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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.
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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.
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This is exactly right under SIA, thanks. Under SIA, almost all the snakes exist in the final scenario, and therefore the limit is 50% as n → infinity.
Under SSA it’s a bit higher than 50%, because we always have a 1⁄36 chance of there being a single red-eyed snake.
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Right, and the correct value is 37⁄72, not 19⁄36, because exactly half of the remaining 70⁄72 players lose (in the limit).