I get that too. More generally, if there are n+1 rounds and on the first round the difference in probability between red and blue is z, then the optimal probability for choosing red is 1⁄2 + z/2n. It has to be close to 1⁄2 for large n, because 1⁄2 is optimal for the game where z=0, and over ten rounds the loss from deviating from 1⁄2 after the first round dominates the gain from knowing red is initially favoured.
Solution: I choose red with probability (written out and ROT13) avargl bar bire bar uhaqerq naq rvtugl.
EDIT: V’z fhecevfrq ubj pybfr guvf vf gb n unys.
I get that too. More generally, if there are n+1 rounds and on the first round the difference in probability between red and blue is z, then the optimal probability for choosing red is 1⁄2 + z/2n. It has to be close to 1⁄2 for large n, because 1⁄2 is optimal for the game where z=0, and over ten rounds the loss from deviating from 1⁄2 after the first round dominates the gain from knowing red is initially favoured.
Sure that’s not 1⁄2 + z/4n?
I think he meant “the difference between the probability of red and 1/2” when he said “the difference in probability between red and blue”.
Er, right, something like that.
That works too.