Inspired by the US equivalent - what is the minimum total vote share that a party could obtain while still winning over 325 seats in the Commons? What is the smallest share of the vote that a party has obtained in the past while still obtaining a majority?

As far as assumptions go, it makes sense to assume a constant, sensible turnout across all seats - otherwise the answer is practically 0% - and that all constituencies are contested by the candidates that have declared for the 2019 election. Has anyone done the maths on this?

  • 4
    If you can pull together data on how many candidates are standing in each seat, that's the hard bit. I just filled in my postal vote and there were something like 10 candidates, so if that's typical then about 5% would be enough. Nov 25 '19 at 19:20
  • @PeterTaylor, that may be the case, but how many of those candidates would be expected to get even 1% of the vote? Nov 26 '19 at 11:58
  • @GeoffAtkins, as I understand the question that is completely irrelevant. Nov 26 '19 at 12:22
  • In theory, yes it would be irrelevant. But in all practical senses, only four or five candidates per seat are likely to retain their deposit, which affects the required percentage needed for the candidate winning the seat. Nov 26 '19 at 12:25
  • Even if they don’t get 5% of the vote and lose their £500, the votes for them still count! It doesn’t affect the percentage needed to win the seat.
    – CDJB
    Nov 26 '19 at 12:27

Assuming the winner in each seat gets 50% of the vote you could win 50% of the seats each with just 50% of the vote in those districts with just 25% of the national vote.

But the UK has multiple parties so MP's often win their own constituency with less than 50% of the vote so not even 25% is required and you could theoretically get into the single digits although that would be highly unlikely.

The worst result that has actually happened was in 2015 when the conservatives got 37% of the vote and 51% of the seats in parliament.

  • With the Brexit Party in the mix, this will be potentially more exaggerated in 2019. Though as they will not contest current Tory seats, it may not be as lopsided as it might have been.
    – Jontia
    Nov 26 '19 at 6:25

Effectively, as few as 8.73% of the total electorate could vote for a party to have a 1 seat majority in the House of Commons.

Calculations used for answer Data taken from:

This data has been simplified in the following ways:

  1. I have assumed 4 candidates per seat, or more accurately, the winning candidate to win 25% + 1 of the votes in their constituency. So the calculation is ((Avg Electorate * 0.7) * 0.25) + 1
  2. I have ignored where seats are not taken (speakers, Sinn Fein, etc).
  3. I have assumed 70% turnout, which is slightly above average from the past two elections, but not grossly so.

Based on this, to win 326 seats in the Commons, a party could achieve this from as little as 8.73% of the electorate voting for them, or 12.48% of all actual voters. This is very theoretical and does not represent the actual situation (where elections are actually decided only on marginal seats).

Even if we were to look at 100% turnout, it would only require 280,000 more votes.

  • 1
    Are you assuming every seat is average size? If the party won in the 326 smallest seats then the vote share would be even less. You could also assume Sinn Fein win all NI seats so the number required for a majority is lower.
    – Richard
    Nov 26 '19 at 12:55
  • Yes, if I had time and access to more data, I could put together a more accurate table factoring in the number of candidates per constituency and the electorate size. However, that these ballpark figures are so far outside the normal range for the number of votes held by the winning party meant that pursuing the data to more accurate extremes (which would only be even lower) felt like a moot point. Nov 26 '19 at 13:00

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