Let $A$ and $B$ be two tennis players, playing against each other. A win of the game is considered when one of the two players is leading by two points (e.g. $A$ scores $5$ points vs. $B$ scoring $3$ points). The probability of a winning a point when $A$ is serving is $\frac{3}{5}$ the probability of a winning a point when $B$ is serving is $\frac{2}{3}$

$A$ serves first and they take turns after each serve. What is the probability of $A$ winning the game.

My progress so far:

I realise that A wins the whole game when the following sequence occurs: $WLW$ (where $W$ denotes a win and $L$ denotes a loss). However, also a game win can occur when we have $WLLLW$ or $WLLLLLW$ etc. If $A$ wins a point and $B$ wins a point, the score is essentially re-set. From here on, I am unsure how to continue...

  • $\begingroup$ Do they take turns serving? $\endgroup$
    – Marc
    Sep 26 '16 at 8:57
  • $\begingroup$ I presume you mean the tennis players are playing a set with no tiebreaker game, and you want the probability of A winning the set? $\endgroup$ Sep 26 '16 at 9:27
  • $\begingroup$ very badly worded! when it says 'a win' does it mean 'win a point' or 'win the game' - do they take turns serving? how many serves do they get? Surely he does not win the game with WLW - he need WLWW or WW - he needs to get one point ahead, then 2 points ahead $\endgroup$
    – Cato
    Sep 26 '16 at 9:47
  • $\begingroup$ The key to it is where you said ' the score is essentially re-set. ' - in that case his probability is also reset - note that if the score is equal, it is always A to serve, when B serves it is always 1-0 or 0-1 effectively. It is P(overall win) = P(wins two points) + P(goes back to square 1) x P(overall win) $\endgroup$
    – Cato
    Sep 26 '16 at 10:25
  • $\begingroup$ @Macr Yes the do take turns (I have edited my question) $\endgroup$
    – Newskooler
    Sep 26 '16 at 11:53

P(wins first point) = 3/5

P(wins second point) = 1/3

so he can win those two points with probability 1/5 and win in first two points - but if he wins the first point, then loses the second point, with probability 2/5 - he is back to square one - if he loses the first point and wins the second point (prob 2/15) - he is back to square one - if he loses two points - he lost (prob 4/15)

so after two points, he either wins, loses, or is back to square one with the same probability of winning - therefore P(winning) = M = P(wins in two points) + M x P(back to a draw after two points)

P(winning) = M = (3/5)(1/3) + (3/5)(2/3)M + (2/5)(1/3)M + 0 x(2/5)(2/3)

M = 1/5 + (2/5)M + (2/15)M

M = 3/15 + (6/15)M+ (2/15)M

M = 3/15 + (8/15)M

(7/15)M = 3/15

M = 3/7

Answer - 3/7


Let E be the event sania winning the match, then $P(E) =0.63$ (given) Since $P(\text{Sania loosing the match } )= P(\text{not } E)=P(E) =1-P(E)=1-0.63 =0.37$


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