Consider a tournament of 8 players, p1..p8.If the player with lower index(pi) always wins, what is the probability that p4 will have a match in finals?
Now there could be many different ways of looking at this. It would depend on the tournament structure. Let me evaluate some of the scenarios for you.
Scenario 1: Elimination matches - probability is 4105
Snenario 2: all 8 players play with each of the seven other players and top two from the group play the final - probability is 0
Scenario 3: players are grouped in two groups of 4 and they play all other players from their group. The group leaders play the final - probability is 470
Detailed analysis
Scenario 1: Is it like a lawn tennis grand slam?? - where every match is an elimination match
In this scanario, there would be two rounds before the final - means P4 will have to win two matches - means he will have to play with any two from P5 to P8.
Now calculation of probability.
First match is with a player from P5 to P8 - probability would be 47
Now, in the second round ie semi finals, it is essential that - firstly atleast one player from P5 to P8 should qualify for semi finals and secondly, this player from P5 to P8 should play with P4.
Probability that any one player from P5 to P8 qualifies - so there are 6 players left (since P4 and one player from P5 to P8 are already matched). In these 6 players, we have P1, P2, P3 and three players from P5 to P8.
Our requirement is any two players from P5 to P8 should play a match with each other in firat round so that atleast 1 player from them would qualify for the semi finals. This can be done in 3 ways.
And total number of ways in which matches can be played between 6 players is 15.
So probability that atleast 1 player from P5 to P8 qualify for semi finals would be 315=15
So now we are in semi finals where we have four players. - P4, any two players from P1 to P3 and one player from P5 to P8. So now probability that P4 plays with the one from P5 to P8 is 13
Hence probability that P4 wins the semi final is 15×13 = 115
Now finally, probability that P4 plays the final in case of elimination structure is 47 × 115= 4105
Too many things to deal with in scenario 1?? - have a look at Scenario 2 below.
Scenario 2 - if all 8 players are put in one group, and they play 7 matches each ie everyone plays with everyone (like in a cricket worldcup, or a football league) and top 2 play the final, then P1 & P2 are certain to play to final. Hence probability of P4 playing the final is 0. - this was easy, isnt it?
Scenario 3 - Players are grouped in two groups of 4 each. They play with everyone from their group and toppers of each group play the final.
Here, for P4 to reach the final, he must be grouped with any 3 players from P5 to P8 - this can be done in 4 ways.
Total number of ways to form the two groups will be 70.
So probability is 470
Now there could be many different ways of looking at this. It would depend on the tournament structure. Let me evaluate some of the scenarios for you.
Scenario 1: Elimination matches - probability is 4105
Snenario 2: all 8 players play with each of the seven other players and top two from the group play the final - probability is 0
Scenario 3: players are grouped in two groups of 4 and they play all other players from their group. The group leaders play the final - probability is 470
Detailed analysis
Scenario 1: Is it like a lawn tennis grand slam?? - where every match is an elimination match
In this scanario, there would be two rounds before the final - means P4 will have to win two matches - means he will have to play with any two from P5 to P8.
Now calculation of probability.
First match is with a player from P5 to P8 - probability would be 47
Now, in the second round ie semi finals, it is essential that - firstly atleast one player from P5 to P8 should qualify for semi finals and secondly, this player from P5 to P8 should play with P4.
Probability that any one player from P5 to P8 qualifies - so there are 6 players left (since P4 and one player from P5 to P8 are already matched). In these 6 players, we have P1, P2, P3 and three players from P5 to P8.
Our requirement is any two players from P5 to P8 should play a match with each other in firat round so that atleast 1 player from them would qualify for the semi finals. This can be done in 3 ways.
And total number of ways in which matches can be played between 6 players is 15.
So probability that atleast 1 player from P5 to P8 qualify for semi finals would be 315=15
So now we are in semi finals where we have four players. - P4, any two players from P1 to P3 and one player from P5 to P8. So now probability that P4 plays with the one from P5 to P8 is 13
Hence probability that P4 wins the semi final is 15×13 = 115
Now finally, probability that P4 plays the final in case of elimination structure is 47 × 115= 4105
Too many things to deal with in scenario 1?? - have a look at Scenario 2 below.
Scenario 2 - if all 8 players are put in one group, and they play 7 matches each ie everyone plays with everyone (like in a cricket worldcup, or a football league) and top 2 play the final, then P1 & P2 are certain to play to final. Hence probability of P4 playing the final is 0. - this was easy, isnt it?
Scenario 3 - Players are grouped in two groups of 4 each. They play with everyone from their group and toppers of each group play the final.
Here, for P4 to reach the final, he must be grouped with any 3 players from P5 to P8 - this can be done in 4 ways.
Total number of ways to form the two groups will be 70.
So probability is 470
