Snake Eyes Paradox

Martin Randall

22 Snake Eyes Paradox

11th Jun 2023

7 min read

22

Disclaimer: this is a personal blog post that I'm posting for my benefit and do not expect to be valuable to LessWrong readers in general. If you're reading this, consider not.

This post analyzes the "Snake Eyes Paradox" as specified by Daniel Reeves (2013). This is a variation on the "Shooting Room Paradox" that appears to have been first posed by John Leslie (1992). This article mostly attempts to dissolve the paradox by considering anthropics (SIA vs SSA) and the convergence of infinite series.

Problem statement

As asked by Daniel Reeves:

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?

The Shooting Room Paradox is similar. As asked by John Leslie, in "Time and the Anthropic Principle (1992):

Thrust into a room, you are assured that 90% of those who enter it will be shot. Panic! But you then learn that you will leave the room alive unless double-six is thrown, first time, with two dice. How is this compatible with the assurance that 90% will be shot? Successive batches of people thrust into the room are successively larger, exponentially, so that the forecast “90% will be shot” will be confirmed when double-six is eventually thrown. If knowing this, and knowing also that the dice falls would be utterly unpredictable even to a Laplacean demon who knew everything about the situation when the dice were thrown, then shouldn’t one’s panic vanish?

The Snake Eyes Paradox has a less extreme probability (90% will be shot vs 50% will be devoured by snakes) but it's the same problem.

Less death please

It can be hard to discuss death rationally, so I will discuss a less violent equivalent:

I am a deity, creating snakes from an unlimited number of disembodied souls. I make the snakes in batches, where each batch is twice the size of the last one: the first batch is one snake, the second batch is two snakes, etc. After gathering each batch, I roll two dice. If I get snake eyes, I give all the snakes in the batch red eyes and stop making snakes. Otherwise I give them all blue eyes and make another batch.
You are an eyeless snake. What is the probability that your eyes will be red?

This article	Reeves	Leslie's Shooting Room
Soul	Person not (yet) picked	Person not (yet) picked
Eyeless snake	Person in game	Person in room
Red-eyed snake	Person about to be devoured by snakes	Person about to be shot
Blue-eyed snake	Person about to get a bazillion dollars	Person about to leave room alive

Some variables for later mathematics:

d = P(rolling a two on two six-sided dice) = 1/36
N = number of batches of snakes created.

Anthropic Probability

The problem asks for the probability of a dice roll, from the perspective of a snake, but the number of snakes depends on what dice get rolled. This means I need to apply a theory of anthropic probability. This is awkward because anthropic probability is an unsolved philosophical problem.

Self-sampling assumption

Under the Self-Sampling Assumption (SSA), I first account for how likely each scenario (N=1, N=2, ...) is from the deity's perspective. Then I discard all scenarios with no snakes, which isn't relevant here, as all scenarios have snakes. Then, in each scenario, I am equally likely to be any of its snakes. So, for example, there is a 1/36 chance of N=1, so I am the only snake that exists, and having red eyes.

Under SSA we can therefore directly calculate the probability of having red eyes as the sum of this series, where N is the number of batches of snakes created:

P(red) = d + (2/3)d(1-d) + ... + (2^(N-1)/((2^N) - 1))d(1-d)^(N-1) + ...

This converges because the Nth term exponentially tends to zero. It looks like it might converge to 19/36. It is definitely >= 50%.

(This section could be improved with an exact answer)

What is it like to live in an SSA world?

In an SSA world an eyeless snake notices that the past looks pretty normal - the results of the dice rolls so far are within a typical Poisson distribution. But the future is surprising, because there is a 19/36 chance of rolling snake eyes, instead of 1/36 as the snake might expect. This is because if the deity rolls snake eyes on a batch, there are fewer snakes, and each snake is more real.

Self-indication assumption

Under the Self-Indication Assumption (SIA), a snake is more likely to observe worlds with more snakes, in proportion to how many snakes there are. So for the first few N the weightings are:

N	Weight
1	d
2	3d(1-d)
3	7d(1-d)²

As N increases, the weight increases exponentially. This diverges, so all of the probability goes to the case where N is infinite. This is not a well-behaved probability distribution, and we can't directly calculate the probability of red eyes.

What is it like to live in an SIA world?

