You wake up feeling dizzy. There's a goat licking your toes. On your nightstand: a plane ticket to Damascus next to a coin showing heads. You can't remember the day, so you pick up your phone. It's Tuesday. You notice a new app: "Endless Snake". It's connected to your bank account. You open it.
Hi and welcome to Endless Snake by Nokia™
Here's how the game works: We have this Excel-file with two columns. The first contains all those countably infinite natural numbers, starting with 1., 2, 3, 4, ... We'll explain the second column later.
We'll start with line 1 and throw two fair and random dice and you'll get a dollar. Next we'll double the number of lines, thus take the next 2 lines, throw the dice and you'll get 2 dollars, then 4 lines and 4 dollars, and so on.
Except if the dice show snake eyes, then we'll take a dollar for each line in that round and start the whole procedure again (so after snake eyes is rolled the game starts over with a 1 line, round 1, as though starting anew) with the next line, then the 2 after that, then 4, 8, and so on until the next snake eyes and we repeat until infinity.
Now about that second column: If you click PASS below, we'll leave it empty, you can't play and we're sad to lose you as a customer. If you click PLAY, our intern will copy & paste PLAY into every line and you'll play!
Fine print: By clicking PLAY you agree to our terms of service. For each occurence of PLAY, you will pay a small fee of 10 cents which will be automatically taken out of your account as soon as the respective lines are up for their dice throw.
Do you click PLAY or PASS?
Argument for PLAY:
As this is just a repeated version of
/dreev/is-the-probability-of-dying-in-the
with precommitment, for each line the probability of snake eyes is just 1/36, so you'll get rich
It's possible, although with probability 0, that snake eyes will never happen
Argument for PASS: :
Due to the doubling of the number of lines, you'll lose more on snake eyes than you gained in the combined previous rounds (after the last snake eyes) and that happens infinitely often
It's possible, although with probability 0, that all dice land snake eyes
1,000
3.00Each iteration of the game (i.e., playing up until you roll snake eyes) will lose you at least $1.10 with probability 1. Playing the game forever does mean you'll always have times when your balance is positive, but they'll become rarer as time goes on, and introducing that kind of volatility in your finances is unwise anyway.
If the amount of money you got was controlled by any particular cell (e.g., you get $1 whenever Cell number 567 is played and rolls not-snake-eyes, lose $1 when it rolls snake eyes, and there's no $0.10 fee for any of the other cells), then it would be beneficial to play the game: In the long run, the amount of money you win would approach infinity, with probability 1. But that's not what this game asks. In this game, your payout on each iteration is the sum of the payouts from all cells, not the payout from any individual cell. While the latter is positive expected value, the former is negative expected value.
@PlasmaBallin Would your answer be different if there was a bug and the game stopped after the first occurance of snake eyes?
@ShitakiIntaki I think the second one is crazy because then the game doesn't even converge, your bank balance is oscillating between negative and positive, like Schrödinger's cat.
After snake eyes is rolled the game starts over with a 1 line, round 1, as though starting anew
Correct! Hope you don't mind that I put this in the description. Needs a whole rework anyways sometime in the future.
@Primer So this version of the game seems to fall in to the Balls-and-Urns paradox since you are the only player and you play all lines, (the ten cent fee making it highway robbery 😭). https://en.wikipedia.org/wiki/Ross%E2%80%93Littlewood_paradox
Not to be confused with the other version of end less snake that keeps recruiting new and distinct players who will each only play a single round. @dreev
@ShitakiIntaki That Ross-Littlewood paradox is amazing. We should do a market on it next ;)
The way I see this Endless Snake: If I would PASS on the opportunity to play all lines, and the probability of snake eyes is the same for each line, then I should also PASS when I get the opportunity to play a single line only. And vice versa if I win 35/36 as one single player in a single round, I should want to play all the lines.
@ShitakiIntaki Ah, thanks, yeah, it's probably just @Lorxus's Infinitely Repeated Snake Eyes that I do still want to play!
Review: Line up an infinite queue of people, still doubling till you hit snake eyes, and every time snake eyes is rolled, kill that group and then just keep going with the next person in line. There's no infinitesimal chance of actually being chosen this time. You're in the queue so you'll definitely be chosen.
