You are a member of a countably infinite population of which every member of the population has been assigned a natural number.
In sequential order, an AI will ask each member of the population if they would like to play a game.
Requiring just one player in the inaugural round, and doubling the number of required players in each subsequent round, once the AI has recruited the appropriate number of players the AI will roll a pair of fair six sided dice which are each truly uniformly random such that there is precisely a 1/36 chance that the roll result will be snake eyes. If the result is snake eyes, the game ends and each player in the final round loses one dollar. If the result is not snake eyes each player in that round wins 1 dollar, and the AI will continue the game and resume recruiting players for the subsequent round of the game.
Today the AI asks if you want to play the game, risk losing a dollar for the chance to win a dollar. Assuming that you are a rational player and will always/ONLY play if you have a positive expected return and will pass if your expected return is zero or negative, do you agree to play?
Is this a paradox?
If you play, you only observe a single dice roll, so you should have a 35/36 chance of winning and only a 1/36 chance of losing.
But wait, each round the number of players doubles and has 1 more player than all the prior rounds combined, the final round being the largest of all the rounds, so theoretically, more likely than not, you will play in the final round and therefore you might expect to lose more often than not if you were Paradox?
So do you Play or Pass when asked if you would like to play the game?
1,000
3.00@Primer Ooh! I'd love to hear more of your thinking on this, and any implications for the original Snake Eyes. This seems like potentially a big deal if we're ending up at consensus on this version at least.
I really like my articulation in the comments here for why I'm voting PLAY. Does it match your own reasoning?
@dreev Sorry, I didn't give this that much thought:
I suppose every player is supposed to pretend to be a rational agent, so if everyone says play, play will be the rational choice. If I'm the only one saying play, there will only be that one round with me in it, and I'll have a decent chance of a winning dollar.
@Primer Making sense. So what's your PLAY/PASS answer as a function of the number of people who are playing? By construction, you say PLAY if there are no other people playing -- that's the one-shot case of winning with 35/36 probability and losing with 1/36 probability.
Does there exist a threshold beyond which your answer changes to PASS?
For me, I'm stuck on PLAY no matter what. (Even for the infinite case. Because the only way the question is coherent in the infinite case is by interpreting it as the limit of the finite cases.)
I believe that to answer this directly we need to pin down the probability distribution. You're assigned a natural number... randomly? For arbitrary n, what's the probability that n is your assigned number?
What we want to mean for the spirit of the question about your win/lose chances is that every natural number has equal probability. Which implies that any particular number has probability zero. So now the answer to the question is 0/0 -- undefined.
But then of course the standard thing to do with an answer like 0/0 is take a limit.
So now the thing to pin down is what happens in the finite version of this game where you might run out of people. Say you're one of a huge number of people, N. The game ends when a group loses or when there are no people left.
At this point we do the Bayesian math to get the win/lose probability in terms of N. Then we take the limit as N goes to infinity. The surprising answer is you win a dollar with probability 35/36 and lose with probability 1/36, same as the one-shot version.
(That's a relief because we never found a reason why the original "YES, who cares about the doubling groups, you still live or die on a single dice roll" argument might be wrong.)
In conclusion, the expected value of this game is +94 cents and you should PLAY.
I'm eager to understand where the NO/PASS bettors get off of that train.
@dreev I might be cavalier to assume that you can instantaneously assign a natural number to everyone (at random). Sounds an awful lot like the axiom of choice.
Great question! I like @DavidPennock's idea of tweaking it to "win $1, lose $2" to avoid the confusing boundary case where you're roughly indifferent.
(But also I'm not sure the gambling aspect is even salient. Fundamentally we have the same question as in the original Snake Eyes paradox: what's the probability of losing? If it's 1/36 then, yes, play; if it's closer to 1/2 then pass.)
