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Best Answer bonanova, 11 December 2013 - 06:40 PM

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23 replies to this topic

#21 bonanova

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Posted 16 December 2013 - 01:02 AM

Agree. Better simply to say simply that in any large ensemble of N outcomes, where 60% will be H, the outcomes can be partitioned into groups that are totally fair (H=T) and totally biased (all H.). The fair partition will be N x 2 p(T) in size.
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#22 The_King_Of_Games

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Posted 24 December 2013 - 07:02 PM

one doubt

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#23 The_King_Of_Games

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Posted 25 December 2013 - 06:23 PM

one doubt

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somebody plz help with this doubt

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#24 bonanova

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Posted 26 December 2013 - 04:14 AM

one doubt

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somebody plz help with this doubt

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