(i will try to answer properly some of the issues raised so dont let the length discourage you again lol)
She made multiple bad errors here. I like her in general because she intelligent but she is very partisan also at times (generally on the correct side of things but not always and it is better to be more objective in order to appeal to more people in general with you logic). Here she screwed up the math but her argument still has some merit that excluding the bottom 6 can be a mistake and at least who is number 10 in reality is a question that easily has over 75% chance to not be the observed number 10 with 6 others left below (at least the 3 of them relevant)
Look here
https://en.wikipedia.org/wiki/Multinomial_distribution
Basically when you have many choices (candidates) each having probability pi (i from 1 to k say) all adding up to 1 if you sampled the population N times you expect the i candidate to have
N*Pi as average result and sd = N*Pi*(1-Pi).
You would then select confidence intervals to report. For example if you wanted to have 95% chance that the true probability was within what was observed+- the error you would be looking for a deviation (radius around say) from the mean that has 5% chance to land you outside the range of reported average +- the error. This is how you can find the error that corresponds to that confidence interval. For example at 95% confidence interval you want to have a chance of only 1/2 of that 5% =2.5% (1 for each direction up down to add up to 5%) to be higher than the avg plus the error. So you are looking at a number x standard deviations away than the average such that the chance to be below avg+x*sd is 97.5% (1-5%/2). This is true at 1.96 standard deviations as can be found by the cumulative probability tables or software packages.
The error therefore is 1.96*(Pi*(1-Pi)*N)^(1/2) and in relative terms (ie divided by the population N you tested) is 1.96*(Pi*(1-Pi)/N)^(1/2).
Now keep in mind i think they define the polling error they quote in general in the polls reported in news at some eg 95% confidence interval as the error you would have if Pi=50% (which is the maximum the above error can be clearly). That would give 1.96/2/N^(1/2)=0.98/N^(1/2)
This is how they arrive at the result for example in this link
https://en.wikipedia.org/wiki/Margin_of_error in the section different confidence intervals.
If they wanted 99% confidence interval it would be 1.29/N^(1/2).
I am not sure what the polls she is talking about are (at what confidence interval so i will assume 95%).
In that case a poll error of 4% would suggest that 0.98/N^(1/2)=4% or N=600 people asked.
If they use 5-6 different polls and take an average it will be an error much smaller than the 4% she used as reference though. Eg see if they used 5 polls of 600 each you have technically N=3000 and then at 95% confidence interval the error is 0.98/3000^(1/2)=1.8% much smaller than the 4% of each poll.
Now when you are talking about small Pi in the tails you clearly as David Sklansky and others noted cannot use again 4% (or the 1.8% if using many polls added up).
You need to then go to 1.96*(Pi*(1-Pi)/N)^(1/2).
For example if N=3000 (5 x600 polls) and Pi=2.6% then the error is only
1.96*(0.026*(1-0.026)/3000)^(1/2)=0.57% not 4% not even 1.8%.
So when she says 2.6+-4% she should be saying 2.60+-0.57%.
You can then ask questions like how likely it is that the 2.2% guy is ahead of the 2.6% making it an injustice to be left out of the top 10 say or even generalize it to ask what is the chance any of the other 6 say are better in reality than the 10th cutoff guy.
That is a harder question to ask. You can answer it easily only if you know the true probabilities in the sense of asking what is the chance that you are in reality higher and you will appear lower in sampling.
In general if you have pi,pj and you want the chance that i gets more voters/samples for them than j you can assume both are normal (central limit theorem) and take the difference (also normal) and that will have a expected value of (pi-pj)*N and the sd of that difference will be the (si^2+sj^2)^(1/2) (see addition or subtraction of normal distributions)
with si^2=pi*(1-pi)*N , sj^2=pj*(1-pj)*N.
So sd=((pi^2*(1-pi)^2+pj^2*(1-pj)^2)^(1/2)*N^(1/2)
I mean if you have 2.60%+0.57% and 2.20+-0.53% then the difference is 0.40%+-0.77%.
So the chance the 2.2% comes ahead of the 2.6% in a poll in real life (ie result 0.52 sd above expected) is 30%.
Now we do not have this here we do not know the true probabilities, these are sample estimates out of say 3000 people. The reverse question is the interesting one. What is the chance that the true order is different than observed in the samples. I think the crude estimate will remain the same (near 30%) but i need to check properly how to answer such reverse question.
Also one can easily see by looking at the accumulation of so many candidates near 2% that the probability someone other than the observed 10th is the real 10th is probably way over 75%.
This is the legitimate criticism of selecting only 10 for debates. Yes you have to cut off at some point because eg 16 is more cluttered debate than 10 that is already big.
You would likely either need much more accurate polls or develop a format of 3 or more debates and include also the bottom 6 or 10 say or so few at a time or maybe with some drawing based on how they scored that would still give them the chance if not to be in all debates to at least be in a few of them (1 or 2 or 3 for the bottom 10 say at least out of 3 or 4 total debates keeping say the top 6-7 fixed always there).
(which you can consider an expansion to follow the logic better of what goofball and DS on first page who already posted earlier to give proper credit)
Last edited by masque de Z; 07-28-2015 at 07:20 PM.