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Probability of Winning Tennis Match

See figure attached below for the question.

I was struggling with part a).

Is there a systematic way of solving for the sample space in this question? (I'm having a hard time determining whether I've truly written down every permutation of winning)

The outcomes that only last 3 or 4 sets it is still fairly clear whether you've obtained all possible outcomes or not, but for the outcomes that require 5 sets I find myself racking my brain to determine whether or not I've actually listed all the possible outcomes.

Is there a way to systematically determine this? For example, if I knew from the beginning that there would be 20 samples in my sample space when I have listed say 18, I would know that I am still missing 2 possible samples.

The only way I could think of simplifying things when generating this sample space is to create a sample space that contains all the games where one particular player ones. Then for the total sample space I would simply invert all the entries so that all the games that the other player could have one are covered as well.

Any ideas?

Re: Probability of Winning Tennis Match

Writing "F" for "Federer wins the set" and "R" for "Roddick wins the set" we can have

"RRR" (Roddick wins in three straight sets)

"FFF" (Federer wins in three straight sets)

"RFFF"

"FRFF"

"FFRF"

(But NOT "FFFR" since that would end at "FFF".

"FRRR"

"RFRR"

"RRFR"

"RRFFF" etc (there are 5!/(2!3!)= 10 ways to order 2 Rs and 3 Fs but we do NOT include FFFRR or FFRFR, etc. There are 4!/3!= 4 ways to get 3Fs in the first four sets so we subtract 10- 4= 6.)

Re: Probability of Winning Tennis Match

Quote:

Originally Posted by

**HallsofIvy** Writing "F" for "Federer wins the set" and "R" for "Roddick wins the set" we can have

"RRR" (Roddick wins in three straight sets)

"FFF" (Federer wins in three straight sets)

"RFFF"

"FRFF"

"FFRF"

(But NOT "FFFR" since that would end at "FFF".

"FRRR"

"RFRR"

"RRFR"

"RRFFF" etc (there are 5!/(2!3!)= 10 ways to order 2 Rs and 3 Fs but we do NOT include FFFRR or FFRFR, etc. There are 4!/3!= 4 ways to get 3Fs in the first four sets so we subtract 10- 4= 6.)

How did you obtain these numbers,

$\displaystyle \frac{5!}{2!3!}$

Can you generalize it somehow so I know how you are getting those numbers?

For example,

$\displaystyle \frac{\text{(Number of sets)!}}{(Number of F Wins)!(Number of R Wins)!}$

That formula seems to work for all cases, correct?

Of course, you would need to remove the cases such as,

FFFRR and FFRFR as you've mentioned.