In an SIA world, an eyeless snake notices that the past is fine-tuned for snakes. The deity has been rolling dice for what seems like forever. It is not possible to describe how many batches there have been in any finite length of time, in any notation. But the future unfolds as the snake expects, with a 1/36 chance of rolling snake eyes, and the creation process following a typical Poisson distribution from this point.

Trying to find a limit for SIA

I can't directly calculate the probability of red eyes under SIA. Instead, let's find a series of finite games and find the limit as the game gets large.

Conditional on the game stopping by batch B

First idea: condition the universe on cases where the deity happens to stop after B batches or less, and calculate the conditional probability of red eyes in that universe. Then take the limit as N tends to infinity. We'll make things even easier by just finding a minimum value for the probability.

For any B we have B possible scenarios:

S1: There is a single snake with red eyes. P(red) = 100%.
S2: There is one snake with blue eyes and two snakes with red eyes. P(red) = 66.7%.
S3: There are 3 snakes with blue eyes and four snakes with red eyes. P(red) = 57.1%.
...
S(R): There are 2^(B-1) - 1 snakes with blue eyes and 2^(B-1) snakes with red eyes. P(red) >= 50%.

Because all scenarios have P(red) >= 50%, the combined P(red) >= 50%. This holds for both SIA and SSA.

(This section could be improved with an exact answer)

Can we take the limit?

Under SSA anthropics, we can absolutely take the limit. It should end up giving the same answer that we (sort of) calculated directly earlier: 19/36.

Under SSI anthropics, we can't take the limit. For any finite B, the chance of the game stopping by batch B is zero. Therefore as we let B tend to infinity, we never improve our ability to

Purple-eyed snakes

See Daniel Reeves treatment of the problem for a comparison. For this limit we first address a different problem.

I am a god, creating snakes, as before. But after N batches I run out of souls. In that case I give all the snakes in the last batch purple eyes and stop making snakes.

The idea here is that we let N tend to infinity, and let the probability of having purple eyes tends to zero. The probability in that limit should then be the same as the probability in the infinite game we started with.

Under SSA this approach works great. P(red eyes) = 19/36. P(blue eyes) = 17/36. P(purple eyes) = 0%.

Under SIA we instead find that as N tends to infinity, the probability of purple eyes increases, to a limit of (I think) about 34/36. So the limit of the finite games does not tell us the probability in the infinite game, where there are no purple-eyed snakes.

(This section could be improved with an exact answer)

Conclusion

Under SSA anthropics, we have shown that there is a defined probability of ending up with red eyes, which is greater than 50%, and seems to be about 19/36.

Under SIA anthropics we were not able to find a probability using any process. It's possible that more advanced techniques will be able to find a probability. I think the most plausible answer is that it is undefined.

It is possible for a snake to observe the universe and decide if it is SSA or SIA.

Resolution of a Manifold Market

The manifold market has a FAQ that specifies that we can use the "purple-eyed snakes" limiting process above, saying:

We can cap it and say that if [there are no red snakes] after N rounds then the game ends and [all snakes have blue eyes]. We just need to then find the limit as N goes to infinity, in which case the probability [the game ends with no red snakes] goes to zero.

As discussed above, under SSA, the probability goes to zero as N goes to infinity. Under SIA it does not. This means the question is implicitly assuming SSA anthropics. As shown above, the limit under SSA anthropics is >= 50%, probably 19/36. Therefore the market should resolve NO. Further investigation may provide a more accurate answer but is not needed to resolve the market.

References

https://halfofknowing.com/shooting.html
https://dreeves.github.io/snakeeyes/snakeeyes.pdf
https://manifold.markets/dreev/is-the-probability-of-dying-in-the
Thanks to helpful commenters here and on Manifold.

AnthropicsParadoxesWorld Modeling

Personal Blog

22

New Comment

25 comments, sorted by

top scoring

Click to highlight new comments since: Today at 5:56 PM

[-]Wamba-Ivanhoe1y101

Calculating the odds of dying when playing the snake-eyes game with a player base of arbitrary size.

For an arbitrary player base of players, the maximum possible rounds that the game can run is then the whole number part of $L o g_{2} (B)$ , we will denote the maximum possible number of rounds as $m$ .

We denote the probability of snake-eyes as $p$ . In the case of the Daniel Reeves market $p = 1 / 36$ .

Let $D_{n} =$ the probability density of the game ending in snake-eyes on round n then $D_{n} = p (1 - p)^{n - 1}$ .

