Question 1208705
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The Boomtown Bears are playing against the Tipton Toros in a baseball tournament. 
The winner of the tournament is the first team that wins three games. 
The Bears have a probability of 1/2 of winning each game. 
Find the probability that the Bears win the tournament. 
Assume there are no ties—every game has a winner.
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<pre>
We have two teams: B (Bears) and T (the other team).


First, from the condition, it is clear that every team has the probability 1/2 to win/(to lose) 
every individual game.


Next, there are not too many cases when the Bear wins the tournament.
It is easy to list and to analyze each of these cases individually.


Case 1. The Bear wins the tournament after the 3 first games.

        It means that the Bear wins 3 of the 3 first games; the other team loses all three games.

        The winning record of the tournament for Bears is BBB : the Bears wins all three games of the first 3 games.

        The probability of such outcome is P(case 1) = {{{(1/2)^3}}}  = {{{1/8}}}.    (1)



Case 2. The Bears wins after the 4 first games (but does not win after 3 first games).

        It means that the Bears wins 3 of 4 games; the other team wins 1 game.

        The winning records of the tournament for the Bears are BBTB,  BTBB,  TBBB.

               Notice that in case 2 "T" can not be in the last position (!)

        
        The probability for the Bears to win in this case is

            P(case 2) = {{{3*(1/2)^3*(1/2)}}} = {{{3/16}}}.    (2)



Case 3. The Bear wins after 5 games (but does not win after 3 or 4 games).

        It means that the Bears wins 3 of 5 games; the other team wins 2 games.

        The winning records of the tournament for the Bears have 5 positions; the last position must be B and can not be T.
        
               So, two "T" can occupy any of 4 remaining positions, and it provides  {{{C[4]^2}}} = {{{(4*3)/2}}} = 6 possible winning records for B.

        Therefore, the probability for the Bears to win in this case is

            P(case 3) = {{{6*(1/2)^3*(1/2)^2}}}  = {{{6/32}}} = {{{3/16}}}.    (3)


It is easy to understand that these three cases exhaust all possibilities 
when the Bears wins the tournament.


Our last step is to add the three found probability (1), (2) and (3) for the Bears to win the tournament


    P = P(case 1) + P(case 2) + P(case 3) = {{{1/8}}} + {{{3/16}}} + {{{3/16}}} = {{{(2+3+3)/16}}} = {{{8/16}}} = {{{1/2}}}.


<U>ANSWER</U>.  The probability for the Bears to win this tournament is  {{{1/2}}} = 0.5.
</pre>

Solved.


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Probably, this answer could be predicted without calculations. 
Indeed, some of the two teams must be the first winning three games, and there is a symmetry between the teams.
So, the probability is 1/2 for each team to win the tournament.