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 $M$ 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 $\sum _{n=1}^m\frac{2^{n-1}}{M}=\frac{2^m-1}{M}$; 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

$\left[\begin{array}{cccccc}\frac{1}{M}& \frac{2}{M}& \frac{4}{M}& \frac{8}{M}& ...& \frac{2^{m-1}}{M}\end{array}\right]$

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 $p\ast (1-p{)}^{n-1}$ 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

$\left[\begin{array}{c}p\\ p(1-p)\\ p(1-p{)}^2\\ p(1-p{)}^3\\ ⋮\\ p(1-p{)}^m\\ (1-p{)}^m\end{array}\right]$

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

$\left[\begin{array}{cc}& \left[\begin{array}{cccccc}\frac{1}{M}& \frac{2}{M}& \frac{4}{M}& \frac{8}{M}& ...& \frac{2^{m-1}}{M}\end{array}\right]\\ \left[\begin{array}{c}p\\ p(1-p)\\ p(1-p{)}^2\\ p(1-p{)}^3\\ ⋮\\ p(1-p{)}^m\\ (1-p{)}^m\end{array}\right]& \left[\begin{array}{cccccc}\frac{p}{M}& \frac{2p}{M}& \frac{4p}{M}& \frac{8p}{M}& ...& \frac{p\ast 2^{m-1}}{M}\\ \frac{p(1-p)}{M}& \frac{2p(1-p)}{M}& \frac{4p(1-p)}{M}& \frac{8p(1-p)}{M}& ...& \frac{p(1-p)\ast 2^{m-1}}{M}\\ \frac{p(1-p{)}^2}{M}& \frac{2p(1-p{)}^2}{M}& \frac{4p(1-p{)}^2}{M}& \frac{8p(1-p{)}^2}{M}& ...& \frac{p(1-p{)}^2\ast 2^{m-1}}{M}\\ \frac{p(1-p{)}^3}{M}& \frac{2p(1-p{)}^3}{M}& \frac{4p(1-p{)}^3}{M}& \frac{8p(1-p{)}^3}{M}& ...& \frac{p(1-p{)}^3\ast 2^{m-1}}{M}\\ ⋮& ⋮& ⋮& ⋮& \ddots & ⋮\\ \frac{p(1-p{)}^{m-1}}{M}& \frac{2p(1-p{)}^{m-1}}{M}& \frac{4p(1-p{)}^{m-1}}{M}& \frac{8p(1-p{)}^{m-1}}{M}& ...& \frac{p(1-p{)}^{m-1}\ast 2^{m-1}}{M}\\ \frac{(1-p{)}^m}{M}& \frac{2(1-p{)}^m}{M}& \frac{4(1-p{)}^m}{M}& \frac{8(1-p{)}^m}{M}& ...& \frac{(1-p{)}^m\ast 2^{m-1}}{M}\end{array}\right]\end{array}\right]$

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

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

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

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

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

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

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

$=\frac{p{\sum }_{i=1}^m\left[\right(1-p{)}^{i-1}\ast 2^{i-1}]}{(2^m-1)(1-p{)}^m+p{\sum }_{i=1}^m\left[\right(2^i-1\left)\right(1-p{)}^{i-1}]}$$=p$

More over with the full Bayes Probability table we can find other conditional probabilities at infinity by taking the limit as the population grows to allow bigger and bigger maximum rounds of m.

So the conditional probability of losing given that you have been selected from an infinite population

Such as the conditional probability of being chosen in the game that has no snake eyes given you were chosen at all. 

Or the conditional probability of dying given that you were chosen in precisely round k

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 $\frac{\infty }{\infty }$.  We cannot assume this term is zero just because of the $\infty$ in the denominator and disregard this universe.

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 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.