*This is part of my quest to untangle the controversy in **The Snake Eyes Paradox**. It's called Variant N because it's how NO bettors intepreted the original.*

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 infinite 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.

When snake eyes is eventually rolled, that's the final group. They're devoured by snakes and we're done.

[EDIT and META-EDIT: I had added a thing here about always rolling snake eyes "even in the finite version of the game" but part of the point of Variant N is that there's no finite/truncated version. There's simply an infinite pool of people.]

What is your chance of dying, given that you're chosen to play *and that snake eyes gets rolled in a finite number of rolls*?

### Argument for ~1/2 aka the frequency argument

We're explicitly guaranteed that we'll roll snake eyes and when we do, that's a final group dying that's 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 1/36 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 and the dice rolls are independent and truly random.

Choosing each group also happens uniformly randomly and without replacement.

What if the question turns out to be self-contradictory or the answer isn't a real number? Then we argue about how to fix it in the comments and do our best to agree on the best possible real number answer.

### Resolution Criteria

This resolves to the mathematically correct probability, Pr(death | chosen). How do we decide what that is? With a council of 2 people who've argued coherently for NO in the original market, 2 people who've argued coherently for YES, plus a mathematician who hasn't bet. See the companion market, Snake Eyes Variant Y, for details. I won't trade in this market.

According to classical probability theory, the answer is unambiguously 1/36. If you try to come up with a nonclassical probability theory that allows for a uniform distribution over the natural numbers (impossible normally), the probability of the original snake eyes game is still unambiguously 1/36, but Variant N becomes 1/2.

The results depends on clarification 3. It's impossible to uniformly select natural numbers. Therefore, it's impossible to select groups of people uniformly at random from an infinite pool as the problem statement says. One way to resolve the self-contradictory problem is just to ditch Clarification 2. Then it actually doesn't matter what probability distribution you use, the result is always 1/36, no matter what. The other way to fix the problem is to reject the notion of classical probability altogether and use a non-standard probability theory where a uniform distribution on the naturals makes sense. Then you find that P(death|chosen) = 1/36, but P(game lasts forever|chosen) = 17/18, and P(death|chosen,finite game) = 1/2.

There's a fuller explanation of how to get those probabilities in this post: https://plasmabloggin.substack.com/p/the-snake-eyes-paradox

After intensive further discussion in @Lorxus's Discord, I now believe it's safe to remove this edit that I'd previously added:

[EDIT: Also, for this to be the true Variant N, we make it explicit that the game always ends in snake eyes. Even for the finite version of the game where it might otherwise be possible to run out of people without rolling snake eyes, we're conditioning on that not happening.]

It suffices to say that we're assuming snake eyes is rolled in a finite number of rolls. There's no finite/truncated version of the game. There's simply an infinite pool of people, as @Primer and other advocates for the ~1/2 answer said all along.

But that means that we need one of two approaches: (1) A not-quite-uniform prior over when you'll be chosen. (2) Nonstandard analysis, where we can assign infinitesimal probabilities. If we go with (1) then 1/36 is correct, even here in Variant N. If we go with (2) then the assumption of snake eyes in a finite number of rolls does change the answer to 1/2. Wild stuff.

@dreev The idea for this variant was to say "Snake Eyes always ends with throwing snake eyes". It's part of the setting, like the number of dice. If somehow we'd construct a formula or run a simulation where the game would not end in snake eyes, we shouldn't use it as it deals with some similar game, but certainly not Snake Eyes N.

@Primer Right, I think the market description here for Variant N does that. I'm really sorry for how much confusion I managed to inject. I've been doing my best! And we sure have come a long way in understanding all this.

And an update: We've found a new way to make the Variant Y vs Variant N distinction. We call the protagonist Euclid (previously just "you" which could be confusing) and we can either ask the question from Euclid's perspective or *Euclid's mom's* perspective -- the difference being that Euclid's mom wants to know the probability of Euclid's death after learning both that Euclid was chosen to play and that the game ended in snake eyes in a finite number of rounds.

Write ๐(p) for Pr(death | chosen) as a function of p, p being the probability of snake eyes on a single roll.

