2014 UNCO Math Contest II Problems/Problem 10

Problem

The March Hare invites $11$ guests to a tea party. He randomly assigns to each guest either tea or cake, but no guest receives both. The guests know that the March Hare always does this, but they never know which guests will receive tea and which will receive cake. The guests decide to play a game. Each one tries to guess who of all $11$ guests will get cake and who will get tea. If one guest has more correct guesses than all the others, that guest wins. When several guests tie for the most correct guesses, then the Dormouse selects one to be the winner by selecting at random one of the guessers who has tied.

(a) All the guests make their guesses at random, perhaps by tossing a coin. What is the probability that Tweedledee, the last guest to arrive, is the winner?

(b) Tweedledum is the first guest to arrive. What is the probability that one or the other of Tweedledee and Tweedledum is the winner?

(c) Suppose that instead of guessing randomly, Tweedledee always makes the guess opposite to Tweedledum’s guess. If Tweedledum guesses that a guest will have tea, then Tweedledee will guess cake. If all the other guests have guessed randomly, what is the probability that one or the other of Tweedledee and Tweedledum is the winner? Your answer should be an explicit number, but partial credit may be given for reasonable formulae.


Solution

(a) $\tfrac{1}{11}$ (b) $\tfrac{2}{11}$ (c) $\tfrac{1023}{5120} = \left(\tfrac{1}{m+1}\right) \tfrac{2^{m+1}-1}{2^m}$,

where there are $m$ guessers, not including Tweedledee and Tweedledum. Here $m = 9$.

See also

2014 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Last Question
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions