Condorcet jury theorem

Condorcet's jury theorem is a political science theorem about the relative probability of a given group of individuals arriving at a correct decision. The theorem was first expressed by the Marquis de Condorcet in his 1785 work Essay on the Application of Analysis to the Probability of Majority Decisions.

The assumptions of the simplest version of the theorem are that a group wishes to reach a decision by majority vote. One of the two outcomes of the vote is correct, and each voter has an independent probability p of voting for the correct decision. The theorem asks how many voters we should include in the group. The result depends on whether p is greater than or less than 1/2:

To avoid the need for a tie-breaking rule, we assume n is odd. Essentially the same argument works for even n if ties are broken by fair coin-flips.

Now suppose we start with n voters, and let m of these voters vote correctly.

Consider what happens when we add two more voters (to keep the total number odd). The majority vote changes in only two cases:

The rest of the time, either the new votes cancel out, only increase the gap, or don't make enough of a difference. So we only care what happens when a single vote (among the first n) separates a correct from an incorrect majority.

Restricting our attention to this case, we can imagine that the first n-1 votes cancel out and that the deciding vote is cast by the n-th voter. In this case the probability of getting a correct majority is just p. Now suppose we send in the two extra voters. The probability that they change an incorrect majority to a correct majority is (1-p)p², while the probability that they change a correct majority to an incorrect majority is p(1-p)(1-p). The first of these probabilities is greater than the second if and only if p > 1/2, proving the theorem.

First lets take the simplest case of n=3, p=0.8. We need to show that 3 people have higher than 0.8 chance of being right. Indeed: 0.8*0.8*0.8+0.8*0.8*0.2+0.8*0.2*0.8+0.2*0.8*0.8=0.896

Now lets take n=9 (the general case works the same). Following the same considerations as in the n=3 case, we need to select all options for majorities and multiply by respective p or q=1-p. So we need to show: ${\displaystyle {\binom {9}{5}}p^{5}q^{4}+{\binom {9}{6}}p^{6}q^{3}+...+{\binom {9}{9}}p^{9}>p}$ i.e. we need to show

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