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One hundred thousand early votes have been counted in a national election. If candidate XYZ has only received 49000 of these votes, does this imply that his opponent will in?

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Unanswerable: how many candidates are there, and how many votes did the others get (two seems to be implied, but no note of spoiled ballot papers)? How many people have not yet voted (fewer that two thousand, a couple of million)? Is there any reason to believe that the early voters are representative of the others, or not? – Mark Bennet Oct 14 '12 at 20:14
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Short answer: no. – Karolis Juodelė Oct 14 '12 at 20:24
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If we are benevolent and assume that there are two candidates, no spoiled votes and early voters are representative, then the error in $\bar x$ is of the order of magntude $\sqrt n \approx 300$. Thus we are off by about $3\sigma$ and there is still a non-negligible chance (in the order of magnitude 1%) that a sufficiently big full population turns the result over. – Hagen von Eitzen Oct 14 '12 at 20:45
Votes to close are out of control on this site. – zyx Oct 18 '12 at 3:31

closed as off topic by mercio, martini, Mark Bennet, Erick Wong, Noah Snyder Oct 15 '12 at 13:23

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It hinges on how similar the early voters are to the total voter population. If they are a perfect random sample (so that counting the early vote is similar to taking 100K random voters after the election and counting their results) then a 49-51 split is unlikely to occur by random chance with that large a sample. But the early vote might disproportionately count people in particular locations, or subgroups of the population that differ from the average.

This came up vividly in some US presidential elections where television networks broadcast their East Coast poll results during the 3-hour time zone difference before the West Coast voting was finished. There is no reason to think the East and West voting patterns would have been completely similar, and the networks were pressured to stop releasing the early results, so as not to influence the Western vote.

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