This is a logic problem about tennis.

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Pat won a tennis set against Stacy with a score of 6-3, where five games were won by the player who did not serve. The reasoning presented concluded that since Pat won six games, she could not have served first, implying Stacy served first. However, it was clarified that both players alternated serving, and the five games won by the receiver could have been split between them. The correct deduction showed that Pat won the last game, indicating she must have served first. Ultimately, the analysis confirmed that Pat served first, aligning with the game's rules and the provided conditions.
  • #31
Terrell said:
##P_r+ S_r = 5##
Yes
Terrell said:
##6 - S_r= P_r##
No. I don't think that is what you meant to write.
Terrell said:
##S_s+ S_r = 3##
Yes.
I fixed up the latex by changing your single dollar signs to double hash signs (#).
 
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  • #32
Terrell said:
... Ss be the no. of games won by Pat where she served. ... S_r be the no. of games won by Pat where she received.
That should be Ss = # games won by Stacy when she served ... Sr # games won by Stacy when she received.

Terrell said:
6 - P_r = P_r
That should be 6 - Pr = Ps, or rewriten, Ps + Pr = 6 (number of games won by Pat). You already have Ss + Sr = 3 (number of games won by Stacy).

As scottdave posted, there are only two possible orderings for who served in a set PSPSPSPSP or SPSPSPSPS, which you responded to previously, but it's not clear if you've followed up on this.
 
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  • #33
rcgldr said:
That should be Ss = # games won by Stacy when she served ... Sr # games won by Stacy when she received.

That should be 6 - Pr = Ps, or rewriten, Ps + Pr = 6 (number of games won by Pat). You already have Ss + Sr = 3 (number of games won by Stacy).

As scottdave posted, there are only two possible orderings for who served in a set PSPSPSPSP or SPSPSPSPS, which you responded to previously, but it's not clear if you've followed up on this.
yeah those were typos. i got a bit lazy typing it. i'll work on this if i still have the brain power since schoolwork deadlines are near.
 
  • #34
I almost forgot about this after burning out on finals week, anyway here is what I got:

##P_s + S_s = 4##
##P_r + S_r = 5##
##P_r + P_s = 6##
##S_r + S_s = 3##

then we get the following clues to deduce from the case1: PSPSPSPSP or case2: SPSPSPSPS:

##P_r - S_s = 2##
## P_s - S_r = 1##

Observing case 2 first, we must have ##P_r + S_s = 5## and due to ##P_r - S_s = 2##, it must be that ##P_r > S_s##. Furthermore, ##P_r## cannot be 5 or 4 since ##5 - 0 \neq 2## and ##4 - 1 \neq 2##. It is reasonable that ##P_r be \leq 3##, but it won't satisfy ##P_r + S_s = 5##. However, for case 1 we can set ##P_r = P_s = 3## and ##S_s = 1## and ##S_r = 2## to satisfy the clues given above. Therefore, Pat served first. Did get it right?
 
  • #35
Yes, you got it right, but having reached this point for case 2
Terrell said:
we must have ##P_r + S_s = 5## and due to ##P_r - S_s = 2##,
you can solve for those two variables and discover an impossibility.
 
  • #36
haruspex said:
Yes, you got it right, but having reached this point for case 2

you can solve for those two variables and discover an impossibility.
because ##P_r## cannot be 7/2 ? Also, in retrospect, I never thought that I need to set up an equation to work on this problem so are there problems that does not need setting up equations? thank you!
 
  • #37
Terrell said:
because ##P_r## cannot be 7/2 ? Also, in retrospect, I never thought that I need to set up an equation to work on this problem so are there problems that does not need setting up equations? thank you!
Equations are just a convenient way of expressing reasoning. The Ancient Greeks managed without. Whether it is worth encoding a problem as equations depends on how succinct the notation is and how complex the problem.
 
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  • #38
Terrell said:
Also, in retrospect, I never thought that I need to set up an equation to work on this problem so are there problems that does not need setting up equations? thank you!
This can be done logically. If Pat wins on Stacy's serve and Stacy win's on Pat serve, then the set's game score is the same as if Pat won on her serve and Stacy won on her serve. The key factor is how many times Pat won on Stacy's serve minus how many times Stacy won on Pat's serve. Assume game score is even and equal to the number of times Stacy won at some point in the set. Consider the case where the game score is Pat 4, Stacy 4, and that Pat wins the last two games, the set ends up Pat 6, Stacy 4 regardless of who served first. Now consider the case where the game score is Pat 3, Stacy 3, and that Pat wins the last three games. If Stacy served first, then Pat won both times Stacy served in the last 3 games with Pat winning on Stacy's serve 2 more times than Stacy won on Pat's serve, but 5 can't be split up so that the difference equals 2. If Pat served first, then Pat only won once on Stacy's serve in the last 3 games, Pat winning on Stacy's serve 1 more time than Stacy won on Pat's serve, and 5 can be split up so the difference equals 1, specifically 3 - 2 = 1.

In tennis when a player wins when the other serves it's called a "break".
A list of possible outcomes for a set won when the winner wins their 6th game:

6 4 => +1 break, doesn't mater who served first
6 3 => +1 break if winner served first, +2 breaks if loser served first
6 2 => +2 breaks, doesn't matter who served first
6 1 => +2 breaks if winner served first, +3 breaks if loser served first
6 0 => +3 breaks, doesn't matter who served first
 
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