You're offered a gamble where a pair of six-sided dice are rolled and unless they come up snake eyes you get a bajillion dollars. If they do come up snake eyes, you're devoured by snakes.

So far it sounds like you have a 1/36 chance of dying, right?

Now the twist. First, I gather up an unlimited number of people willing to play the game. I take 1 person from that pool and let them play. Then I take 2 people and have them play together, where they share a dice roll and either get the bajillion dollars each or both get devoured. Then I do the same with 4 people, and then 8, 16, and so on.

At some point one of those groups will be devoured by snakes and then I stop.

[Alt. phrasing: We keep going until one of those groups is devoured by snakes, then the game stops.]

Is the probability that you'll die, given that you're chosen to play, still 1/36?

### Argument for NO aka the frequency argument

Due to the doubling, the final group of people that die is slightly bigger than all the surviving groups put together. So if you're chosen to play you have about a 50% chance of dying! ๐ฌ ๐

### Argument for YES aka the one-fair-roll argument

The dice rolls are independent and whenever you're chosen, whatever happened in earlier rounds is irrelevant. Your chances of death are the chances of snake eyes on your round: 1/36. ๐

### Clarifications and FAQ

The game is not adversarial; the dice rolls are independent and fair.

Choosing each group also happens uniformly randomly and without replacement.

[Retracted: An attempt to define a truncated game to take a limit of.]

[Retracted: Blah blah conditional probability.]

[Retracted: Whether it's possible for everyone to survive.]

[Retracted: More blah-blah on conditional probability.]

[Retracted: Ruling out "undefined" as an answer.]

[Retracted: Nitpicking "at some point".]

[Retracted: Yet more blah-blah on conditional probability.]

### Resolution criteria

For an official resolution we'll write up a proof (or "proof") that the answer is 1/36 and a proof that the answer is ~1/2 (really anything greater than 1/36 would be fine) and then recruit some mathematician(s) to make an independent judgment on which is correct. Or maybe we'll just reach consensus in the comments?

UPDATE: @Lorxus has accepted the role of adjudicator. Here are Lorxus's terms, with some comments of mine in brackets:

0. [For a fee of 1% of market volume at the time of acceptance, M$1308, now paid], I will continue to uphold my commitment not to bet on this market. More importantly, now that @dreev has passed arbitration to me, I will read the mathematical arguments of both sides and use my best professional judgement to diligently determine soundness and mathematical correctness.

1. The YES bettors and the NO bettors must reach their separate consensuses as to which writeup to send me. I will only accept a single writeup per side, and will only accept these writeups in the form of a PDF or text file. If you choose to use LaTeX, send me the PDF and not the raw TeX. This must be sent to me before 2023-09-30 at 23:59 EST. I will treat any failures in this regard as a blank entry.

2. The YES bettors and the NO bettors must each choose at most 3 champions per side. The role of these champions will be threefold: to contact me to send me the official writeup of their side, to answer reasonable mathematical questions about the writeups that I pose to them, and to pick formal holes in the other side's writeup to present to me. These champions must be selected and the outcome communicated to me before 2023-09-30 at 23:59 EST. I will treat any failures in this regard as a forfeit.

3. All bettors of this market agree that my judgement will be final, binding, and at my sole discretion. [And other terms about harassing Lorxus but I, @dreev, am insuring against such and believe we don't need to worry about this. Everyone just be nice, ok??]

4. If only one such writeup is sound and mathematically correct, I will choose it as the winner. If only one side's champions can find flaws in the writeup of the other's, or only one side can resolve any flaws I point out in their writeup, I will choose it as the winner. If somehow neither writeup is sound and mathematically correct, and this difficulty persists equally for both sides through questioning, I will say so, and the market will resolve N/A, making everyone including me unhappy. If both writeups are approximately sound and mathematically correct, then I will move to cross-questioning, and I will determine the winner by subjective severity of the worst irresolvable flaw any champion can pick in either writeup. If at that point I have reached the conclusion that the question was fundamentally and irresolvably ill-posed, or both sides have found comparably good flaws, I will arbitrate that the market be resolved to 30-day average of market price. [Resolve-to-MKT is fraught and we may need to hash this out better to ensure fairness -- @Lorxus mentions "anti-spiked" MKT price? -- but I'm nearly sure it will be moot!]

