Snake Eyes Paradox

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

For an arbitrary player base of $B$ players, the maximum possible rounds that the game can run is then the whole number part of $Lo{g}_2\left(B\right)$, 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

$=\sum _{n=1}^m{D}_n$

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

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

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

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

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  $D_m+S_m=(1-p{)}^{m-1}$

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

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

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

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

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

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

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 $n$ 

$\sum _{n=1}^m{P}_n{D}_n=\sum _{n=1}^m{2}^{n-1}p(1-p{)}^{n-1}=p\times \frac{1-2^m(1-p{)}^m}{1-2(1-p)}$ 

 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 $n$.

$\left(\sum _{n=1}^m{D}_n{T}_n\right)+S_m{T}_m=\left(\sum _{n=1}^{m-1}{D}_n{T}_n\right)+(D_M+S_m)T_m=\left(\sum _{n=1}^{m-1}{D}_n{T}_n\right)+(1-p{)}^{m-1}{T}_m$

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

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

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

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

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

$\frac{E\left(Lose\right)}{E\left(Played\right)}=\frac{{\sum }_{n=1}^m{P}_n{D}_n}{\left({\sum }_{n=1}^m{D}_n{T}_n\right)+S_m{T}_m}=\frac{p\times \frac{1-2^m(1-p{)}^m}{1-2(1-p)}}{p(2\times \frac{1-2^{m-1}(1-p{)}^{m-1}}{1-2(1-p)}-\frac{1-(1-p{)}^{m-1}}{1-(1-p)})+(1-p{)}^{m-1}(2^m-1)}$

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

$P\left(Lose|Played\right)=\frac{P(Lose\cap Played)}{P\left(Played\right)}=\frac{P\left(Lose\right)}{P\left(Played\right)}=\frac{E\left(Lose\right)}{E\left(Played\right)}$

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 $p$.

$P\left(Lose|Played\right)=p$  $\forall$  $n\in N$   therefor the limit of our conditional probability is  $\underset{n\to \infty }{lim}p=p$

Because I'm not a real mathematician, I'm not going to find the actual limit, but just show that the limit is at least 50%.

Note that 1 + 2 + 4 + ... + 2^(n-1) = 2^n - 1.  Therefore, if we have a bunch of blue-eyed groups of size 1, 2, 4, ..., 2^(n-1), and one red-eyed group of size 2^n, then the overall fraction of snakes that are red-eyed is 2^n / (2^n + 2^n - 1), which, if we divide the numerator and denominator by 2^n, comes out to 1 / (2 - 1/(2^n)).  This is slightly above 1/2, and the limit as n -> ∞ is exactly 1/2.

It would be useful to have a summary (at the top or bottom) that this post is mostly about comparing SSA and SIA, and readers familiar with the differences can skim most of it.  

IMO, the key is to remember that probability is in the map (or the agent's head, if you prefer), not the territory.  Unless you're talking about God (who famously doesn't play dice, but kind of seems to), all probabilities are 1 or 0 - that is the universe you're in, or it isn't.  Your assignment of probability is based on your prediction of which universe you're in (ignoring logical uncertainty for this comment).  Which means you really need to ask about what future experience OF THE AGENT is being predicted by the probability calculation of the agent.

From the experiment-runner's perspective, half of players die, and any given player AFTER the experiment, without knowing what group the player is in, has 50% chance of death.  From a PLAYER's perspective, they have a 1/36 chance of death if the experiment is still open, and a 0% chance of death if it's finished and they're alive to answer the question.

I feel like this is one of the cases where you need to be very precise about your language, and be careful not to use an "analogous" problem which actually changes the situation.

Consider the first "bajillion dollars vs dying" variant. We know that right now, there's about 8B humans alive. What happens if the exponential increase exceed that number? We probably have to assume there's an infinite number of humans, fair enough.

What does it mean that "you've chosen to play"? This implies some intentionality, but due to the structure of the game, where the number of players is random, it's not really just up to you. 

NOTE: I just realized that the original wording is "you're chosen to play" rather than "you've chosen to play". Damn you, English. I will keep the three variants below, but this means that the right interpretation clearly points towards option B), but the analysis of various interpretations can explain why we even see this as a paradox.

A) One interpretation is "what is the probability that I died given that I played the game?", to which the answer is 0%, because if I died, I wouldn't be around to ask this question. 

B) Second interpretation is "Organizer told you there's a slot for you tomorrow in the next (or first) batch. What is the probability that you will die given that you are going to play the game?". Here the answer is pretty trivially 1/36. You don't need anthropics, counterfactual worlds, blue skies. You will roll a dice, and your survival will entirely depend on the outcome of that roll.

C) The potentially interesting interpretation, that I heard somewhere (possibly here) is: "You heard that your friend participated in this game.  Given this information, what is the probability that your friend died during the game?". The probability here will be about 50% -- we know that if N people in total participated, about N/2 people will have died.

