HMMT 十一月 2018 · 冲刺赛 · 第 31 题
HMMT November 2018 — Guts Round — Problem 31
题目详情
- [ 17 ] David and Evan each repeatedly flip a fair coin. David will stop when he flips a tail, and Evan will stop once he flips 2 consecutive tails. Find the probability that David flips more total heads than Evan.
解析
- [ 17 ] David and Evan each repeatedly flip a fair coin. David will stop when he flips a tail, and Evan will stop once he flips 2 consecutive tails. Find the probability that David flips more total heads than Evan. Proposed by: Anders Olsen 1 Answer: 5 Solution 1: We can find the values of the functions D ( h ) and E ( h ) , the probabilities that David and h 1 Evan, respectively, flip exactly h heads. It is easy to see that D ( h ) = 2 . In order to find E ( h ) , we note that each sequence must end with the flips HTT (unless Evan flips only 2 heads). We disregard these flips for now. Then there are h prior places we can include an extra tails in the sequence, one h +1 h between each pair of heads. There is a 2 probability of this happening with no extra tails, h 2 h h 1 probability with 1 extra tail, 2 probability with 2 extra tails, and so on. This sum is 2 ✓ ◆ ✓ ◆ h h X h 3 h +1 n 2 2 = 2 . n 4 n =0 We divide by 8 to account for the probability of getting HTT to finish our sequence to get that h 3 E ( h ) = . h +1 4 Our answer is ! 1 1 1 n X X X 3 1 E ( n ) D ( m ) = = . n +1 8 5 n =0 m = n +1 n =0 Solution 2: Since we only care about the number of heads, we think of this as a “survival” game 1 where they flip a single head each round, such that David has a chance of flipping another head and 2 3 Evan has a chance of flipping another head. (If they don’t get to flip another head, they lose.) David 4 wins if and only if when at least one of David and Evan loses, David does not lose but Evan loses. The 1 3 5 probability that at least one of them lose each round is 1 · = , and David wins this round with 2 4 8 1 1 1 1 probability · = , so the overall probability is . 2 4 8 5