Winning in a tennis match DSE 2011

[aditi5000]

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

[AJ]

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.

[aditi5000]

Perfect thank you AJ
 

[aditi5000]

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 :(

[aditi5000]

Am i making a silly mistake in the above...

[AJ]

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.

[aditi5000]

ok yes.... Thank you so so much!!!!!

[AJ]