Archimedes, Bernoulli and Cauchy are at it again. This time everyone wants the last slice of pizza and they want a fair way of deciding who gets it. They decide on the following rules:
- On the count of three, each person displays either ONE fingers or TWO fingers.
- If all three display the same number of fingers, then they throw again.
- If they do not all pick the same number, then two will match each other while the third person will have his own number. The third person wins and gets the piece of pizza.
Assume that each person will choose ONE or TWO with equal probability and independently of the others.
- Let \(N\) be the number of shoots necessary to determine a winner. Find \(\mathbf{E}(N)\) and \(\mathbf{SD}(N)\). Hint: No infinite sums are necessary. Express \(N\) in terms of a geometric random variable
- Now suppose that Archimedes and Bernoulli decide to cheat and they make a pact with each other.Archimedes will pick ONE with probability \(2/3\) and Bernoulli will pick TWO with probability \(2/3\).
- Show that this does indeed improve their chances of winning.
- What are \(\mathbf{E}(N)\) and \(\mathbf{SD}(N)\) now?
- Having made this pact, Archimedes wants to cheat Bernoulli as well. He will consider choose ONE with a probability \(p\) which may be different than\(2/3\). (Continue to assume that Bernoulli picks ONE with probability \(1/3\) and Cauchy picks ONE with probability \(1/2\)).
- Write Archimedes’ probability of winning the slice of pizza in terms of p. What value maximizes his chances ?
- Now suppose that Archimedes doesn’t care about winning, he just wants this to be over with. Write the expected number of games necessary to have a winner, \(\mathbf{E}(N)\), in terms of \(p\). What value of \(p\) minimizes \(\mathbf{E}(N)\)?