>half of the threads are about philosophy >on the literature

>>9481538
I don't know why the history boards needs to be shitted up with it as well. You guys can keep it since this place would be slower than /3/

>>9481538
The correct answer is who cares.

The average person playing this game would probably go again if it was lower than .5. You wouldn't have a calculator and formula sheets with you, and even if you did, it doesn't matter because the problem has no relation to reality anyway.

tl;dr just cut the knot with a sword

t. Alexander the Great

Who in the heck are you quoting OP?

>>9481538
0.618033988749895

Let x be the cutoff point for switching to a new number, and let E(x) be the expected value of the final number for cutoff x. E(x) = (probability of switching) * (expected value of switching) + (probability of not switching) * (expected value of not switching)
A number chosen uniformly from [0,1] would have expected value 1/2, so this is the expected value if we decide to switch numbers. In the case that we don't switch, our number is randomly chosen from [x,1], so the expected value would be (1 + x) / 2.
Thus, we can say:
E(x) = x/2 + (1 - x)(1 + x)/2
We can find the maximum of E by setting the derivative equal to zero and solving for x, yielding x = 0.5 for a maximum expected value of 0.625.
We're not done yet, though: If I know my opponent can achieve an expected value of 0.625, then I have to switch if my first number is less than that. That brings our cutoff, x, to 0.625. We plug x back into E to find the new expected value, which yields 0.617. Now we have to switch if the first number is less than x = 0.617. Iterating this process, we quickly converge to 0.61803..., which actually turns out to be 1/phi.
So our final answer is 0.61803..., which is remarkably close to your estimate of 0.6.