The latest things I believe:

Taking the limit of a truncated game, ๐(p) = p for both Euclid and his mom.

Using a not-quite-uniform prior over Euclid's position in the queue -- same thing, ๐(p) = p.

Using a uniform infinitesimal prior (warning: very deep math rabbit hole here involving hyperreals), we have ๐(p) = p for Euclid but ๐(p) = 1/2 (for at least some values of p?) for Euclid's mom.

For quite a while I didn't know there was anything one could do besides #1 -- taking the limit of a truncated game. To me that's still a perfectly cromulent solution, but if we want to reject it because we don't agree on a reasonable truncated version or if the existence of different truncated versions means we don't trust the limit, then, ok. Move on to #2, which I think is where team NO from the original Snake Eyes has landed. We fix the problem that Pr(chosen) = 0 by saying Pr(chosen) is not quite zero. The game's not truncated, Euclid is just ever-so-slightly more likely to be chosen sooner than later.

The amazing thing with doing that: no matter what that prior, no matter how close it is to uniform, it shifts the probability mass exactly the amount it needs to to make ๐(p) = p. Roughly, it's because the knowledge that Euclid was chosen to play is evidence that it took a looooong time to roll snake eyes. Half the players end up dead -- that's a given -- but Euclid is more likely in the early, lucky first half.

(That makes sense for an extremely nonuniform prior, like if you're sure you'll be chosen first, then, sure, ๐(p) = p trivially. But then as you push that prior further and further towards uniform? That's what makes games in which Euclid is chosen be longer and longer ones. The math all works out such that it stays perfectly balanced: Euclid always has probability 1-p of being in the lucky, early half!)

Of course now the door is open for skeptics to reject that nonuniform prior. When/if the game has ended and we have some finite number of people who played and half of them are dead, it feels like we don't have a fair basis to believe Euclid was special and more likely in the early half than the later half of recruits. I'd sure have been suspicious myself. But it's true even in the limit as our prior approaches uniform. And that limit seems harder to object to than the one we used for the truncated game.

But if we do want to reject it, that's what brings us to #3. I don't understand #3 yet and team NO has not wanted to go there but if we do go there, that's when Variant Y vs Variant N (aka Euclid's probability vs his mom's probability) matters. The frequency argument (see market description) is wrong for Euclid because, in the infinitely unlikely event that he's chosen from the infinite queue, the game is probably literally infinitely long. Snake eyes is never rolled and infinitely many people all survive.

@dreev How could solution (1) work in variant N? You're reasoning as if this was regular SE or variant Y.

In the language of the discussion in the discord, this is bounded death.

@Primer Oh, yes, at this point it's fine to call it my personal opinion that bounded escape is reasonable and bounded death is unreasonable.

(Recap: If snake eyes were never rolled, no one would die. In the unbounded game that has probability zero -- rolling not-snake-eyes forever -- but if hypothetically it happened, an infinite number of people would all survive. Bounded escape matches that. If snake eyes isn't rolled before running out of people then all the people *that there are* all survive.)

Point being, we don't agree on the truncated game to use, so we're not going to use a truncated game at all. Scratch #1 off the list.

Here's @Primer's original proposal for Variant N:

You're offered a gamble where a pair of fair, 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, so Snake Eyes always ends with throwing snake eyes.

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

I believe this version needs more clarification. The game always ends in rolling snake eyes? That's true in the limit with fair dice, but what are we taking the limit of?

(Without taking a limit, the answer to the problem as stated is technically 0/0 โ undefined โ due to the zero probability of being chosen out of an infinite pool of people. Which is fine; the standard thing to do in that case is take a limit.)

So we need to specify a finite version of this game. Does "Snake Eyes always ends with throwing snake eyes" still apply in the finite version? The problem statement doesn't say anything about the finite version so we have to fill in those details. And this is the crux of the original version. If we say that for the finite version the game can end with no snake eyes and no deaths, we have Variant Y. For Variant N we need every finite version to end in snake eyes. And the way to do that without violating the fairness of the dice is to condition on snake eyes happening.

