# Multiple dependent proportions problem - Election

#### Jonallard

##### New Member
Hello,

I have spent a few hours trying to untangle this problem by my knowledge of statistics appears to be too limited for me to come to a solution. I hope you can help me with this.

On election night, the results are coming in, but only poll by poll. Assuming each poll is an independent sample of the whole riding (population), what is the probability of each candidate of winning?

Say with 6 polls out of 154 reporting (3.9%), the results are:
• Corbeil, 40%
• Garneau, 35%
• Drabkin, 12%
• Roy, 8%
• Carkner, 5%

What is each candidates' probability of winning? (being the one with the highest proportion of votes)
I've tried computing the margin of error for each proportion, but I can't figure out how to deal with all proportions being dependent.

Thank you in advance

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#### Dason

##### Ambassador to the humans
You might want to look into the multinomial distribution.

#### Jonallard

##### New Member
I don't see how it applies... multinomial dist. is how a group of events behaves when each of them has a fixed probability of happening, but here, I have a sample of proportions I want to make inferences about. How would you get to the result?

Nevertheless, thanks because I didn't know about multinomial distribution, but I just don't know if it applies here.

#### BGM

##### TS Contributor
Maybe I try to give a mathematical formulation for you.

Let $$X_i$$ be the number of votes that the $$i$$-th candidate have in the remaining $$154 - 6 = 148$$ polls,
$$i = 1, 2, 3, 4, 5$$

Assuming there are $$k$$ votes in each poll.

Then $$X_i \sim \mathrm{Binomial}(148k, p_i)$$ marginally,

and $$(X_1, X_2, X_3, X_4, X_5) \sim \mathrm{Multinomial}(148k, p_1, p_2, p_3, p_4, p_5)$$ jointly.

The total number of votes that the $$i$$-th candidate have , $$Y_i = X_i + \hat{p}_i \times 6k$$
where $$\hat{p}_i \times 6k$$ is the number of votes that we already known in that 6 polls.

The probability that the $$i$$-th candidate win $$= \Pr\{Y_i > Y_j, i \neq j, j = 1, 2, 3, 4, 5 \}$$

I am not sure if there is any multi dimensional normal approximation can help.

Anyway, you can always get the mean vector and variance-covariance matrix of those $$X_i$$ and get the corresponding estimate by replacing the $$p_i$$ by $$\hat{p}_i$$

#### Jonallard

##### New Member
Thank you, that is very helpful. How do I use the mean, variance and covariance to get, say $$P(Y_i > Y_j)$$ for one $$i$$ and $$j$$? I mainly don't know what to do with covariance.

#### Jonallard

##### New Member
Well, I'm stuck here. The multinomial model is too complicated for me and I don't know how to use it (huge factorials, multiple dimensions, can't understand what's going on and how to calculate probability based on an inequation rather than equation, and this for several variables, no approximation, etc.), and I can't even do something with inference for proportions, the interrelations confuse me too much. Basically, I'm all confused.

Any idea on how to actually do it?

#### BGM

##### TS Contributor
http://en.wikipedia.org/wiki/Multinomial_distribution

$$E[X_i] = 148kp_i, Var[X_i] = 148kp_i(1-p_i), Cov[X_i, X_j] = -148kp_ip_j, i \neq j$$

After obtaining the estimated mean vector and variance-covariance matrix,
you may try the multivariate normal approximation; and your problem will be turn to a
problem to calculate the multivariate normal probabilities.

#### rft01

##### New Member
Google "Comparing Non-Independent Proportions" by Berry & Hurtado. They have an example similar to the voter survey problem here. They're with SAS. The paper gives formulas and also SAS macros.