HMMT 十一月 2010 · GEN2 赛 · 第 10 题

HMMT November 2010 — GEN2 Round — Problem 10

[ 7 ] Justine has a coin which will come up the same as the last flip of the time and the other side of 3 3 the time. She flips it and it comes up heads. She then flips it 2010 more times. What is the probability that the last flip is heads?

解析

[ 7 ] Justine has a coin which will come up the same as the last flip of the time and the other side of 3 3 the time. She flips it and it comes up heads. She then flips it 2010 more times. What is the probability that the last flip is heads? 2010 3 +1 Answer: Let the “value” of a flip be 1 if the flip is different from the previous flip and let 2010 2 · 3 it be 0 if the flip is the same as the previous flip. The last flip will be heads if the sum of the values of ( ) ( ) ( ) 1005 2 i 2010 − 2 i ∑ 2010 1 2 all 2010 flips is even. The probability that this will happen is . 2 i 3 3 i =0 We know that ( ) ( ) ( ) ( ) ( ) ( ) 1005 2 i 2010 − 2 i 2 i +1 2010 − (2 i +1) ∑ 2010 1 2 2010 1 2

= 2 i 3 3 2 i + 1 3 3 i =0 ( ) ( ) ( ) ( ) 2010 k 2010 − k 2010 ∑ 2010 1 2 1 2 = + = 1 k 3 3 3 3 k =0 and ( ) ( ) ( ) ( ) ( ) ( ) 1005 2 i 2010 − 2 i 2 i +1 2010 − (2 i +1) ∑ 2010 1 2 2010 1 2 − = 2 i 3 3 2 i + 1 3 3 i =0 ( ) ( ) ( ) ( ) ( ) 2010 k 2010 − k 2010 2010 ∑ 2010 − 1 2 − 1 2 1 = + = k 3 3 3 3 3 k =0 1 2010 2010 1+( ) 3 +1 3 Summing these two values and dividing by 2 gives the answer = . 2010 2 2 · 3 Theme Round