For example, in the 1-player 1-round version, I have a 100% chance of death. I'm definitely chosen to play and we're assuming that snake eyes is rolled.

So I guess then I'd argue that the original can't be Variant N because "in the 1-player version you die with 1/36 probability" is the very first thing specified in the original. But maybe there's another way to construct Variant N? Or something else I'm missing?

@dreev We can just play a game with dice and snakes, see if we die or not, and then condition on the game having been snake eyes.

I don't think we need a "finite version". There is a random process which determines how many players there will probably be. The ratio between winners and losers will be around 1:1.

Didn't you do a simulation of the original? If so, you could just go back there and discard any game which didn't end in snake eyes. Just like when you take a random number for the dice, you discard any number >36.

As the one who came up with variant N, I feel like I have to clarify that this is not the variant I suggested. Especially the 2nd to last paragraph should read

At some point one of those groups will be devoured by snakes and then I stop, so Snake Eyes always ends with throwing snake eyes.

And furthermore, there are no other clarifications in variant N. I consider this market's variant to be N-y and would prefer it to be changed to original N.

The original variant N always ends in snake eyes. If you played a game with a finite or infinite number of rounds which did not end in snake eyes, you didn't play variant N, but some other game we don't care about.

@Primer I see what you mean. Yeah, I went down a bit of a rabbit hole questioning how to interpret "assume we eventually roll snake eyes" and I think your point is that the true Variant N just does in fact condition on snake eyes being rolled before running out of people.

So all my hand-wringing about the two possible interpretations is really just arguments for why the other interpretation is more reasonable *in the original*. For Variant N we should just agree that we're conditioning on rolling snake eyes before running out of people.

PS: I've now edited this into the description, in brackets so it's clear what changed. Do we agree on this being the true Variant N now?

@dreev I appreciate the work you're doing here!

I had given my original phrasing (in the comment in the original market) quite a bit of thought. It's supposed to be as close to the original phrasing as possible. Thus I disapprove of additions and alterations. "Snake Eyes always ends in snake eyes" totally captures that point: any calculation we do and any simulation we run is only variant N if it ends in snake eyes. No need to condition on this, it's an intrinsic part of the whole setup.

@Primer I think what you're describing *is* conditioning on rolling snake eyes. Another way to conceptualize conditioning on something (thanks to @FlorisvanDoorn) is to revert time and start over if the thing doesn't happen. So we don't need to rig the dice to make sure the last group dies (I was wrong to suggest as much in the past!). Instead we roll fair dice every time and if we run out of people with no deaths, we pretend the whole thing never happened and start over. Keep starting over until we do roll snake eyes before running out of people.

(For example, in the 1-round, 1-player version, that means that that one person is definitely dying.)

I believe that that's what it takes to have a proper Variant N. But let's keep hashing this out until we're satisfied. If at this point we need a separate market that's fine, or I can edit this one again if no one objects?

@dreev Do we need to condition on 2 dice being thrown (instead of no dice, or 5 dice)? I think we don't, that's just how the game works. The same applies for ending in snake eyes. It doesn't need to be another conditional.

@Primer Ah, this is key, yes. We don't say we're conditioning on 2 dice between thrown, we just throw 2 dice. Now try to do the same for "we roll snake eyes before running out of people". It contradicts the assumption of fair dice! See what I mean?

This is what conditional probability is made for. When we condition on an event E, we're not constraining the process like we are when we say how many dice we're rolling or what snake-related things we do to people based on the rolls. Rather, we just throw out all possible worlds where E didn't happen to... happen.

Consider this problem: We flip a coin that has a 49% chance of landing heads, a 50% chance of landing tails, and a 1% chance of landing on its side. Assuming it didn't land on its side, what's the probability it landed heads? We're not saying it couldn't land on its side, we're just saying it happened not to. We can find the probability empirically by flipping it lots of times and discarding flips where it lands on its side.

I'm not sure what answer you get without conditional probability. Maybe the question doesn't make sense? It's like saying "the coin has a 1% chance of landing on its side just kidding it has a 0% chance of that".

I guess the Snake Eyes version of that is "the last group might not die just kidding the last group totally dies".