Consider now the second variant with snakes and colors. Before the god starts his wicked game, do snakes exist? Or is he creating the snakes as he goes? The first sentence "I am a god, creating snakes." seems to imply that this is the process of how all snakes are created. This is important, because it messes with some interpretations. Another complication is that now, "losing" the roll no longer deletes you from existence, which similarly changes interpretations. Let's look at the three variants again.

A) "What is the probability you have red eyes given that you were created in this process?" -- here the answer will be ~50%, following the same global population argument as in variant C of the first variant. This is the interpretation you seem to be going with in your analysis, which is notably different than the interpretation that seems to be valid in the first variant.

B) If snakes are being created as you go with the batches, this no longer has a meaning. The snake can't reflect on what will happen to him if he's chosen to be created, because he doesn't exist.

C) "Some time after this process, you befriended a snake who's always wearing shades. You find out how he was created. Given this, what is the probability that he has red eyes?" -- the answer, following again the same global population argument, is ~50%

In summary, we need to be careful switching to a "less violent" equivalent, because it can often entirely change the problem.

Is it epistemically isomorphic to sleeping beuaty problem?

It depends.

If croupier choose the players, then players learn that they were chosen, then croupier roll the dice, then players either get bajillion dollars each or die, then (if not snake eyes) croupier choose next players and so on - answer is 1/36.

If croupier choose the players, then roll the dice, then (if not snake eyes) croupier choose next players and so on, and only when dice come up snake eyes players learn that they were chosen, and then last players die and all other players get bajillion dollars each - answer is about 1/2.

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

The statement above is only true if the stopping condition is that "If we get to the batch $B$ we will not roll dice but instead only make snakes with red eyes", or in other words $B$ must have been selected because it resulted in red eyes.

Where $B$ is selected as the stopping condition independent of the dice result, the fate of batch $B$ is still a dice roll so there exists a $B+1$ scenario. Any scenario stopping less than $B$ must have stopped because snakes with red eyes were made, however batch $B$ could result either in $2^{B-1}$ snakes with red eyes with probability $d=1/36$ or $2^{B-1}$ snakes with blue eyes with probability $(1-d)=35/36$.   The fact that snakes are more likely to be blue eyed in batch $B$ than they are to be red combined with the exponential weight of batch $B$ is not with out consequence.

Note that sequence of weights provided the table are the summation terms required to find expected number of snakes created.   A sum product of the series of the number of players in each scenario { 1, 3, 7, ... , $2^n-1$ } and the frequency of terminating at a given scenario due to death {$d(1-d{)}^{i-1}$}  for $i=1\to n$.   Note: however that the sum of the scenario frequencies is $d\times {\sum }_{i=1}^n(1-d{)}^{i-1}=1-(1-d{)}^n$.  This sum is less than $1$, so we need to recover the missing $(1-d{)}^n$ probability which comes from the scenario where all snakes have blue eyes, our $B+1$ scenario/term required to find the expected number of snakes.

Let’s consider a scenario where we skip the doubling. So, we have a long line of people, people play one at a time, 50% chance they win a prize, 50% they are eliminated, game ends after first elimination.

It’s quite clear that in expectation, more than 50% of people who played lose the game even though it’s a fair coin (we have a 50% chance of 100% of players losing). This will be explained momentarily.

Imagine the players as a team who score 1 point when they win and lose 1 point when eliminated, with the ability to stop at any time, but who have adopted this stopping strategy.

This strategy of stopping means you come out behind with 50% chance, even with 25% chance and ahead with 25% chance.

If we now look at the percent of wins vs. losses, we can see that we expect more losses than wins on average (though the calculation is more complicated).

The point is that you should only expect an average of half the coin flips to be heads when you adopt some kind of uniform stopping rule, such as always stopping after N rounds. Otherwise, you can pick a stopping rule that makes these kinds of trade-offs.

So you are more likely to lose than win. However, it allows you to maintain uncapped potential profit, whilst your losses are capped at 1.

We can note consider snake eyes from a team perspective. It’s similar, but more extreme. This time your team almost guaranteed to lose one point, though there is an infinitesimal chance that you score infinite points.

This sounds like free money from the perspective of the house, but it’s not, it’s just a martingale.

Are all snakes equally real (in proportion to their likelihood of existing)? I'm assuming yes in the post, but that leads to weirdness like undefined probabilities, which we'd prefer to avoid. Also, because there are infinite snakes, there would be a zero probability of being any particular snake, or (under SIA) being in any particular batch of snakes.

If we distribute realityfluid to the snakes in differing amounts, to avoid that, then we can probably avoid the undefined probabilities. The probability of having red eyes will depend on the formula for distributing realityfluid. Then instead of saying "the probability is undefined" we can say "the probability depends on this undefined formula".