5. I will communicate with all champions through the medium of their collective choice, with a Discord group chat as the default. Conditional on @dreev , at least one >2k YES holder, and at least one >2k NO holder replying to this comment to accept these terms by 2023-09-21 at 23:59 EST [which has now happened!], I will render my judgement by 2023-10-31 at 23:59 EST. Should I fail to do so, I will return the 1308 marbles to @dreev and render my judgement by 2023-12-31 at 23:59 EST.

tl;dr: I got paid 1% of market volume to come Be A Mathematician at this paradox. To both YES and NO: pick <= three champions and start putting your writeups together. I'll read both, ask some probing questions, and puzzle out which side has the better (sounder/more formally correct/more philosophically compelling) arguments based on champion arguments, existing resolution criteria, and my understanding of math. Be kind to me here, because if you break this mathematician you're not getting another.

Writeups (not too late to add more!):

YES by Wamba Ivanhoe (before combining with mine)

Related markets:

Two-round version via simulation (resolved to 1/36)

Ball-and-urn version that tries to be fully non-anthropic (resolved to ~1/2)

Golden-brown pancakes (simple conditional probability sanity check)

Snake Eyes Variant Y (what I meant this market to be)

Snake Eyes Variant N (how NO bettors here interpreted this market)

## Related questions

# ๐ Top traders

# | Name | Total profit |
---|---|---|

1 | แน5,659 | |

2 | แน4,830 | |

3 | แน3,224 | |

4 | แน2,623 | |

5 | แน1,084 |

The below is my official reasoning and ruling as arbiter of the Snake Eyes market.

> At some point one of those groups will be devoured by snakes and then I stop.

We must understand this as a ground assumption. We are therefore in ~the N scenario and should more naturally generalize from it; thankfully arguments already made and accepted mean that we need not worry about attempt to take limits of truncated games or bounded-escape/bounded-dies.

> Is the probability that Euclid will die, given that he has been chosen to play, still 1/36?

We must understand "given" here to mean a conditionalization of some kind on a finite round, especially given the previous supposition of "at some point". We however acknowledge that the full horror of the (meager) possibility of infinitely long games which misleadingly seem to become much more likely upon having been chosen can be made manifest under the right conditions.

Additionally, the question is not asked of what an outside observer thinks, but rather what the probability is that Euclid - that is, *you* - will die.

Finally, Team YES made excellent and valid arguments to the tune that whether we take the limit as a nonuniform prior over what round Euclid is chosen in goes towards a uniform distribution over the positive integers, or we take the limit of suitably defined finite games the probability of his death works out to 1/36; if we instead use nonstandard analysis and have a(n infinitesimal) uniform prior over all positive integers instead, the probability that Euclid will die is still 1/36, regardless of what his mother might expect to see.

Additionally, Team YES consistently made fewer errors of reasoning, found more in Team NO, and patched their own errors consistently more thoroughly.

Given the above, it is my official opinion that the probability is in fact 1/36. Accordingly I direct that the Snake Eyes market resolve YES.

@Lorxus Thank you so much for the huge amount of work adjudicating this! It was all massively edifying for me and I think we made a ton of progress on actual consensus. My only tiny complaint with this verdict is that it puts any weight on how well team YES and team NO performed, rather than being solely about the mathematical truth of the matter. But that's a tiny complaint indeed because if both teams had been perfect mathematicians, we'd still be (to use Lorxus's estimate) 98% YES. (See my previous comment.)

@dreev Sounds like resolution result will be announced this evening.

Will the market close before the results are announced? I prefer to prevent after the fact trading.

@ShitakiIntaki Repeating from the Discord:

how about if i leave the market open so anyone can jump on the the alpha (finance version of "alpha") when @Lorxus's verdict is out? i'll even abstain myself, as penance for my sins

The "sins" there referring to a possible accusation from @Primer that this has been a rugpull (but maybe they were just making a Big Lebowski reference?). My response to that:

i feel like i've been doing the opposite -- shoving rug back under people as hard as i can! like the clarifications team NO objected to, all undone!

extremely nonrhetorical question: what else would you do in my shoes? besides the math, i really want to learn how to run a market better than i did this one.

PS: Primer clarified that it was indeed just a Big Lebowski reference ("I just want my rug back, man")! <3

@peterpumpkin What's your thinking in betting so heavily on NO at this point? I've now spent 179 hours and counting on this problem and this market and its derivatives and am feeling very confident that YES is the mathematically correct answer. Resolution is out of my hands, of course, but still.

Btw, I just added another long comment summarizing the state of things in the Variant N market (now alternatively called the "Euclid's mom" variant, for reasons that are explained over there). In short, I now believe that *even for Variant N* the answer is 1/36, unless we go down a hyperreal -- to use the math term -- rabbit hole.

