Correct solutions were received from alum Nathan Pauli, Philippe Fondanaiche, Steve Prowse, Nancy Schwarzkopf, and a philosophical Brian Laughlin. A large number of incorrect solutions were submitted.

In the case of evenly matched teams, the probability is 2/3.  This can be arrived at by use of a decision tree and summing of some infinite series, or by the following nifty analysis.   

The only way for a score of 19-19 to result in a winning score of 21-19 is for the eventual winning team to reach a score of 20-19 and be serving the ball.  Call this Event A.  Let's call P the probability that the final score is 21-19, p the probability that the winning team wins any rally, and q = 1 - p the probability that the losing team wins any rally.  We must assume, of course, that p and q are constant throughout the play of the match.  

Once Event A occurs there are two possible ways the match can proceed. 

(a) The team in the lead and serving wins the next point, consequently ending the match.  This occurs with probability p.

(b) The team in the lead and serving loses the next point, and consequently gives up the serve.  It must then immediately win back the serve on the next rally, since otherwise the score would reach the forbidden 20-20.   This puts us right back into event A again.  This occurs with probability pq.  

Therefore P = p + pq P.  Solving for P gives P = p/(1 - pq).  In the particular case that both teams are evenly matched, p = q =  1/2, and P = 2/3 as stated.  

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