The sum of the probability density of the games which end with snake eyes is

$= m \sum n = 1 D_{n}$

$= m \sum n = 1 p (1 - p)^{n - 1}$

$= p m \sum n = 1 (1 - p)^{n - 1}$

$= p \frac{1 - (1 - p)^{m}}{1 - (1 - p)}$

$= 1 - (1 - p)^{m}$

The sum of the probability densities of the games ending in snake-eyes is less than $1$ which means that the rounds ending in snake-eyes does not cover the full probability space.

Since there are finitely many dice rolls possible for our population there is always the possibility that the game ends and everyone won. This would require not rolling snake-eyes $m$ times and has a probability density of $S_{m} = (1 - p)^{m}$ which is exactly the term we need to sum to $1$ giving us full coverage of the probability space.

Note: That $D_{m} + S_{m}$ are the probability densities of the final round ending in a loss or ending in a win and must there for equal $(1 - p)^{m - 1}$ the probability of getting to round $m$ by the players winning the prior $m - 1$ rounds.

Here we verify that $D_{m} + S_{m} = (1 - p)^{m - 1}$

$D_{m} = p (1 - p)^{m - 1}$

$S_{m} = (1 - p)^{m}$

$D_{m} + S_{m} = p (1 - p)^{m - 1} + (1 - p)^{m}$

$= p (1 - p)^{m - 1} + (1 - p) (1 - p)^{m - 1}$

$= [p + (1 - p)] (1 - p)^{m - 1}$

$= (1 - p)^{m - 1}$

Denote the number of players in each round $n$ as $P_{n} = 2^{n - 1}$

Denote the total number of players chosen in a game that ends at round $n$ as $T_{n} = 2^{n} - 1$

The probability of losing given that you are chosen to play can be computed by dividing the expected number of players who lost (red-eyed snakes) by the expected total number of players chosen.

First we will develop the numerator of our desired conditional probability. What is the expected number of players who lose in a game, given that cannot go beyond $m$ rounds? This will be the sumproduct of the series of players in each round $n$ and the series of probability densities of the game ending in round $n$

$m \sum n = 1 P_{n} D_{n} = m \sum n = 1 2^{n - 1} p (1 - p)^{n - 1} = p \times \frac{1 - 2^{m} (1 - p)^{m}}{1 - 2 (1 - p)}$

Now we will develop the numerator of our desired conditional probability. What is the expected total number of people chosen to play? This will be the sumproduct of the series of total players in a game ending in round $n$ and the series of probability densities of the game ending in round $n$ .

$(m \sum n = 1 D_{n} T_{n}) + S_{m} T_{m} = (m - 1 \sum n = 1 D_{n} T_{n}) + (D_{M} + S_{m}) T_{m} = (m - 1 \sum n = 1 D_{n} T_{n}) + (1 - p)^{m - 1} T_{m}$

$= (m - 1 \sum n = 1 p (1 - p)^{n - 1} (2^{n} - 1)) + (1 - p)^{m - 1} (2^{m} - 1)$

$= p (m - 1 \sum n = 1 2^{n} (1 - p)^{n - 1} - m - 1 \sum n = 1 (1 - p)^{n - 1}) + (1 - p)^{m - 1} (2^{m} - 1)$

$= p (2 m - 1 \sum n = 1 2^{n - 1} (1 - p)^{n - 1} - m - 1 \sum n = 1 (1 - p)^{n - 1}) + (1 - p)^{m - 1} (2^{m} - 1)$

$= p (2 \times \frac{1 - 2^{m - 1} (1 - p)^{m - 1}}{1 - 2 (1 - p)} - \frac{1 - (1 - p)^{m - 1}}{1 - (1 - p)}) + (1 - p)^{m - 1} (2^{m} - 1)$

Combining the numerator and denominator into the fraction we seek we now have

$\frac{E (L o s e)}{E (P l a y e d)} = \frac{\sum_{n = 1}^{m} P_{n} D_{n}}{(\sum_{n = 1}^{m} D_{n} T_{n}) + S_{m} T_{m}} = \frac{p \times \frac{1 - 2^{m} (1 - p)^{m}}{1 - 2 (1 - p)}}{p (2 \times \frac{1 - 2^{m - 1} (1 - p)^{m - 1}}{1 - 2 (1 - p)} - \frac{1 - (1 - p)^{m - 1}}{1 - (1 - p)}) + (1 - p)^{m - 1} (2^{m} - 1)}$

Note: that since it is impossible to lose unless you play the conditional probability is equal to the above ratio.