Finally, I'm confident in this answer even without all the clarifications I tried to add, from back when taking a limit of a truncated game was the only approach I'd conceived of. Team NO was reasonable to reject those attempted clarifications, but ultimately it doesn't help their case.

I'm about to remove those clarifications now, which should make the problem statement a lot cleaner.

@dreev I think the market should resolve no, though a fair argument could be made for NA. The fact that it is going to resolve yes is not important to me (play money). To me the puzzle with this paradox is not whether the probability of dying is 1/36, the question is what is wrong with the argument for 1/36 given that half population dies and therefore the probability of dying is not 1/36. I looked in on the discord discussion and it was mostly irrelevant nonsense. If the debate was higher quality I would've been more inclined to join in.

@peterpumpkin Have you followed any of the debate in Lorxus's Discord? [edit: sorry, reading comprehension fail; you said it was too noisy in the Discord, which is fair]

Here's my characterization of where we ended up, framed in terms of what would be needed for a NO victory:

Condition on snake eyes being rolled in a finite number of rolls. This is what Euclid's mom does and I don't think it makes sense for Euclid but I did accidentally say "at some point we roll snake eyes" instead of my intended "keep rolling UNTIL we roll snake eyes". Those sound the same to me but, ok, this is one prereq for a NO victory! Conditioning on snake eyes getting rolled in finite time is key. Importantly, it turns out not to be ok to condition on eventual snake eyes by virtue of its probability being one! Normally that's totally fine and the not-fine-ness in this case is at the heart of the paradox.

Reject the nonuniform prior solution. This is a big ask since it seems to be a more robust solution that avoids the controversy with how to define a truncated game. And team NO quite reasonably agreed that Pr(chosen) can't be exactly zero. But ok, no nonuniform prior.

Either use nonstandard analysis or take the limit of a truncated game in which Euclid knows the max number of rounds, N, and so knows he's more likely to be doomed (should he be chosen) as round N bears down.

It's a tight needle to thread and, if I've characterized things correctly, it's why @Lorxus considered resolving to 98% YES instead of full YES. It's not *impossible* to argue for NO but there are multiple solid YES arguments. I think the tie-breaker was how thoroughly @ShitakiIntaki attacked every mathematical challenge Lorxus and team NO threw at us. Team NO also did brilliant work (I learned a lot from them along the way) and were crucial in reaching the consensus that we can't fully fairly take limits of truncated games.

But they haven't engaged with the nonuniform prior argument yet. It would be great if any other NO believers are game to do that. And it's so wild! I would never have believed it. I encourage people to step through the proof in section 4 of our write-up or try Bartha & Hitchcock's version (section 2). I think it leads inexorably to 1/36 and I really, really want to understand where the flaw is if there is one. I'm willing to bet a ton of mana that no one will find one!

@dreev I think Peter was not impressed with the discussion on Discord based upon their response above.

I looked in on the discord discussion and it was mostly irrelevant nonsense. If the debate was higher quality I would've been more inclined to join in.

I invite you into this poll to answer about your reading of the word "given".

I am sorry if this has already been discussed a lot but the comment thread is way to long. How is this going to be resolved? I was thinking about writing a proof that gives us an answer almost surely (excluding all events with Lebesgue measure 0) to avoid the non-converging sum issue that seems to be the problem in the current yes proof. But the description says deadline for proofs has passed.

@AlexbGoode deadline has passed, but nobody stops you from contacting the champions.

Currently Fintan is trying to explain yes mistakes, you can write me in discord kongolandwalker and i will pass your notes to Fintan.

Your subjective probability that you will die *when you get drafted in*, during any given round in an ongoing game, is 1/36. the Probability that a random participant in a *completed* snake paradox game is dead is 1/2. These sound like paradoxical takes but they aren't, and the resolution of this market depends on how it interprets the inherent ambiguity in the question asked.

Apologies if the below discussion has already beaten this to death but it seems obvious to me that this is the entire source of the confusion

@MilfordHammerschmidt the question says that the person "is chosen", not "chosen into round N". That is why No party defends the answer 0.52... . (Not 0.5, because no possible game could have exactly 0.5 ratio. All of games have higher ratio, but different chances of happening. The 0.52 is achieved as a result of convergent infinite sum).

@KongoLandwalker Ok, fine, but I'm just making the point that there is no logic or maths ambiguity here; the question is entirely determined based on what you're asking about and at what point in the process you're estimating this probability.

