Is the probability of dying in the Snake Eyes Paradox 1/36?
172
closes 2100
90%
chance

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.

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

Argument for NO

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! 😬 🐍

Argument for YES

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. 😅

Clarifications and FAQ

  1. The game is not adversarial and the dice rolls are independent and truly random.

  2. Choosing each group also happens uniformly randomly and without replacement.

  3. The question is about the unrealistic case of an unbounded number of people but we can cap it and say that if no one has died after N rounds then the game ends and no one dies. We just need to then find the limit as N goes to infinity, in which case the probability that no one dies goes to zero. [ANTI-EDIT: This is back to its original form since a couple people had concerns about whether my attempts to clarify actually changed the question. A previous edit called the answer "technically undefined" in the infinite case, which I still believe is true. The other edit removed the part about how the probability of no one dying goes to zero. That was because I noticed it was redundant with item 5 below.]

  4. We're asking for a conditional probability: given that you're chosen to play, what is the probability that you die? I.e., what fraction of people chosen to play die [in expectation]?

  5. Importantly, in the finite version it's possible for no one to die. But the probability of that approaches zero as the size of the pool approaches infinity.

  6. What if "the fraction of people chosen to play who die in expectation" is different from the conditional probability? Answer: we're talking about the conditional probability. We're treating this as a decision theory problem: assuming you want to play the one-shot version, do you still want to play the doubling-groups version?

  7. What if the most correct answer is "undefined"? If we went by just the title of this market, that would be NO, but from the beginning we clarified that NO requires the probability to be strictly greater than 1/36, which "undefined" is not. I failed to consider the possible answer of "undefined" when creating this market! So if that's the answer it's going to be hard to have a satisfactory resolution but I think N/A will be the least unsatisfactory in that case.

  8. "At some point one of those groups will be devoured by snakes and then I stop" has an implicit "unless I roll non-snake-eyes forever". I.e., we are not conditioning on the game ending with snake eyes. The probability of an infinite sequences of non-snake-eyes is zero and that's the sense in which it's correct to say "at some point snake eyes will happen" but non-snake-eyes forever is possible in the technical sense of "possible".

  9. Via David Pennock: The word "given" is being used in the sense of standard conditional probability: what is the probability that you die in the case where you happen, by random chance, to be chosen. "Given" here does not mean being somehow forced to be chosen. (Corollary: The probability of being chosen is zero in the infinite population case.)

(FAQs 6-9 were added as clarifications based on questions in the comments. NO bettors have objected to this so we plan to adjudicate this as if those weren't there.)

PS: To expand on FAQ8, "probability zero" and "impossible" are mathematically distinct concepts. For example, consider flipping a coin an infinite number of times. Every infinite sequence like HHTHTTHHHTHT... is a possible outcome but each one has probability zero.

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. [UPDATE: We now have at least one NO champion (Kongo Landwalker) and at least one YES champion (me). Please comment if you're game to join us!]

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.

Candidate writeups:

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Sort by:
Lorxus avatar
Lorxus Mathfox

https://discord.gg/RtM3QU4n I have set up some rudimentary channels in my discord for the purposes of this market.

1 reply
Primer avatar
Primerpredicts NO

@Lorxus Thanks!

dreev avatar
Daniel Reevesbought Ṁ380 of YES

Nice intuition pump from Greta Goodwin:

  • Snake Eyes 1: The same as before. The game continues until some group dies.

  • Snake Eyes 2: Flip survival and death. If the dice come up snake eyes then that group survives, otherwise they get eaten. The game continues until some group survives.

If you think the probability of death is 1/2, you should be equally (un-)happy to play either of those versions. But search your feelings, you prefer the first one.

22 replies
Primer avatar
Primerpredicts NO

@dreev I'm pretty indifferent to those two options.

DeadRhino avatar
hipredicts YES

@Primer want to play a game? We simulate 10000 games with a 35/36 chance of dying, and for every person who lives i will give you 2 mana, and for every person who dies, you give me 1 mana.

Primer avatar
Primerpredicts NO
MartinRandall avatar
Martin Randall

@dreev There are many numbers between 1/2 and 1/36, and people can have uncertainty over philosophy, so I don't think this shows much.

@DeadRhino @Primer I think neither of you have the infinite bankroll needed for this offer.

DeadRhino avatar
hipredicts YES

@Primer What do you think of the following python code (I haven't run it yet):

from random import *

alive = 0

dead = 0

for _ in range (10000):

rounds = 0

while randint(1,36) == 36:

rounds +=1

dead += 2**rounds

alive += 2**rounds - 1

print(dead)

print(alive)

DeadRhino avatar
hipredicts YES

@MartinRandall most games will end after a few rounds.

Primer avatar
Primerpredicts NO

@DeadRhino Thanks for writing code a non-coder can understand, at least in part.

