# Winning in a tennis match DSE 2011

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## Winning in a tennis match DSE 2011

 Hi, q35 dse 2011 p(GAME ENDS IN 6)=P(win 4 OUT OF 6) = 6c4 (0.5)^6 P(GAME ENDS IN 7)=P(win 4 OUT OF 7)  = 7c4 (0.5)^7 how are the probabilities equal? please help me with this. broke my head with this one :P
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## Re: Winning in a tennis match DSE 2011

 I might be wrong but I did it like this: Two players are A and B cases when game will end in 6 matches: these would be like AAABBA ..and BBBAAB .. in this fix the last A or B .. Now, in each of these first five can be rearranged in 5!/3!2! ways ... So, No. of cases when game will end in 6 matches= 5!/3! = 5*4 Now, cases when game will end in 7 matches... are cases when first six matches went like these AAABBB ... and any rearrangement of it... So, no of cases= 6!/3!3! = 5*4 Both equals.
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## Re: Winning in a tennis match DSE 2011

 Perfect thank you AJ
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## Re: Winning in a tennis match DSE 2011

 In reply to this post by AJ Hey in cases when game will end in 7 matches... it would be like AAABBB ... and any rearrangement of it... ie 6! and the last match could be won either by B or A right so 6!/3!3! * 2 ??? = 40 :(
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## Re: Winning in a tennis match DSE 2011

 Am i making a silly mistake in the above...
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## Re: Winning in a tennis match DSE 2011

 U dont have to multiply by 2.. If the first 6 are in this way.. game will end in 7th watever may happen... When we were doing for game ending in 6 matches... we had to multiply be 2 bcuz first five were having different arangement.. like if B wins in 6th then we need BBBAA kinda setting... and if A wins AAABB kinda. and thats why we doubled. Here, either A wind or B wins last match.. initial arrangements would be same.