Are Manifold users voting for Trump or Biden in the 2024 election? [RESOLVES BY %]
Nov 9


This question resolves by percentage. Both Trump and Biden votes will count to the resolution.


Trump receives 23 votes, Biden receives 77 votes.

Then Trump resolves 23%, Biden resolves 77%.

Hopefully that clears up how this question will resolve.

This question resolves at the percentage of this poll:

Get Ṁ600 play money
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To clear the resolution criteria for this question, it resolves on the % of votes for the 2 candidates only (not American/not voting wont count to this total).

So hypothetically if I resolved this today (Poll closes on election night):
There are a total of 34 votes (12 Trump, 22 Biden).
This question would resolve 65% Biden and 35% Trump (if I can avoid rounding and go into decimals, I will. if not normal rounding rules apply).

Let me know if this muddies the water, I am happy to try to clarify further.

This market resolves to the percent for Trump and Biden in the poll, not 100% either way, is that correct?

bought Ṁ150 Trump

@thepurplebull oh that's a good point.. a bit unclear but it does sound that way

@thepurplebull that is correct, resolves based on the % difference

@Bayesian I mean, obviously Manifold is mostly for Biden, so this market would be pointless if it wasn’t resolving proportionally

@thepurplebull Many markets are pointless, and I thought at first this was one. Haws could you clarify proportionality in the description? The title obviously makes it not sound that way but as long as description is clear it’s fine

@Bayesian I'll flesh out the description a bit to hopefully satisfy the resolution criteria.

those are great changes, I appreciate it :p

I'll vote but not for either of them.

@HankyUSA Your vote works about as well as it does on this chart to be honest. hopefully that's what you wanted.

@StagDragon What about not voting for Joe Biden or Donald Trump won't "work"?

@HankyUSA there are three possible outcomes of your vote:

1 - it has zero effect on the outcome of the election;

2 - it is the one that the major candidate you dislike the least would have needed to win;

3 - it is one of those that causes your preferred candidate to finish in the top two.

I'd say 1 is ~8 orders of magnitude more likely than 2, and 2 is maybe ~12 OOM more likely than 3.

Twelve orders of magnitude? It's one trillion times more likely that HankyUSA's preferred candidate gets into the top two by one vote, than that their least favorite between Trump and Biden is one vote away from winning?

If you think that, then I'd assume you think that there's less than a one in a trillion chance HankyUSA's candidate wins the election overall. (If not, I suspect there's a inconsistency between some of your numbers here.)

Without even knowing who that candidate is, I'll absolutely take a bet at those odds. Or actually, since you'll presumably never have a trillion units of any currency, I'll give you much better odds than a trillion to one - $1,000,000 to $1. Assuming you don't have $1,000,000 either, I'll just take as much as you do have, and we can arrange a deal for you to pay me a sustainable fraction of your salary for the rest of your life, or until you've paid me more than $1,000,000 (inflation-adjusted) in total.

@Zane I like to put my money where my mouth is. However, I don't have $1M, and I want a bet to settle all at once, especially given the marginal utility of money is not constant (losing $1M is worse than winning $1M is good).

I say I pay you $10,000 if you win, and you pay me $100 if both the following happen: I win AND a mutually agreed public random number generator gives a result matching a 1 in 10,000,000 probability. If my finger arithmetic is right, this is 1000x higher expected value for you than what you proposed.

The bet is conditional on me knowing your actual identity and having some way to be decently confident you'd pay up, and of course I offer similar terms on my side.

The exact terms I'd bet on are a bit more similar to the conversation I'd had with @HankyUSA , with tweaks. On the one hand, who his candidate is doesn't matter for my bet, so this is strictly better for you than if it depended on his vote. Also, I don't just lose if a third party candidate wins – I also lose if they finish second, which is again strictly better from your point of view.

On the other hand, I never said anything about Trump and Biden specifically; the bet is about the Republican and Democratic candidates. Trump or Biden kicking it doesn't affect the two-party system. So I win if the GOP and the Dems are the parties of the top two candidates in the electoral college. Furthermore, we'll screen off faithless electors and other shenanigans; we'll look at the popular vote winner reported by credible news organizations and attribute the electoral votes in each jurisdiction the usual way. This means the winner takes all everywhere, except Maine and Nebraska; Maine and Alaska use ranked choice voting and we'll use that. So the bet resolves as soon as we know who won the popular vote in enough jurisdictions that no other candidate can reach number 2 - meaning, as soon as both the GOP and Dems have more EVs called for each of them than there are EVs left to call.

The bet is off if something goes wrong with the election as a whole. This means the bet is on if and only if, at 12:00 Eastern on November 6, at least one credible news organization has declared a set of states which combine for at least 270 EVs to have been won by a specific candidate (not necessarily the same in every state), or to be "too close to call". I expect that if there is some serious disruption of the election, media will not be reporting states as won by any candidate or "too close to call".

How do these terms sound to you?

The resolution condition (does a candidate who is not a Republican or a Democrat get into the top 2 in the electoral college, as long as the election is not disrupted) sounds good.
I'm not willing to tell people my real name at these stakes. But if you can't trust me to pay, I'll just send you 1 mana today (a tenth of a cent) instead, still for your $10,000 if I win. Bruno Parga seems to be your real name, and I'll trust you on your word.

@Zane deal.

Apparently the minimum amount of mana you can send someone is 10; I'm still fine with that. Sent.

@Zane received. If i lose and either of us is no longer active here come election time, I'm easy to reach elsewhere.