probability

[SHIKHA]

In a two team basketball play off, the winner is the first team to win ''m'' games.
Counting separately that the number of play offs requiring ''m,m+1,m+2.... 2m-1games'' , show that the total number of different outcomes (sequences of wins and losses by one
of the team) is

2[(m-1 C m-1)+ (m C m-1)+...+(2m-2 C m-1)]
C means combination...
 
how to go about this question..

[anon_econ]

Remember that if m games are required, the mth one must be won by someone who has already won m-1 games, with the other one having won less than m games (otherwise he would already have won).
So the number of games required can vary from m to 2m-1, & in each case we have to choose the number of ways in which the winning team could have won m-1 games earlier. Finally we multiple by 2 bcoz either of the teams could have won.
Think about this and look at the solution.

[SHIKHA]

"Remember that if m games are required, the mth one must be won by someone who has already won m-1 games"....

i got ur point but y we r putting strictness while solving  ..y cant we say that a person need to win m games out m, m games out m+1..n so on..

 

[anon_econ]

If a person has won m games, he has already won. There is no need to play the (m+1)th match. So the last game played must be the 'deciding factor', that is, it must be the mth win for one of them.