The one-fair-roll argument still seems correct. Your probability of death is the probability of snake eyes when the dice get to you. One independent dice roll. What about the frequency argument? We try to analyze this in our cross-exam responses but I'm not sure how convincing it is. The dead/chosen ratio is ∞/∞ which could be anything so why not 1/2?
This version is nonsensical. At the end, with probability 1, you have to pay countably infinite number of dollars, and you receive countably infinite number of dollars (whose frequency is expected to be 35 times higher), but this just doesn't make sense.
If this stops after a finite number of rolls (but unknown to me), then I'd be very happy to take this.
@FlorisvanDoorn Thanks for your input! I was trying to get rid of some ambiguities in Snake Eyes. Sorry if I escalated with the infinities. I'm open to ideas to make this more approachable.
For argument's sake, we could just stop at Graham's number rolls of snake eyes or TREE(3) total dice rolls, whichever comes first.
@ShitakiIntaki Thanks for your opinion!
I'm trying to find out where the paradoxa differ. Does the argument
If no snake eyes is thrown ever (with probability 0 but not impossible), number of affected players is infinite, we can show in the limit it's 1/36
not apply here? If so: Why not? I was trying to leave that part intact.
@Primer the fact that snake eyes has the definite end tied up to a loss, and your lottery has not, is already the difference. You created a more complicated case.
@KongoLandwalker But the 1/36 argument - infinitely many non-snake-eyes - should still be possible here, shouldn't it? Maybe somehow it's no longer the dominant case here, but I want to understand if, why and when that argument no longer applies.
@Primer the heuristic that it is incredibly unlikely that you are chosen to play arises because the game ended before it got to you, infact a majority of the finite games end before selecting you to play.
This version of the game you are not only not unlikely to play but you play every single round.
If you observe only a single roll you are not likely to observe the Snake Eyes roll, but if you play all the rounds you should expect to see snake eyes get rolled.
the heuristic that it is incredibly unlikely that you are chosen to play arises because the game ended before it got to you, infact a majority of the finite games end before selecting you to play.
Hmm... ok, so in this view we're not one of the players in Snake Eyes (SE), but we're just in the pool of potential players, and then we get the news that it's our turn with the dice? And then we're asked the question if we want to play? Depending on who and when we are when we're trying to calculate our odds, 1/36 might make more or less sense.
I remember the discussion in SE going there (Anthropics, as well as how to interpret "given you are chosen to play"), but I don't think this has been unambiguously clarified in SE. This seems to be a crux, I'll call it "crux: given".
Would your answer here be different if the game stopped after the first time the dice show snake eyes?
Would your answer in SE be different if you wouldn't play a single player in one round, but as every player? Like suppose the players in SE don't know whether they should play or not, so they all ask you, and you can only give the same Yes or No to all of them.
@KongoLandwalker and to keep that in line with the original problem there would be some mechanism, which only allows you to exit the game immediately after Snake Eyes. Because the original game can only finish at those moments.
If that is not done, the the series are divergent and you cannot expect neither positive into infinity nor negative balance.
The current poll state is just an additional level of complication.
@KongoLandwalker Maybe. I'm not sure. It gets rid of
the argument of having a two-round game not ending in snake eyes
the problem about conditioning on being chosen
Anthropics
the possibility to chose a controversial game state as a starting point for a limiting process
Explanation of payout:
Imagine regular old Snake Eyes, but you win $0.90 instead of a bajillion and you lose $1.10 instead of dying. And you play all rounds yourself and after snake eyes is thrown it starts again and repeats forever. You need to commit beforehand whether to play all rounds, or none at all.
@KongoLandwalker Put it another way: In Snake Eyes, team Yes starts with a round probably not ending in snake eyes and takes the limit. The natural starting point here is taking a whole Snake Eyes game which ended in snake eyes and take the limit.
@Primer do i lose 1.1 multiplied by something? Or just 1.1 per snake eyes?
I think you wanted to multiply the punishment by the analogue of "doubling" value.
Some series do not converge, so the multiplicators have to be clearly defined.
@KongoLandwalker Yeah, you win or lose per line. In first round you play for 0.9/1.1, second round for 1.8/2.2, third round 3.6/4.4,next 7.2/8.8, etc.
@KongoLandwalker Doesn't
Except if the dice show snake eyes, then we'll take a dollar for each line in that round
cover that?