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Let $n$ coins, where at least one of them is a fair coin. Each one of the $n$ coins is tossed - Prove the probability to get even number of "Heads" is $\frac{1}{2}$.

I'd be glad for a direction.

Thanks.

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Lat $A$ be one of the fair coins. Let $p$ be the probability that the number of "heads" among the rest (i.e., the coins $\ne A$) is even. Then the probability for a total number of even "heads" is $$P(A\text{ tails})P(\text{rest even})+P(A\text{ heads})P(\text{rest odd})=\frac12p+\frac12(1-p)=\frac12.$$

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  • $\begingroup$ Can you explain $\frac12p+\frac12(1-p)$? $\endgroup$ – AlonAlon Nov 27 '14 at 17:59
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    $\begingroup$ @AlonAlon: This is Bayes's formula, conditioning on whether $A$ turned up heads (the second term) or tails (the first term). $\endgroup$ – Ted Shifrin Nov 27 '14 at 18:03
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Toss a coin a million times. Tally the number of heads you obtain. Divide by the number of toss. The result should be close to 1/2, proving that the probability of heads is 1/2.

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    $\begingroup$ This is not answering the question. You're tossing a fair coin here! $\endgroup$ – Ted Shifrin Nov 27 '14 at 18:03
  • $\begingroup$ Also the question specifically says "prove". $\endgroup$ – Eff Nov 27 '14 at 18:04

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