@MilfordHammerschmidt What about agreeing to play before the game starts? To me it feels like if I'll agree to play, my outcomes look like this:

`P(not chosen) + P(chosen, dead) + P(chosen, alive) = 1`

`P(chosen, dead) = P(chosen, alive)`

And I'm confused about at what point during the game `P(chosen, dead)`

stops being equal to `P(chosen, alive)`

@roma Same thing. You're naturally thinking "but 50% of participants have died *when the game ends*", and yes, that's correct, but *when the game starts*, your subjective probability of dying is still 1/36.

@roma To reveal the issue here it might be helpful to think about what you'd bet. Say a particular SEP run has ended, and someone asks you to bet on the possibility that a random participant in that run is dead. You'd bet at at least 1:1 odds, because the game always ends on a loss. But say someone is running a variant of the SEP, and they offer you to get drafted, with a reward of 50$ if you win, and a fine of 50$ if you lose. That's a good deal - you'll play that game forever depending on bankroll - because the game has a 1/36 chance of ending on your round each time you play.

@MilfordHammerschmidt Yeah, I'm with Roma, despite also believing you're 100% right! Which is to say, this needs way more explanation than just "depends if the game ended". For starters, with probability 1 the game will end in a finite number of rounds, right? So something bizarre has to happen for your probability to change just by conditioning on something you're already certain of.

Another way to put it: Take the perspective of a candidate player (we've started calling him "Euclid" instead of "you"). Where *exactly* does the reasoning of "eventually the game will end and he'll be in a set of people half of whom are dead" go wrong? Or if you're on team NO, where *exactly* does the reasoning of "if he's chosen to play he'll get one fair dice roll, independent of any other rolls" go wrong?

That's what a resolution of this paradox has to do, I feel.

Where

exactlydoes the reasoning of "eventually the game will end and he'll be in a set of people half of whom are dead" go wrong?

I'm in the "set" of [USA citizens + Indian citizens], and yet there is not a 66% "chance" I am Indian, when you condition on a bunch of other information of which I am aware. If all you knew was that I was in the above set, then indeed, that might be a fair estimate.

In general, if you could phrase this paradox in terms of a bet you don't know how to price, that would go a long way in terms of cutting to the heart of any confusion.

@MilfordHammerschmidt Definitely. I mean, we're debating a probability, so turning it into a bet is straightforward:

You join a pool of players for the protocol in the description here. But instead death or a bajillion dollars, I just charge you $1000 if you roll snake eyes and pay you $500 if you roll anything else. If you're not chosen to play (never roll the dice) then no money changes hands.

Team YES would pay up to $450 to play that game. Team NO would have to *get* paid $250 to play it.

I'm in the "set" of [USA citizens + Indian citizens], and yet there is not a 66% "chance" I am Indian, when you condition on a bunch of other information of which I am aware. If all you knew was that I was in the above set, then indeed, that might be a fair estimate.

Oh, when you put it this way I actually want to play devil's advocate for the NO side for a minute! The problem says people are chosen uniformly from the pool. So from the perspective of a completed game, there's no possible evidence either way for you being in the lucky vs unlucky half of the chosen players.

Team YES has to make one of two counterarguments:

That's all rock solid for completed games. But! What about the infinitesimal chance of rolling not-snake-eyes an infinite number of times? We have a messy 0 ร โ issue here but intuitively, the 100% survival rate for an infinite number of people skews the expected frequency despite its 0% probability of happening.

Uniform selection from an infinite pool is impossible. Nearly uniform, sure. Arbitrarily close to uniform even, but not quite uniform. And then, amazingly, whatever deviation from uniform you make, it always works out to be exactly enough to make your American-vs-Indian argument work perfectly. If you make a huge deviation from uniform and are likely chosen super early, then you're very likely safe. If you make a minuscule deviation then it's going to take nearly forever for you to be chosen. Long enough that if you're chosen at all it's because the game lasted nearly forever. Which means the non-uniform distribution puts us most likely in the early, lucky half.

What I absolutely love about this is that, if you'd given the above arguments to me at the outset, I'd have reacted like so:

I will be very surprised if working that out by taking a limit yields 1/36. (Working it out without limits I didn't even know was a thing.)

Surely you can make the answer be anything if you can have any crazy prior. Are you going to tell me that if I have a prior like "I happen to know I'll be chosen on a prime numbered round" that I'm going to magically end up having a 1/36 chance of across all the prime numbered round I might be picked on? Don't be daft.

So, yeah, I'm extremely sympathetic to team NO here, but now believe the math does not turn out in their favor, very much contra my own initial expectations.