Point being, I don't see anything else for Variant N to do but condition on rolling snake eyes before running out of people. Or in the infinite version, condition on not rolling snake eyes forever. (Which seems superfluous but, fascinatingly, is not.)

We don't say we're conditioning on 2 dice between thrown, we just throw 2 dice. Now try to do the same for "we roll snake eyes before running out of people". It contradicts the assumption of fair dice! See what I mean?

Yeah. But this is why we have an unlimited amount of people in variant N (and not in variant Y):

We don't say we're conditioning on rolling snake eyes before running out of people, we just never run out of people.

(I'm aware that maybe we can't solve this as easily by taking a limit of some naive finite case. That's where it might get mathematically interesting.)

@Primer It's so gratifying the steady progress we're making on this. Thank you! So I think this brings us to FAQ3: arguing about how to make Variant N have a real-number answer. Approaches:

Define a finite version to take a limit of.

Do nonstandard analysis I guess? (HT @Lorxus)

I don't understand 2 yet (I'm eager to hear if anyone else does!). For 1 I'm fairly sure that the only thing to do is condition on rolling snake eyes before running out of people.

@dreev There is no FAQ 3 in the original variant N.

All this seems you're trying to turn variant N into variant Y. They are different.

@Primer The point of Variant N is that the answer is ~1/2. The reason for FAQ3 here in Variant N is that the original asks "Is the answer 1/36 (YES) or is it >1/36 (NO)?" with no allowance for non-real-number answers like "undefined". (Nor answers less than 1/36, for that matter, but thankfully that's moot!) Hence "take a limit or whatever we need to do to get a real-number answer".

@dreev Still, the FAQ interfers negatively with the variant's intent. The problem you're referring to could have easily been handled by asking "Is the answer 1/36 (Yes) or not (No)?" without unnecessarily introducing additional, unclear or problem-altering phrasings.

@Primer Maybe we just want Variant U as well, for the version to which the answer is "undefined"? Variant N is meant to be the version closest to the original that has an answer of >1/36. My latest thinking is that in Variant N we have to be explicit about conditioning on rolling snake eyes before running out of people and then taking the limit as the number of people goes to infinity. Do you have a different way?

@dreev I think mathematically, they're all undefined.

Variant N can maybe be solved by interpreting the game as a conditional, but that shouldn't be part of the problem statement.

@Primer Oh, huh, shall we debate Variant Y in the comments there? That one seems mathematically defined to me.

I can't tell if there's more to hash out with Variant N. We agree that the answer is ~1/2. I guess I've lost track of what the disagreement about conditionals was about. What's your ideal statement of Variant N at this point? I think it should have the constraint that it be mathematically defined and have a real-number answer.

PS: Oh yeah, and to review my argument for why we need a conditional probability, it's because otherwise these two statements contradict each other:

The dice are fair.

We roll snake eyes before running out of people in the finite version.

Maybe you're not disputing that, just saying that that's one way to get an answer, not part of the problem statement?

What's your ideal statement of Variant N at this point?

The original one I put in the comments at the original market.

oherwise these two statements contradict each other:

The dice are fair.

We roll snake eyes before running out of people in the finite version.

Then maybe the second statement is just wrong: there is no finite version?

The game always ends in snake eyes

The dice are fair

Those statements live on equal footing. If you desperately want to find out what happens in a hypothetical other setting, why not start with the game ending in snake eyes and try to make the dice fair in a limiting process?

@Primer Roger that, I've added your original as a new top-level comment here for reference. I believe the edit I previously added in the market description here makes it match. The key is conditioning on rolling snake eyes before running out of people.

At least I'm still stumped on any other way other way to do it. I don't understand your new question above, about starting with the game ending in snake eyes and trying to make the dice fair in a limiting process. Can you elaborate?

@dreev The dice being fair and the game ending in snake eyes both are intrinsic properties of the paradox.

There's no way in which treating one as more fundamental than the other and conditioning on the other is justified.

Here's an idea: We can start with alternating between a "random" dice throw and a predetermined "snake eyes" and then take the limit of adding more and more random dice throws between the snake eyes.