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More heads?



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Let B have n coins and let A have n+1 coins.

Keep in mind that coin tosses are independent events. Thus, tossing them simultaneously is not material to the outcome.

Suppose A keeps one of his coins in his hand while all the other 2n coins are tossed simultaneously. By symmetry, the expected number of heads tossed by B will equal the expected number of heads tossed by A. Now A tosses his remaining (n+1)st coin. It will be heads with probability of 1/2.

The probability that A obtains more heads than B (obtains) is 1/2.



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