Special Topics Pr(b wins) 3-49 28 Finallv, we have Pr(A wins)= 1-Pr(B wins)-Pr(C wins) 55.4% Amazingly, the worst shooter has the best chance of winning, and the best shooter has the worst chance of winning Of course, an explicit assumption in this analysis was that B and C are both shooting to kill, unlike A in the first round. If B and C have no such requirement, then the problem is underspecified; there is no definite mathematical solution. Every gunfighter might reason that he is better off not shooting and the whole lot might go toast smores over a campfire 3 Penney-Ante Let's play a game! We repeatedly flip a fair coin. You have the sequence HHT, and I have the sequence HTT. If your sequence comes up first, then you win. If my sequence comes up first, then I win. For example, if the sequence of tosses is THTHTHHT then you win. This problem is tricky, because the game could go on for an arbitrarily long time. Draw enough of the tree diagram to see a pattern, and then sum up the probabilities of the(infinitely many) outcomes in which you win A partial tree diagram is shown below. All edge probabilities are 1 /2Special Topics 5 Pr (B wins) = 3 4 · x = 9 28 ≈ 32.1% Finally, we have: Pr (A wins) = 1 − Pr (B wins) − Pr (C wins) ≈ 55.4% Amazingly, the worst shooter has the best chance of winning, and the best shooter has the worst chance of winning! Of course, an explicit assumption in this analysis was that B and C are both shooting to kill, unlike A in the first round. If B and C have no such requirement, then the problem is underspecified; there is no definite mathematical solution. Every gunfighter might reason that he is better off not shooting and the whole lot might go toast smores over a campfire. 3 Penney-Ante Let’s play a game! We repeatedly flip a fair coin. You have the sequence HHT, and I have the sequence HTT. If your sequence comes up first, then you win. If my sequence comes up first, then I win. For example, if the sequence of tosses is: TTHTHTHHT then you win. This problem is tricky, because the game could go on for an arbitrarily long time. Draw enough of the tree diagram to see a pattern, and then sum up the probabilities of the (infinitely many) outcomes in which you win. A partial tree diagram is shown below. All edge probabilities are 1/2