$P (L o s e | P l a y e d) = \frac{P (L o s e \cap P l a y e d)}{P (P l a y e d)} = \frac{P (L o s e)}{P (P l a y e d)} = \frac{E (L o s e)}{E (P l a y e d)}$

And then we thank WolframAlpha for taking care of all the algebraic operations required to simply this fraction, see alternate form at : https://www.wolframalpha.com/input?i2d=true&i=Divide%5Bp*Divide%5B1-Power%5B2%2Cm%5DPower%5B%5C%2840%291-p%5C%2841%29%2Cm%5D%2C1-2%5C%2840%291-p%5C%2841%29%5D%2Cp%5C%2840%292*Divide%5B1-Power%5B2%2Cm-1%5DPower%5B%5C%2840%291-p%5C%2841%29%2Cm-1%5D%2C1-2%5C%2840%291-p%5C%2841%29%5D-Divide%5B1-Power%5B%5C%2840%291-p%5C%2841%29%2Cm-1%5D%2C1-%5C%2840%291-p%5C%2841%29%5D%5C%2841%29%2BPower%5B%5C%2840%291-p%5C%2841%29%2Cm-1%5D%5C%2840%29Power%5B2%2Cm%5D-1%5C%2841%29%5D

We find that the conditional probability of losing given that you are chosen to play from a arbitrary population of players is simply $p$ . The size of the population was arbitrarily selected from the natural numbers so as the population goes to infinity the limit will also be $p$ .

$P (L o s e | P l a y e d) = p$ $\forall$ $n \in N$ therefor the limit of our conditional probability is $lim n \to \infty p = p$

[-]Primer1y21

"The sum of the probability densities of the games ending in snake-eyes is less than 1 which means that the rounds ending in snake-eyes does not cover the full probability space."

This is contradicted by the problem statement: "At some point one of those groups will be devoured by snakes", so there seems to be some error mapping the paradox to the math.

[-]Wamba-Ivanhoe1y63

I posit that the supposition that "At some point one of those groups will be devoured by snakes" is erroneous. There exists a non-zero chance that the game goes on forever and infinitely many people win.

The issue is that the quoted supposition collapses the probability field to only those infinitely many universes where the game stops, but there is this one out of infinitely many universes where the game never stops and it has infinitely many winners so we end up with residual term of . We cannot assume this term is zero just because of the $\infty$ in the denominator and disregard this universe.

[-]dreeves1y10

"At some point one of those groups will be devoured by snakes" is erroneous

I wouldn't say erroneous but I've added this clarification to the original question:

"At some point one of those groups will be devoured by snakes and then I stop" has an implicit "unless I roll 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".

It sounds contradictory but "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.

So I think it's correct to say "if I flip a coin long enough, at some point I'll get heads" even though we understand that "all tails forever" is one of the infinitely many possible sequences of coin flips.

[-]Primer1y10

I think that's what makes this a paradox.

[-]Wamba-Ivanhoe1y30

Going back to each of the finite cases, we can condition the finite case by the population that can support up to m rounds. iteration by the population size presupposes nothing about the game state and we can construct the Bayes Probability table for such games.

For a population that supports at most $m$ rounds of play the probability that a player will be in any given round $n <= m$ is $\frac{2^{n}}{M}$ and the sum of the probabilities that a player is in round n from $n = 1 \to m$ is $m \sum n = 1 \frac{2^{n - 1}}{M} = \frac{2^{m} - 1}{M}$ ; We can let $M = 2^{m} - 1$ because any additional population will not be sufficient to support an $m + 1$ round until the population reaches $2^{m + 1} - 1$ , which is just trading $m$ for $m + 1$ where we ultimately will take the limit anyhow.

