same I win! Let's first focus on the subtree shown in bold. note that if two heads are flipped in a row, then you are guaranteed to win eventually. The sum of the probabilities of all your winning outcomes in this subtree is 4+16+64 1/4 The upppermost subtree marked same is the identical to the one shown in bold, except that each outcome probability is reduced by 1/2, because it is one edge farther from the root. Thus, the sum of your winning outcomes in this subtree is 1/6. Similarly, the sum of your winning outcomes in the next subtree marked same is 1/12, and so forth. Overall, your probability of winning is 31-1/2 3 In fact, as long as you pick a sequence first, and I pick a sequence second, I have at least a 2/ 3 probability of winning the game How is this possible? We know that each sequence of length three is equally likely,so therefore there should be no advantage to one sequence over another, and certainly no disadvantage to picking a sequence first. However, as we've already seen, this does not appear to be true6 Special Topics H H H H H T T T T T T T H H you win (eventually) pr = 1/4 you win pr = 1/16 I win! I win! you win pr = 1/64 etc. etc. same same Let’s first focus on the subtree shown in bold. Note that if two heads are flipped in a row, then you are guaranteed to win eventually. The sum of the probabilities of all your winning outcomes in this subtree is: 1 4 + 1 16 + 1 64 + ... = 1 4 · 1 1 − 1/4 = 1 3 The upppermost subtree marked same is the identical to the one shown in bold, except that each outcome probability is reduced by 1/2, because it is one edge farther from the root. Thus, the sum of your winning outcomes in this subtree is 1/6. Similarly, the sum of your winning outcomes in the next subtree marked same is 1/12, and so forth. Overall, your probability of winning is: 1 3 + 1 6 + 1 12 + ... = 1 3 · 1 1 − 1/2 = 2 3 In fact, as long as you pick a sequence first, and I pick a sequence second, I can always have at least a 2/3 probability of winning the game. How is this possible? We know that each sequence of length three is equally likely, so therefore there should be no advantage to one sequence over another, and certainly no disadvantage to picking a sequence first. However, as we’ve already seen, this does not appear to be true