Shouldn't it be

while randint(1,36) =! 36:

Also why is there that (10000)? It was 10000 games, not 10000 rounds.

DeadRhino avatar
hipredicts YES

@Primer

for

while randint(1,36) =! 36:

This increase the round count with a 35/36 chance, so theres a 1/36 chance of dying, but we want a 35/36 chance of dying (so our chance of continuing is 1/36)

Also why is there that (10000)?

The

while randint(1,36) == 36:

line keep increasing the round until death, so 10000 games are actually played.

Primer avatar
Primerpredicts NO

@DeadRhino There should be a new round in 35/36 of cases. And the round needs to increase until survival, not until death.

DeadRhino avatar
hipredicts YES

@Primer

  1. there is a new round, the while loop keeps repeating itself until it gets a random number besides 1.

  2. Doesn't the game end after a group of people die?

Primer avatar
Primerpredicts NO

@DeadRhino

  1. Not a coder, so I don't quite understand, but nevermind, I suppose the crux will be:

  2. The reversed game ends after a group survives. It's right there:

The game continues until some group survives.

DeadRhino avatar
hipredicts YES

@Primer

  1. ok

  2. oops, didn't read the terms fully, can we not bet?

Primer avatar
Primerpredicts NO

@DeadRhino Ok. But then I don't really understand your investment in this market.

  • 35/36 chance of living + ends in dying

and

  • 35/36 chance of dying + ends in living

seem pretty symmetrical to me.

roma avatar
Romapredicts NO

@Primer @DeadRhino

Can I suggest another game you could simulate?

- You simulate one game.
- In each round, you generate an extra random number to tell whether or not @Primer is picked from the pool to play in this round. The probability should rise with the batch size but never reach 1. Could be something like P(picked) = batch_size / (batch_size + 100000).
- If @Primer is picked, this can be last round you simulate as you only want to know @Primer 's fate.
- If the game ends, but @Primer weren't picked in any of the rounds, you'll simulate another game.

roma avatar
Romapredicts NO

This seems like a correct simulation of the paradox, except for the probability of being picked from an infinite pool, which I guess should be 0 for any batch size. Not sure whether this is what causes the paradox somehow.

ShitakiIntaki avatar
Wamba Ivanhoepredicts YES

@roma I agree, you don't need to simulate any rounds except for the round the person of interest was selected. You certainly don't need to simulate whole games because the conditional does not condition on the game ending, it only conditions on selection.

If they were not selected during the game, then the game does not satisfy the conditional.

Primer avatar
Primerpredicts NO

@ShitakiIntaki If you leave out the part where you play in a game that will end in snake eyes, you end up simulating two dice throws 🤣

roma avatar
Romapredicts NO

@Primer That's why so many people convinced of YES. If you try to do everything as described in the paradox, including looking at it from the perspective of a single random player, you do end up simulating two dice throws :)

ShitakiIntaki avatar
Wamba Ivanhoepredicts YES

@roma @Primer at the time of selection, there is no reason to believe you were selected for the final round with credence greater than 1/36.

After the game has ended, it is true that if you don't know which round you were selected in, then your credence should be approaching 1/2 that you were selected for the final round and therefor believe that you lose 1/2 the time.

My reading of the conditional is that we are asking for credence at selection, not after the game has ended.

I belive the paradox is that we agree the game ends with probably of "1" so it shouldn't matter if you add a conditional that is, for most practical purposes, "guaranteed" to occur, however inclusion of this "gauranteed" event does matter since it is not independent of the selection event. (This dependence arises from the zero measure of being selected. I don't know that it is directly related but heuristicly I am think of it being like a Dirac delta function at infinity.)

Primer avatar
Primerpredicts NO

@roma Well... that's boring 😂

I really like the fact that after months, the paradox (and the people participating in this market) still provides interesting thoughts:

The Yes solution means that the doubling of players has absolutely no consequence. Your chances are the same as if there were just a single player in each round, or if you started with Graham's number of players and halfed the player count each round. 1/36 all the way. That was always implied by the Yes solution, but only now has entered my conscious mind.

Primer avatar
Primerpredicts NO

@ShitakiIntaki

After the game has ended, it is true that if you don't know which round you were selected in, then your credence should be approaching 1/2 that you were selected for the final round and therefor believe that you lose 1/2 the time.

My reading of the conditional is that we are asking for credence at selection, not after the game has ended.

Yeah, I pretty much agree. But that's just Sleeping Beauty, isn't it?

Credence of coin landing heads 1/2

=

Credence of dice snake eyes 1/36

Credence of observing heads 1/3

=

Credence of observing snake eyes 1/2

I took the paradox to ask how an agent should behave, or rather bet. That's 1/3 for Sleeping Beauty and ~1/2 here.

ShitakiIntaki avatar
Wamba Ivanhoe

@Primer for the reasons stated above, I think it matters when you ask them.

Both are rational answers to the same question asked at different times.