The horizontal axis of the Bays Probability table now looks like this

$[\begin{matrix} \frac{1}{M} & \frac{2}{M} & \frac{4}{M} & \frac{8}{M} & . . . & \frac{2^{m - 1}}{M} \end{matrix}]$

The vertical axis of the Bays Probability table we can independently look at the odds the game ends at round n for n<=m. This can be due to snake eyes or it can be due to reaching round m with out rolling snake eyes. For the rounds $n = 1 \to m$ where snake eyes were rolled the probability of the game ending on round $n$ is $p * (1 - p)^{n - 1}$ and the probability that a reaches round $m$ with out ever rolling snake eyes is $(1 - p)^{m}$ . The sum of all of these possible end states in a game that has at most finite m rounds is $(1 - p)^{m} + \sum_{n = 1}^{m} p (1 - p)^{n - 1}$ which equals =1

So we have m+1 rows for the horizontal axis

$⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} p p (1 - p) p (1 - p)^{2} p (1 - p)^{3} ⋮ p (1 - p)^{m} (1 - p)^{m} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

So the Bayes Probability Table starts to look like this in general.

$⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} [\begin{matrix} \frac{1}{M} & \frac{2}{M} & \frac{4}{M} & \frac{8}{M} & . . . & \frac{2^{m - 1}}{M} \end{matrix}] ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} p p (1 - p) p (1 - p)^{2} p (1 - p)^{3} ⋮ p (1 - p)^{m} (1 - p)^{m} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ & ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{p}{M} & \frac{2 p}{M} & \frac{4 p}{M} & \frac{8 p}{M} & . . . & \frac{p * 2^{m - 1}}{M} \frac{p (1 - p)}{M} & \frac{2 p (1 - p)}{M} & \frac{4 p (1 - p)}{M} & \frac{8 p (1 - p)}{M} & . . . & \frac{p (1 - p) * 2^{m - 1}}{M} \frac{p (1 - p)^{2}}{M} & \frac{2 p (1 - p)^{2}}{M} & \frac{4 p (1 - p)^{2}}{M} & \frac{8 p (1 - p)^{2}}{M} & . . . & \frac{p (1 - p)^{2} * 2^{m - 1}}{M} \frac{p (1 - p)^{3}}{M} & \frac{2 p (1 - p)^{3}}{M} & \frac{4 p (1 - p)^{3}}{M} & \frac{8 p (1 - p)^{3}}{M} & . . . & \frac{p (1 - p)^{3} * 2^{m - 1}}{M} ⋮ & ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \frac{p (1 - p)^{m - 1}}{M} & \frac{2 p (1 - p)^{m - 1}}{M} & \frac{4 p (1 - p)^{m - 1}}{M} & \frac{8 p (1 - p)^{m - 1}}{M} & . . . & \frac{p (1 - p)^{m - 1} * 2^{m - 1}}{M} \frac{(1 - p)^{m}}{M} & \frac{2 (1 - p)^{m}}{M} & \frac{4 (1 - p)^{m}}{M} & \frac{8 (1 - p)^{m}}{M} & . . . & \frac{(1 - p)^{m} * 2^{m - 1}}{M} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

The total probability of losing is equal to the sum of the diagonal where i=j

$\frac{p}{M} + \frac{2 p (1 - p)}{M} + \frac{4 p (1 - p)^{2}}{M} + \frac{8 p (1 - p)^{3}}{M} + . . . + \frac{p (1 - p)^{m - 1} * 2^{m - 1}}{M} = \frac{p}{M} m \sum i = 1 [(1 - p)^{i - 1} * 2^{i - 1}]$

Total probability of being chosen is the sum of the diagonal and all the cells below the diagonal.

$m \sum i = 1 [\frac{2^{i - 1} * (1 - p)^{m}}{M}] + m \sum i = 1 i \sum j = 1 [\frac{2^{j - 1} * p (1 - p)^{i - 1}}{M}]$

$= \frac{(2^{m} - 1) (1 - p)^{m}}{M} + \frac{p}{M} m \sum i = 1 [(2^{i} - 1) (1 - p)^{i - 1}]$

So the conditional probability of losing given that you have been selected is

$\frac{\frac{p}{M} \sum_{i = 1}^{m} [(1 - p)^{i - 1} * 2^{i - 1}]}{\frac{(2^{m} - 1) (1 - p)^{m}}{M} + \frac{p}{M} \sum_{i = 1}^{m} [(2^{i} - 1) (1 - p)^{i - 1}]}$

$= \frac{p \sum_{i = 1}^{m} [(1 - p)^{i - 1} * 2^{i - 1}]}{(2^{m} - 1) (1 - p)^{m} + p \sum_{i = 1}^{m} [(2^{i} - 1) (1 -}$