to end up}\\
\text{with effective votes that meet the owner's stipulation. With online vote weight $c$ we have}\\~\\
\dfrac{c N_o}{c N_o + N_r} = \dfrac{1}{10}\\
10cN_o = cN_o + N_r\\
9cN_o = N_r\\
c = \dfrac{N_r}{9N_o}
\)

So take the number of online votes for each artist and multiply it by \(\displaystyle c\) above and add to it the number of rl votes they received.
Use that number for comparison among the artists to determine the winner.