Two equally skilled teams play a four games tournament. What

Probability of one team winning exactly three games, 4c3.

4!/(3!*1!) = 4 options
Each option has a probability of \((1/2)^4\) = 1/16

Probability of T1 winning exactly 3 games is 1/4
Probability of T2 winning exactly 3 games is 1/4

Probability of one team winning all four games:

\((1/2)^4\) = 1/16

Combining the probability of all four scenarios:

1/4 + 1/4 + 1/16 + 1/16 = 5/8


Alternatively you can find out the probability of either team winning exactly two games:

4!/2!*2! = 6 * 1/16 = 3/8

The question asks for "the probability of the tournament ending at the fourth game with a winner" ...

... if I'm not wrong, it implies that the final match determines the winner of the tournament. That means that at the end of the 3rd match, the score card is 2-1 in favor of either team.

First, we have to determine what we're looking for. If we are looking for a probability that after playing 4 games there will be a winner than we must consider a case of a tie. The statement does not tell us that the teams should either win or lose. It implies there might be ties.
Therefore:
1) The total possible options are WLT: 3*3*3*3=81

2) In order to be a winner, a team can do 4W, 3W+L&T, 2W+L+T, 2W+2T, 1W+3T
a)4W=1
b)3W=4!/3!=4 and we have T or L as options=4*2=8
c)2W+L+T=4!/2!=12
d)2W+2T=4!/2!*2!=6
e)1W+3T=4!/3!=4

3)Total for one team to become a winner is 1+8+12+6+4=31

4)Total for any of two to become a winner is 31*2=62


5)The probability is 62/81