

P1, P2, . . . , P8 are eight players participating in a tournament. If i < j, then Pi will win
the match against Pj. Players are paired up randomly for first round and winners of
this round again paired up for the second round and so on. Find the probability that P4
reaches in the final. How to get the answer???
Guys, answer is coming 4/35.


Is answer correct.. I am getting
4/35*1/3
may b some mistake.. please check it..
players are .. 1,2,3,4,5,6,7,8
final round will be third round... so, player 1 has to win 1st and second round..
this required that in both rounds he is paired with one of (5,6,7,8)
and therefore, also that ..one of these shud also reach second round.. so that 4 can be paired with that....
for first round.. probability of such pairing = 4/7* 3C2/6C2.. (4/7 for P4 getting paired with 5,6,7,8 out of all 7...and second part cuz we want a pair from 5,6,7,8 as well..so that winner reaches second round where P4 can defeat it) ...
this comes out to be 4/35...
now in second round.. we have total 4 ppl.... Player 4.. someone from (5,6,7,8) and two players from (1,2,3) ...
for reaching finals he needs to get paired with some1 frm (5,6,7,8) .... and probability of that is.. 1/3 ...
so my answer become 4/35*1/3 ..
Thats what I got! :/


A.J ur probabilities r not correct according to me..
I did it by folding it up from round 2.
In round 2, the following groups must be there:
4,5
4,6
4,7
(4,8 can't be there bcoz 8 could not hv won in round 1).
For 4,5 to be there in round 2, we must hv had either 5,6 or 5,7 or 5,8 in round 1 (bcoz 5 must have won).
For 4,6 to be there in round 2, we must hv had either 6,7 or 6,8 in round 1.
For 4,7 to be there in round 2, we must hv had 7,8 in round 1.
Also, 4 must have been with 5,6,7 or 8 in round 1.
So the possibilities for round 1 if 4,5 r to be there in round 2 are:
4,6 and 5,7
4,6 and 5,8
4,7 and 5,6
4,7 and 5,8
4,8 and 5,6
4,8 and 5,7
Possibilities if 4,6 r to be there in round 2 are:
4,5 and 6,7
4,5 and 6,8
4,7 and 6,8
4,8 and 6,7
Similarly if u have 4,7 in round 2, u can hv:
4,5 and 7,8 or 4,6 and 7,8.
Now, in round 1, if u know 2 of the teams, the other 2 can be chosen in 3 ways. Also the total no of ways of choosing the 4 teams is 105. So probability of getting any 1 of the above mentioned possibilities is 3/105.
Now u just hv 2 do the final calculation. For 4,7 in round 2, probability is 6/105*1/3. Add the probabilities for cases 4,6 & 4,5 in round 2..u get 4/35.
I know this is a very bad explanation, i dint know how else to explain it. The basic thing is that it has to be folded backwards.


Hey Vasudha
Thanks yaar
You are the champ

