Wamba-Ivanhoe

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... (read 414 more words →)

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... (read 465 more words →)

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 $(1-d)=35/36$.... (read more)