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We have the problem:

Problem 2 Alice and Bob have 2n+1 coins, each coin with probability of heads equal to 1/2. Bob tosses n+1 coins, while Alice tosses the remaining n coins. Assuming independent coin tosses, show that the probability that after all coins have been tossed, Bob will have gotten more heads than Alice is 1/2.

How would you solve this problem using only bayes theorem and/or law of total probability and not using binomial distribution?