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by Q_is_4_Quantum
975 days ago
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Given access to repeated uses of a coin of unknown bias "p" (which is not 0 or 1) you can (eventually) always generate a new coin flip with bias given (exactly) by: 1. 1/2 (i.e fair - von Neumann) 2. p^2 3. p^2/(p^2+(1-p)^2) 4. sqrt(p) Number 4 really surprised me, I learned it from this paper: http://www.math.chalmers.se/~wastlund/coinFlip.pdf But you can never generate the biases: 5. 2p 6. 4p(1-p) Although... if you change the game to allow a quantum coin then 5. and 6. are possible (a paper of mine: https://arxiv.org/abs/1509.06183) |
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