Talk:Condorcet's jury theorem

Latest comment: 4 years ago by ToBeFree in topic Proof disputed

Untitled edit

Needed: some discussion of the history, a citation, perhaps a proof. --best, kevin [kzollman][talk] 17:52, 9 August 2006 (UTC)Reply

True information does not necessarily imply the true answer edit

The article currently includes these statements:

Any answer based on true information correctly processed must have a greater than fifty percent chance of being correct. The only way to go below fifty percent chance is to have either incorrect information or badly processed information.

This is incorrect. True information could easily point to the wrong answer. For example, suppose a certain innocent individual is accused of a crime. DNA evidence is found at the crime scene that matches the defendant's. Correctly processing this information may mean that defendant is probably guilty, but in this example, the conclusion is wrong. Or suppose in a primitive culture, a rain dance is performed five times and it rains each time. Correct processing of this information (in the absence of any information about the true causes of rain) leads to the conclusion that performing the rain dance brings rain, but the conclusion is incorrect. Anomalocaris (talk) 21:23, 21 December 2007 (UTC)Reply

Proof? edit

I believe that the 'proof' presented here is wrong. It only takes into account the borderline cases where we are one vote away from a switch, and does not consider the relative probabilities of being in each of these cases. For example, if it is very improbable (let us say 0 for sake of argument) that we are just below the cutoff, and very probable (let us say 1 for sake of argument) that we are just above the cutoff, then adding two voters clearly increases the chance of a wrong verdict as long as they are not correct with probability 1. The probabilities of being just below or above the cutoff are obviously not 0 and 1, but this nonetheless demonstrates the invalidity of the proof provided.

I believe the article would be greatly improved by the addition of any of the following:

-A proof taken from the 'Essay on the Application of Analysis to the Probability of Majority Decisions', reference 1 at time of this writing.

-A discussion/use of the binomial distribution.

-A discussion/use of the central limit theorem. This is somewhat touched upon in the section on asymptotics (where they talk about how the probability grows as a linear function of N^(1/2)), but this section should be expanded and its ideas referenced in the proof.

The text of the proof as it currently stands is as follows:

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:
- m was one vote too small to get a majority of the n votes, but both new voters voted correctly.
- m was just equal to a majority of the n votes, but both new voters voted incorrectly.
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.

--- Adding to the above, I believe there is a relatively easy way of showing the first proof is correct, although the wording above is not very accurate. The alternative proof example seems incorrect altogether, for example: (9 choose 8) > (9 choose 9), and the proof assumes it is the opposite.

— Preceding unsigned comment added by 141.226.218.88 (talk) 02:16, 15 January 2019 (UTC)Reply

Proof disputed edit

213.57.154.128 believes that there is a mathematical error in the proof; please see Special:Diff/903254835 and Special:Diff/903250726. ~ ToBeFree (talk) 15:39, 24 June 2019 (UTC)Reply