Problem Solving

Two equally skilled teams play a four games tournament. What is the probability of the tournament ending at the fourth game with a winner?(Note : a team cannot win all the 4 games )

gmatter2012 wrote:Two equally skilled teams play a four games tournament. What is the probability of the tournament ending at the fourth game with a winner?(Note : a team cannot win all the 4 games )

Here, we have 2 teams (equally skilled). So, the probability of win for each team is 1/2 for each game.
Exhaustive set of events for any team - (W,L,L,L), (W,W,L,L), (W,W,W,L)
If tournament ends at 4th game, it means we have a clear winner. A score of 3-1 (4-0 is not considered as is written in the question)
Permuting, the number of ways = 4!/3! + 4!/2!2! + 4!/3!= 4+6+4 = 14 ways
Number of ways in which we have a decisive winner = 8 ways
So, the required Probability = 8/14 = 4/7

IMO Ans should be 3/16 ...

Let the two teams be A & B.
For the game to end in the 4th game, we have following combinations -

Assuming A wins the tournament -
A-B-A-A
B-A-A-A
A-A-B-A

There are 3 ways in which A can win the tournament

Similarly, for B there are 3 ways to win as above.

Total = 3+3 = 6 ways

Reqd probability = 6/16 = 3/8

Plz share OA.

Goodjob adthedaddy! Nice and simplified sol.
Expound further on my blunders in solving this problem after OA.

Bravo good job both of you !! you are the only ones brave enough in this community to even attempt this !! well done !! Hats off to both of you .

well let the teams be A and B

there are 4 games in the tournament and if the last one decides the winner then

let - - - - ( Four dashes represent the 4 games )
winning score for A is A = 3 and B = 1 as A= 2 and B = 2 will result in a draw , so for A to win he must win 3 of the 4 games.

- - - A : here A wins , so that means the three games before the 4th game A must have won 2 times , and the final score must be A = 3 and B=1 , as the possibility of A winning all games has been has not been allowed in the question, so out of 3 space's in how many ways can we fill 2 A's ? yes that's 3C2 = 3

so total ways A can win is 3C2 *1 ( Last game A won , so only 1 way )= 3

( as shown by adthedaddy too)


similarly for B, suppose B wins then - - - B , so the 3 games before the 4th game must have 2 B's for B to finally win , that's 3C2 again so = 3

So total ways B can win with it winning the last game = 3c2 *1 = 3

Hence for the last game to decide the winner 3+3= 6 favorable outcomes

total outcomes for the 4 games = 2*2*2*2= 16 ( as there are two outcomes for each game )

hence probability = 6/16 = 3/8

Bonus Question :

All the conditions remaining same as before , if the possibility of a team to win all the 4 games were incorporated( that was not allowed earlier ), how would the probability change.( Note : all the 4 games of the tournament were played )

Bonus Question :

All the conditions remaining same as before , if the possibility of a team to win all the 4 games were incorporated( that was not allowed earlier ), how would the probability change.( Note : all the 4 games of the tournament were played )

Probability of one team winning all the four games = (1/2)^4 = 1/16

adthedaddy wrote:

Bonus Question :

All the conditions remaining same as before , if the possibility of a team to win all the 4 games were incorporated( that was not allowed earlier ), how would the probability change.( Note : all the 4 games of the tournament were played )

Probability of one team winning all the four games = (1/2)^4 = 1/16

I think you didn't understand the Bonus question.
The fourth game should decide the winner and a team can also win all the 4 games ( this was restricted earlier )
so
---A now the 3 spaces can be filled in 3C2 or 3C3= 3+1 = 4
( also note the question says that the four games must be played , so even if A or B won 3 games in a row and won the tournament, the fourth game still will have to be played )
same for B
number of ways B can win the tournament, with it winning the last game ---B = 3C2 +3C3 = 3+1 = 4

so favorable cases 8
total cases 16 , hence probability = 8/16 = 1/2


if you look at your diagram which you made earlier

A-A-B-A
B-A-A-A
A-B-A-A
A-A-A-A

you will find the fourth case has now been allowed , now the same for B , hence total 8 Cases

Hope its clear .