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An interesting question on Politics SE spawned a mini discussion about the number of candidates that could potentially tie for US President based on the Electoral College.

The break down of the college is:

  • 3 - Alaska, Delaware, District of Columbia*, Montana, North Dakota, South Dakota, Vermont, Wyoming
  • 4 - Hawaii, Idaho, Maine**, New Hampshire, Rhode Island
  • 5 - Nebraska**, New Mexico, West Virginia
  • 6 - Arkansas, Iowa, Kansas, Mississippi, Nevada, Utah
  • 7 - Connecticut, Oklahoma, Oregon
  • 8 - Kentucky, Louisiana
  • 9 - Alabama, Colorado, South Carolina
  • 10 - Maryland, Minnesota, Missouri, Wisconsin
  • 11 - Arizona, Indiana, Massachusetts, Tennessee
  • 12 - Washington
  • 13 - Virginia
  • 14 - New Jersey
  • 15 - North Carolina
  • 16 - Georgia, Michigan
  • 18 - Ohio
  • 20 - Illinois,Pennsylvania
  • 29 - Florida, New York
  • 38 - Texas
  • 55 - California

(*) DC is not a state, but it's treated as one for the purpose of the presidential election (23rd Amendment to the US Constitution).

(**) All but 2 states are winner take all. 2 states (Nebraska and Maine) award the winner of each congressional district 1 vote (Nebraska 3 and Maine 2) and the over all winner for the state 2 votes. So potentially 4 winners could come out of Nebraska and 3 out of Maine assuming the winner of the over all vote in the state did not win any districts, or some combination of those.

Based on this what is the most candidates that could tie for 1st place and how would that work?

Assume that all electors are faithful and vote for the candidate that they are pledged.

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  • $\begingroup$ Sorry I am unfamiliar with the tags here so feel free to set the proper flags $\endgroup$ Commented Dec 9, 2016 at 17:11
  • $\begingroup$ Just for further clarity, D.C. is also winner take all for its 3 electoral votes. I'm not sure this question isn't simply a math question. $\endgroup$ Commented Dec 9, 2016 at 17:34
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    $\begingroup$ Technically speaking, the answer is actually the same as the number of electoral votes, since each elector has the right to cast their vote for anyone, regardless of which party or candidate their state awards electors to. Electors who vote contrary to whom their state awarded them to are known as faithless electors. $\endgroup$ Commented Dec 9, 2016 at 23:02
  • $\begingroup$ Depending upon how you mean "tie", the answer is potentially infinite. It takes an absolute majority to win the electoral college. You could have 270 candidates, and give one of them 269 votes and the other 269 candidates 1 vote each, and it will be the case that nobody wins the electoral college. Every candidate loses (and then Congress gets to appoint whomever they want), and in that sense it's a 270-way tie. Assuming no faithless electors and not counting zero-vote candidates, the answer is 53. $\endgroup$ Commented Dec 10, 2016 at 0:38
  • $\begingroup$ What happens if an elector dies just before the vote? Are they substituted? $\endgroup$ Commented Dec 10, 2016 at 5:27

3 Answers 3

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Well... The upper bound is

9, because California must be included in the winning configuration, and 538/55 = 9.

Here is one potential solution

CA(55)
TX(38) + NJ(14) + AK(3)
FL(29) + IL(20) + AR(6)
NY(29) + PA(20) + IA(6)
OH(18) + GA(16) + VA(13) + KY(8)
MI(16) + NC(15) + WA(12) + AL(9) + DE(3)
AZ(11) + IN(11) + MA(11) + TN(11) + KS(6) + NM(5)
MD(10) + MN(10) + MO(10) + WI(10) + CO(9) + MS(6)
SC(9) + LA(8) + CT(7) + OK(7) + OR(7) + NV(6) + UT(6) + WV(5)
With all other Electoral votes going to tenth or lower places. The top three electoral vote receivers are then supposed to go to the House of Representatives to be decided... good luck to them.

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  • $\begingroup$ Damn, ninjaed. Well done. $\endgroup$ Commented Dec 9, 2016 at 17:35
  • $\begingroup$ Nice avoidal of the NE/NH (democratic?) debacle. $\endgroup$ Commented Dec 9, 2016 at 17:42
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    $\begingroup$ Check the electoral votes for your second candidate. I think he/she won. $\endgroup$ Commented Dec 9, 2016 at 17:53
  • $\begingroup$ @MichaelSeifert - You are right however interestingly this solution would work for the linked question of how complicated choosing the top 3 could be $\endgroup$ Commented Dec 9, 2016 at 17:55
  • $\begingroup$ Oops, my bad, fixed $\endgroup$ Commented Dec 9, 2016 at 17:55
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An upper bound on the number of tied candidates is

$\lfloor538/55\rfloor = 9$. Since California gets 55 votes, it is not possible for the tied candidates to have fewer than 55 votes apiece.

The question is whether or not this bound can be realized. The answer appears to be

Yes. Here is the allocation I found:

- Candidate A: California (55)
- Candidate B: Texas, Missouri, Oklahoma (38 + 10 + 7)
- Candidate C: Florida, Illinois, Iowa (29 + 20 + 6)
- Candidate D: New York, Pennsylvania, Nevada (29 + 20 + 6)
- Candidate E: Ohio, Georgia, Michigan, West Virginia (18 + 16 + 16 + 5)
- Candidate F: North Carolina, New Jersey, Washington, Massachusetts, Vermont (15 + 14 + 12 + 11 + 3)
- Candidate G: Virginia, Arizona, Indiana, Tennessee, Alabama (13 + 11 + 11 + 11 + 9)
- Candidate H: Maryland, Minnesota, Wisconsin, Colorado, South Carolina, Connecticut (10 + 10 + 10 + 9 + 9 + 7)
- Candidate I: Kentucky, Louisiana, Oregon, Arkansas, Kansas, Mississippi, Utah, Hawaii, Idaho (8 + 8 + 7 + 6 + 6 + 6 + 6 + 4 + 4)

The remaining 43 electoral votes would go to a tenth candidate.

As an aside: This problem appears to be a variant of the partition problem or the bin-packing problem.

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As stated in other answers,

the upper bound is $9$.

To achieve this bound,

the $9$ tied candidates can each have at most $59$ electoral votes, since $\left \lfloor{538/9}\right \rfloor = 59$.

One way to achieve this result would be to start with:

$\begin{array}{c|l} Candidate\phantom{.}A: & CA(55), HI(4) \\ Candidate\phantom{.}B: & TX(38), AZ(11), AR(6), ID(4) \\ Candidate\phantom{.}C: & FL(29), IL(20), IA(6), ME(4) \\ Candidate\phantom{.}D: & NY(29), PA(20), KS(6), NH(4) \\ Candidate\phantom{.}E: & OH(18), GA(16), MI(16), NM(5), RI(4) \\ Candidate\phantom{.}F: & NC(15), NJ(14), VA(13), MA(11), MS(6) \\Candidate\phantom{.}G: & WA(12), IN(11), TN(11), MD(10), MN(10), NE(5) \\Candidate\phantom{.}H: & MO(10), WI(10), AL(9), CO(9), SC(9), OK(7), WV(5) \end{array}$

Now,

at this point there are $66$ remaining electoral votes, and all of them except for $OR(7)$ will go to $Candidate\phantom{.}I$. Those are: $KY(8), LA(8), CT(7), NV(6), UT(6), AK(3), DC(3)$, $DE(3), MT(3), ND(3), SD(3), VT(3)$, and $WY(3)$. Oregon votes instead for $Candidate\phantom{.}J$.

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