1. The problem statement, all variables and given/known data There are 8 athletes, each numbered from 1 to 8. They all play a game against each other. Thus in each round, a group of two is randomly made. In a group, a person with lower number always wins. eg, a match between 3 and 7 would be always won by 3. The losers are eliminated. Thus in the second round, only 4 are present, and in the third, 2. Find the probability that athlete 4 reaches the finals 2. Relevant equations - 3. The attempt at a solution Athlete 1 will always be victorious, because he cannot be defeated by others. But in the finals, he'd face a person, and I have to find the probability that he is athlete 4. For no. 4 to go to the second round, Lets say sample space is 8C2. No. of combinations that'd lead him to win is 4-5,4-6,4-7,4-8, i.e. 4 combinations. His probability of going to second round is 1/7. In the next round, along with number 4, there are two possibilities. Lets consider all no. below 4 to be X and above 4 to be Y. In the second round, athletes can be present in such order, 4,X,Y,Y --- 4,X,X,Y [Since atleast one X would always be present (1). If all the three are X, probability of 4 going to finals becomes 0] In the first case, probability of 4 winning is 2/4C2=1/3 and in second, 1/6. So probability of 4 going to finals is 1/7*(1/6+1/3)=1/14. But according to the book, this is wrong since the answer given is 4/35. Can someone help?