Does this variant have a mathematical monkey wrench with the choosing process -- how there's no such thing as a uniform distribution over an infinite set? I'm thinking we just can't math this directly in the infinite case so we have to take a limit. Which means we have to pin down what happens in the finite case where we run out of people. And for that I don't think we have much choice: if we never roll snake eyes before running out of people then no one dies.
So I think we're back to the original Snake Eyes with the clarification about taking the limit of the finite game. Which means the answer is yes, play. You always have a 1/36 chance of losing no matter how big the pool of people and no matter how close to certain it is that the game ends with more losers than winners.
Intuitively, my latest understanding of how that can be true is that the "everyone survives" case gets less and less likely but given you were chosen it stays fairly likely. In the limit, I have a 0% chance of being chosen to play in this game. But if, somehow, I am, then the otherwise 0% chance of rolling non-snake-eyes forever becomes, conditionally speaking, possible again.
The chance of no one dying always factors into the calculation, even in the infinite case.
A mind-bending consequence of all that is what @DavidPennock calls the knife-edge argument. Suppose the pool is so big that I'm 99.99999% sure that the game will end in snake eyes. Or jump to the infinite case and say I'm 100% sure it will. My probability of losing is still 1/36, fine. But now if I condition on this thing I'm already certain of then suddenly my probability of losing jumps to ~1/2? Crazypants. I still feel uncertain about what it implies for the play/pass question.
Hmm, have we already computed Pr(only non-snake-eyes rolled | you are chosen) as a function of the number of rounds, n, and snake eyes probability p? I'm Bayesing it out now and getting (1-p)^n / Pr(chosen) which, referring to my original writeup (and assuming p<1/2 which Mathematica says matters somehow but that's fine for our purposes) works out to be 1-2p or 17/18 in the limit as n goes to infinity. Wild.
So it really is the case that rolling non-snake-eyes an infinite number of times in a row, conditional on being chosen to play, is likely. Specifically it happens with probability 17/18. I love this so much.
And I think we've finally resolved @DavidPennock's knife-edge curveball! No matter how sure I am unconditionally that the game will end in snake eyes, I believe with probability strictly between 1-2p and 1-p (i.e., 17/18 to 35/36 or 94% to 97%) that everyone survives โ no snake eyes ever โ conditional on me being chosen. So it makes perfect sense that if I actually condition on the game ending in snake eyes, even in the infinite case when it seemingly has to end in snake eyes, then that totally changes my probability of death. ๐ฅ
I think I'm finally fully confident that my answer for this poll is PLAY. The rough thinking is that I'm pretty sanguine that if I'm somehow chosen then we'll roll non-snake-eyes literally forever. All the math works out to perfectly comport with the original reasoning: You get one dice roll, it's fair, you only die with 1/36 probability no matter the group-choosing shenanigans!
Ok, maybe not fully confident until a few smart people are actually persuaded by this...
Ha, ironic update. I don't have to convince @ShitakiIntaki of this. Repeating an excerpt from my update in the original market:
Oops, wait, very excited but also very embarrassed to belatedly realize that @ShitakiIntaki already had all this in the YES writeup that we've been working on merging into a single document! @ShitakiIntaki has really been light years ahead of me on all this all along. Really grateful for all this.
@DavidPennock I agree.
@ShitakiIntaki Would you consider changing the odds? Else there also might be people answering "play" if they are indifferent, because "why not" (I'm aware you specified otherwise, but a loss of 2 would fit with intuition, whereas "you're a rational agent..." doesn't).
Great poll!
@Primer I already Stated that an indifferent gambler would pass rather than waste their time on an expected value of zero.
@DavidPennock if your credence is exactly 1/2 the ev is exactly zero, you could be indifferent but might as well save yourself the time it takes to play, but if you have a credence of 1/2 minus epsilon of winning you are approaching an expected value of zero from the negative side and should no longer be indifferent. No?
might as well save yourself the time it takes to play
And miss out the thrill of gambling while everyone else plays? I have infinite time, I'll play when EV is zero ๐