Bertrand's Ballot Theorem

Bertrand's Ballot Theorem
Sorry for the vertical video, this was originally posted on Tiktok!
Read more:
en.wikipedia.o...
Corrections:

[@JK-fj2eh]

"But what about bad paths that never come up to cross the x-axis?" From the setup of the problem this is impossible- we know Alice wins, so at some point it must cross the x-axis and we can do the mirror trick.

[@fibbooo1123]

I love combinatorics! A great proof too- it makes use of the slight subtly that a tie is also a fail (in the visual language of the proof, a bad path is any that touch the axis, not just those that go beneath it) but a second rewatch made that clear. Great job!

[@angel-ig]

Brilliant proof!

[@SuperYoonHo]

nice sir!

[@q335r49]

I stumbled upon this problem trying to think about whether Bill and Ted was doomed to hell or purgatory of the grim reaper kept on extending tournament lenghts (best 3 out of 5, best 4 out of 7 ...) In that movie, bill and Ted's bogus journey 😆

[@shayneweyker]

Doesn't it matter in what order the votes arrive to be counted? Consider: If the votes from subgroup of voters that as a group favor one candidate Alice are counted before a different subgroup that favor the other candidate Bob there mere fact that the votes are counted in this order will have biasing effect that will make early predictions of the winner flawed.

[@zacharysilver3632]

I think any probability question asked in good faith assumes randomness. Obviously, we could always start the tallying by opening Bob's vote, in which case Bob will always be ahead at at least one point.

[@shayneweyker]

@@zacharysilver3632 I guess I'm thinking like a political/social scientist. In many if not all elections in a close race some precincts will be more favorable to one candidate than another. If you start counting votes and making predictions based on which precinct's votes come in before others, and if the first reporting precincts are more favorable to a particular candidate, then that candidate will be favored in early predictions based on only those favorable precincts' votes. I suppose the people making predictions could decide not to start making predictions until a good number of precincts report that pre-election polling suggests those precincts' as a group would be roughly equal in number between favoring candidate A and candidate B.

[@plasmaballin]

At the beginning of the video, it stated that the ballots are counted in a random order. Obviously, this is not true in a real election, since certain places tend to count votes more quickly than others, and large batches of ballots are reported together, so the theorem might not apply in real life. So yes, it does matter, but that doesn't affect the theorem, since the theorem assumes a random order.

[@elisecastor5534]

Very cool proof.