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It is interesting to look at how other 3rd party candidates have fared, in past elections. 1992, for example, has Ross Perot, who got 18.9% of the popular vote. But not how that was different. Perot was the wild guy saying crazy things. He was the populist. He was the one going against the status quo. So his outsider views combined with his outsider status to make him attractive for independent voters.The current cycle is a bit different. The wild man saying crazy things, the populist, won the Republican nomination. And even Clinton has borrowed from Sanders some progressive populist ideas. Essentially the mainstream parties have co-opted crazy, and they’ll split the core independent vote.

I'll illustrate two ways of doing the calculation, the first one is the direct way, and will give you the correct answer. The second is just a crude approximation, which is less computationally intensive.There are 200 people in party X and 300 people who are not.Number of ways of choosing 31 people from party X: [math]\binom{200}{31}[/math]Number of was of choosing 29 people not from party X: [math]\binom{300}{29}[/math]Number of ways of choosing 60 people with exactly 31 people from party X: [math]\binom{200}{31}\binom{300}{29}[/math]Number of ways of choosing 60 people with more than 30 people from party X:[math]\sum_{n=1}^{30}\binom{200}{30+n}\binom{300}{30-n}[/math]Total number of ways of choosing 60 people: [math]\binom{500}{60}[/math]Probability of choosing 60 people with more than 30 people from party X: [math]\frac{\sum_{n=1}^{30}\binom{200}{30+n}\binom{300}{30-n}}{\binom{500}{60}}[/math]The answer is 0.0348151For the second method, we're going to assume that 500 is very large number and if we randomly pick one person at a time 60 times, then it is very unlikely that we'll pick the same guy twice. Let's denote a random variable X so that X=1 means someone belongs to party X, and X=0 means not from party X. Now Pr(X=1)=0.4 since 40% of the people belong to party X. From this distribution, we're going to pick n = 60 random samples and take the average. [math]\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i[/math] to get the fraction of variables that are one. We want to know what is the probability [math]Pr(\overline{X} > 0.5)[/math]. Now according to the http://en.wikipedia.org/wiki/Central_limit_theorem, we know that [math]\overline{X}[/math] is approximately distributed as a normal distribution with mean [math]\mu[/math] and variance [math]\sigma^2/n[/math]. Here [math]\mu = 0.4[/math] is the mean, and [math]\sigma^2 = 0.4\cdot 0.6^2 + 0.6\cdot 0.4^2 = 0.24[/math] is the variance of the distribution for X.So we basically want to know what is the total probability that a sample from a normal distribution with mean [math]0.4[/math] and variance [math]0.004[/math] is greater than [math]0.5[/math], which is the same as the cumulative probability of a samaple being more than [math]0.1/\sqrt{0.004} = 1.58113[/math] in a standard normal distribution. We can use the Q-function to calculate this probability.The crude approximate answer in this case, is 0.0569

Multistage sampling technique involves , subsampling within the cluster,If you want to sample voters in different states ,then you need to begin with dividing the state into clusters of equal sizes, like for example i live in UP, and it has 80 Lok sabha seats ,so what i do is , i divide the state into 80 zones, each representing a lok sabha seat , and make clusters of 10 seats each , now since BJP won 70 out of 80 seats in UP , the rest 10 seats should be merged proportionally with each cluster , to get an clear and unbaised view, Now this right there is two stage sampling, also you can take random samples within each clusters and conduct a survey on peoples opinion . like out 10 you may select only 4 seats , and within those 4 seats you can conduct a Survey on 1000 people per seat.Hope this helps.