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Probability in a sport tournament

  1. Oct 1, 2007 #1
    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?
  2. jcsd
  3. Oct 1, 2007 #2


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    In the first round, the number of possible pairings is not 8C2, but rather 7*5*3*1. To see this, ask, who is #1 paired with? There are 7 possibilities. Once #1's opponent has been chosen, pick a player who's left (#2, say, if #2 is not #1's opponent; otherwise, #3), and ask who that player's opponent is. There are 5 possibilities. And so on.

    Now, in that first round, what are the odds that #4 draws #5, 6, 7, or 8, and not #1, 2, or 3?

    Say #4 draws #5. To survive the next round, we must not have #1, 2, and 3 paired up with #6, 7, and 8. What are the odds that this does NOT happen?
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