4 thoughts on “Coin flip experiment

  1. Ah, I don’t think that is the minimum for a single experiment.

    The mapping {1,2}->{0,1},{3,4,5,6}->{00,01,10,11} and so on will give

    \sum _{j=1}^{\infty } \left(\sum _{i=\sum _{k=0}^{j-1} 2^k}^{\sum _{k=0}^j 2^k-1} \frac{j}{2^i}\right) \approx 1.26569.

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