Flip twice, calling it heads if they are the same and tails if they are different.
Its not hard to see that this won't work. if the probability a coin comes up heads is P(H)=k and tails is P(T)=1-k, then the probability of "same" or "different" are given as:
P(S)=P(H)2+P(T)2
P(S)=1-2k+2k2
P(D)=P(H)P(T)+P(T)P(H)
P(D)=2k-2k2
and so the amount more common "same" is over "different" is
P(S)-P(D)=1-4k+4k2
=(1-2k)2
always positive. Thus, if you are ever presented with this situation, call "same".
As a side note, since P(T)-P(H)=1-2k is less than 1, using "same" and "different" has a smaller gap between the good strategy and the bad strategy (squaring a positive number less than one pushes it closer to zero). So it is possible to "converge" to a fair coin by using nested versions of the "same" and "different" strategy. Specifically, you could get within any epsilon of a fair coin if you were given the value of k.
Alright, for the actual answer to the problem (at least the only answer I know of):
Flip the coin twice, discard the result of both sides come out the same, use the second result if they come out different
Since the probability they come out different is 2k(1-k), it will on average take 1/(2k(1-k)) tries to get a result using this method. I suppose a better method might take less flips, if anybody has one feel free to post it in the comments.
2 comments:
Sally's friend's brother purposed:
Flip frisbee until it comes up tails, hand to opponent, he flips until comes up tails, whoever takes less flips, 'wins'.
Of course in the event of a tie, neglect results and try again.
Commenting on that: God that could be a LOT of flips.
Yeah, that should work. I would expect it won't be all that many flips on average, but I would expect it would be more than mine usually. Anyway, good alternate solution.
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