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Suppose a country with 'E' electorates and 'V' voters in each electorate, were to hold an election. Each vote is independent of all others, and has a 50% chance of being for party A and a 50% chance of being for party B. Let 'E' and 'V' both be odd, and the party with a majority of votes in a particular electorate win that seat, and the party which wins the majority of seats win the election overall.

What is the probability, as a function of E and V, that the party which wins the election overall loses the popular vote?

For instance, I can calculate as follows with E=V=3: There are 9 voters, so 2^9 total possibilities. For a party to win the election overall but lose the popular vote, it must win exactly 2 votes in 2 seats, and 0 votes in the third seat. The number of ways this could happen is 2 (parties which could win overall) * 3 (seats in which the winning party could get 0 votes) * 3 (electors which could vote against the winning party in the first seat won by them) * 3 (electors which could vote against the winning party in the second seat won by them). Thus the overall probability is 27/256.

Is there a general formula for calculating this probability as a function of E and V?

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    Doesn't your example come to $\frac {54}{256}$ to account for the two parties? Have you tried sampling the problem? I'd do that first...computers will have no problem running scenarios at whatever scale you are interested in...if nothing else, that should give you a good feel for the answer. – 2017-01-10
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    It comes to 54/512, which simplifies to 27/256. I've tried running a spreadsheet, which has given me some results. – 2017-01-12
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    Ah, yes. My mistake. The constant limit described below is interesting. Wouldn't have predicted that. – 2017-01-12

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This seems like a somewhat difficult question to answer analytically, but numerically I get the following results when increasing E and V together. In the case of ties within a district, I split the electoral votes between the two parties. In terms of deciding the results of the election, I break the tie arbitrarily in favor of one of the two parties (consistently).

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Interestingly enough, the results seem to converge around 20%. The zig-zag pattern accounts for a local effect which suggests that the even E/V numbers are more robust to different election decision styles.

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    The convergence to a constant in the neighborhood of 20.5% appears to be given by a paper of Feix, Lepelley, Merlin, and Rouet, “The Probability of Conflicts in a U.S. Presidential Type Election,” Economic Theory, 2004. I found this citation in the powerpoint presentations of Nick Miller at http://userpages.umbc.edu/~nmiller/ELECTCOLLEGE.html on the electoral college; I sadly don't have access to the original paper. – 2017-01-11
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    Thanks guys. I wonder what happens if a third layer is added, ie the election is held in 'C' countries, each of which has 'E' electorates, each of which has 'V' voters. Seems to be a probability of around 0.096 when C=E=V. – 2018-01-11