Is the following a correct reformulation of your problem?
Say that a magician’s pure strategy is a function on finite bitstrings which returns either “go on” or “stop” with a positive integer n and a frequency f of zeros.
A magician’s mixed strategy is a probability distribution over magician’s pure strategies. (Btw, I’m not sure what kind of sigma-algebra is suitable here.)
A player’s mixed strategy is a probability distribution over infinite bitstrings.
A magician’s mixed strategy together with a player’s mixed strategy define, in the obvious way, a probability distribution over outcomes “magician wins” (i.e., the magician’s prediction was within 1%) and “player wins”.
You’re claiming that there is a magician’s mixed strategy s.t. for every player’s mixed strategy, magician wins with probability at least 0.99.
I think you can slightly simplify to say: the magician has a probability distribution over strategies, such that for every sequence the magician wins with probability 0.99.
Is the following a correct reformulation of your problem?
Say that a magician’s pure strategy is a function on finite bitstrings which returns either “go on” or “stop” with a positive integer n and a frequency f of zeros.
A magician’s mixed strategy is a probability distribution over magician’s pure strategies. (Btw, I’m not sure what kind of sigma-algebra is suitable here.)
A player’s mixed strategy is a probability distribution over infinite bitstrings.
A magician’s mixed strategy together with a player’s mixed strategy define, in the obvious way, a probability distribution over outcomes “magician wins” (i.e., the magician’s prediction was within 1%) and “player wins”.
You’re claiming that there is a magician’s mixed strategy s.t. for every player’s mixed strategy, magician wins with probability at least 0.99.
I think you can slightly simplify to say: the magician has a probability distribution over strategies, such that for every sequence the magician wins with probability 0.99.
Yes, indeed.