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I'm ready to resolve this as NO. I confused and reconfused myself so many times trying to get clarity in my own head about this and, if the answer converges to ~1/2 here, what exactly is the difference between this and the original? Huge thanks to commenters and others outside of Manifold for helping me get there. Here's my current understanding:

In this version there's no conditional probability. In the original Snake Eyes paradox we take the perspective of one of the balls in the urn. We want to know the probability that it ends up deflated given that it ends up outside the urn. But ending up outside the urn is strong Bayesian evidence that we exhausted the urn without ever rolling snake eyes and deflating any balls.

(Specifically, and very surprisingly to me, the probability of never rolling snake eyes at all, given that an arbitrary pre-identified ball was chosen, is 17/18 -- even in the limit as the number of balls in the urn goes to infinity. The unconditional probability of never rolling snake eyes of course goes to zero.)

Back to this version, there's no such conditioning or pre-identifying a ball. We just have a random variable defined as follows:

1. Start with n balls in the urn

2. Pick a group of them, initially 1, doubling each time

3. Deflate them with probability 1/36

4. If deflated, leave them outside the urn and stop

5. If not deflated, leave them outside the urn and go to step 2

6. Also stop if you run out of balls

When that process terminates, pick a ball at random from outside the urn. The random variable is 1 if the ball is deflated and 0 if inflated.

What is the expectation of that random variable?

We can simulate that for finite n. For the n=3 case we have 3 possible states when the process ends:

1. We picked 1 ball and deflated it

2. We picked 1 ball, left it inflated, picked 2 balls, and deflated them

3. We picked 1 ball, left it inflated, picked 2 balls, and left them inflated

Case 1 has 100% of the balls outside the urn deflated and happens with probability 1/36.

Case 2 has 2/3 of the balls outside the urn deflated and happens with probability 35/36*1/36.

Case 3 has 0% of the balls outside the urn deflated and happens with probability 35/36*35/36.

As n increases, that last case, where all the balls end up outside the urn and inflated, happens with smaller and smaller probability while including a more and more astronomical number of balls when it does happen. But for the above process, it doesn't matter how astronomical a number of balls there are. We just pick one of them and get an inflated ball. And in the limit that just doesn't happen.

In the n=3 case we can multiply out those probabilities along with the probability of getting a deflated ball in each case -- 1, 2/3, and 0 -- and we get 4.6% which is a bit more than 1/36. In the limit this converges to around 52%.

New fun question: I don't think we need anthropics for the original Snake Eyes but is that a useful way to see the difference between this and the original? In the original you take the perspective of one of the balls and condition on you having been chosen. Which, in some variants, is like conditioning on your own existence. I'm still thinking about this!

are we resolving no?

@jacksonpolack I said I'd defer to @MartinRandall but I believe that his NO in the original Snake Eyes implies a NO here, so if no one is arguing for YES here, then NO should be a safe resolution.

I keep confusing myself about this and want to run my own simulation to be totally sure.

Relatedly, I noticed a new ambiguity in this version: If n isn't one less than a power of two then you may have a non-empty urn but too few balls for the next round. The given algorithm isn't clear about what to do in that case -- take a smaller final group, or end the game? I'm fairly sure this isn't a real monkey wrench (I think @ShitakiIntaki even proved it) and that we can just stipulate that n is in fact one less than a power of two.

@dreev I concur with ~50%. I would also accept Undefined (eg, from an ultrafinitist). Both resolve the market NO.

You should probably have these resolve to @Lorxus 's opinion or something, I'm not a mathematician.

Could I pay someone mana to simulate this for n=1, n=2, ... up to, say, n=1000 or however far seems reasonable to be sure what it's converging to? If not (or maybe regardless, for replication reasons) I'll plan to do so myself. Sanity check: for n=1 we know we should get 1/36.

@dreev Q1 does it have to be simulated rather than just calculated? If so, how many simulations required for each value of n? Perhaps both calculated and simulate a few different n values 50 times as a check is better?

Q2 is it better to do it for n=1,2,3,4,7,8,15,16,31,32, ... 1048575, 1048576

Graph of convergence from rounds 1-200 (not simulated, done mathematically)

i made a simulator, i'll graph it now

btw the number of iterations of each game scales with log2 n, not n, so n can be a lot bigger than 1000 right

the x axis is log10(n) and the y axis is probability

@jacksonpolack @DeadRhino @ChristopherRandles Thanks, y'all! Yeah, I just meant simulating as an extra check on the math so we can be confident about the resolution here. Do we have a closed form in terms of n and p?

@dreev

I think this is a closed form

is this resolving

>We stop when we roll snake eyes and deflate the last batch of balls, or the urn is empty, whichever happens first.

The order appears to be snake eyes then deflation then selection of a random ball.

With n=7 i.e. 7 balls in the urn, we could have 1 out of 1 deflated or 2 out of 3 deflated or 4 out of 7 deflated (these are all over 0.5) or 0 out of 7 but this possibility of less than 0.5 gets very small with large n.

With n=8 we again have possibilities of 1 out of 1 deflated or 2 out of 3 deflated or 4 out of 7 deflated but we also have 1 out of 8 deflated and well as the 0 out of 8.

This is going to cause our proportion deflated to oscillate between being high for n=2^x-1 and relatively low for n=2^x+y where x is high and y is small (both x and y +ve integers). The wavelength of the oscillations is going to get longer. So is our function going to tend to a limit?

I think it will tend to a limit:

With 1024 balls there are 9 possibilities of over 0.5 1 possibility of 1 in 1024 and 1 possibility of 0 giving > 0.41{wrong}

with 1048576 balls there are 19 possibilities over 0.5 1 possibility of 1 in 1048576 and 1 possibility of 0 giving a value > 0.45{wrong}

So the oscillations do get smaller, after 1048576 the range is between 0.45{wrong} and 0.523. This clearly isn't heading to 1/36.

Original snake eyes problem was "in anticipation" not after game had ended like here. The two are different.

do we fix n, calculate, and take the limit, or are we saying infinitely many balls in urn? basically is this conditional on game ending or not?

In the limit as n goes to infinity -- i.e., an urn with an unlimited number of balls -- what is the probability that that ball is deflated?

you fix n, calculate, and then take the limit. which is what 'given n ... then lim as n goes to infinity' means.

@jacksonpolack Urn w/ large finite number of balls is very different from urn w/ infinite number of balls

does this resolve at close?

I have 2 questions

1. "Now randomly choose a ball from outside the urn."

   1. If we are allowing balls to be outside the urn, what is the point of the urn in the first place?

2. "what is the probability that that ball is deflated?"

   1. Isn't this just 0?

@DeadRhino surely we have to say outside the urn to ensure we are not selecting from in the urn which still has infinite balls in it.

It is balls outside the urn that are deflated, I think, so the chance is ~0.5 as we are looking at an ended game from after it has finished

@ChristopherRandles I made a mistake, I didn't realize the ball was chosen after the game was played.

The argument for Yes asks us to do Anthropic reasoning by assuming the reference class of the balls...

Is this different from the original snake eyes paradox?

To me it seems like it isn't! We'll discuss in the comments.

I tend to agree that this is similar to Snake Eyes.

I'm so confused you see this the same way. Would have lost all my mana in the market "Will Daniel agree this is analogous to Snake Eyes?".

It's certainly similar, but I'm pretty confident that when you write out the math you end up with >1/2, does anyone disagree?