advanced-tiebreaker-solutions - SMT 2013 Advanced Topics...

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SMT 2013 Advanced Topics Tiebreaker Solutions February 2, 2013 1. Andrew flips a fair coin 5 times, and counts the number of heads that appear. Beth flips a fair coin 6 times and also counts the number of heads that appear. Compute the probability Andrew counts at least as many heads as Beth. Answer: 1 2 Solution: Consider the three possible cases right before Beth flips her last coin (at this point Andrew and Beth have each flipped the coin 5 times): 1. Andrew has more heads than Beth. 2. Beth and Andrew have the same number of heads. 3. Beth has more heads than Andrew. Let x be the probability of case 2. As we’ll see, we won’t actually need to compute x . Then the first and last cases are symmetric, so they must each have probability 1 - x 2 . Now, consider what happens when Beth flips her last coin in each of these cases. Case 1 will satisfy the problem regardless of the result of this flip, case 2 will satisfy it half the time (only when Beth flips tails), while case 3 will never satisfy the problem.
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