Optimum Poaching Strategy in Doubles
It is beneficial for the server's partner to occasionally poach the return of service. Not only does this increase the number of points you win, but if you fake a poach or poach on every serve or every first serve, the visual distraction and the induced mental uncertainty hinder the receiving team. In summary, for reasonable values for the probabilities of returning a serve and for the net player winning the point on a ball hit to him, it is optimum to poach about 40% of the time. The best strategy for the receiver is to hit every service return crosscourt unless the net player is crossing more than 40% of the time, in which case, the net player should go down the alley every time. Although the increase in the percentage of points won when following the optimum strategy is only 7%, as can be seen from the other tennis analysis on this site, a 7% change in the probability of winning every point has a very substantial effect on the probability of winning the match.
If you are not poaching, the receiver will return every serve crosscourt, and of those that are successfully returned, the probability that the server's team will then win the point is about 0.5 If you poach, your probability of winning an otherwise successful service return rises to perhaps 0.8. If you were to poach every time however, the returner would return down the alley, and would win the point on almost every successful return. Clearly then, there is an optimum frequency of poaching that lies somewhere between 0.00 and 1.00. Similarly, for each frequency that a server's partner is poaching, there is an optimum frequency of returning crosscourt or going down the line that minimizes the probability that the serving team wins the point. Below I make reasonable assumptions for the probabilities that a receiver can return a serve crosscourt or down the alley and the probabilities that the server's partner can put away a return hit to him.
A fake is made by the server's partner by jumping a step or two the instant you hear the sound of the serve and then quickly moving back to your original position. A poach is made by waiting until the returner's racket is in motion, and probably until the instant that the receiver's racket is hitting the ball. By starting to move at this time, the receiver cannot adjust and you can put away about three-quarters of the returns that are hit crosscourt. The analysis requires considering the four possible actions: fake or poach by the server's partner and the return either crosscourt or up the alley. For each of these actions, I estimate the probability that the server's team wins the point.
Fake, Crosscourt Return
Probability of 0.2 that the return is missed.
Probability of 0.8 that the return is made. In this case there is then a probability of 0.5 that the server's team wins the point. Thus the probability of winning the point via this route is 0.8 x 0.5 = 0.4
Total probability of winning the point is 0.2 + 0.4 = 0.6.
Poach, Crosscourt Return
0.2 that return is missed.
0.8 that return is made, but 0.75 probability of winning = 0.6
Total = 0.8
Fake, Alley Return
0.4 that return is missed. (The error frequency when returning up the alley is surprisingly large.)
0.6 that return is made, but 0.75 probability of winning = .45
Total = .85
Poach, Alley Return
0.4 that return is missed
0.6 that return is made, 0.00 probability of winning = 0.00
Total = 0.4
Let P be the the probability that you will poach. Then (1-P) is probability that you will fake. Similarly let C be the probability of of the receiver hitting a crosscourt return. Then, (1-C) is probability of going up the alley.
The expected number of points when each team is randomly deciding to poach-fake and hit crosscourt or down the alley is then the probability of each of the four possible actions times the probable point gain for each of these outcomes. For example, the first action, fake and a crosscourt return occurs with probability (1-P) x C, and the probable point gain from this is 0.6, thus this term is (1-P) x C x 0.6. Adding all four such contributions gives the probable points won per serve with a probability P that the net player poaches and a probability C and the service returner tries to hit a crosscourt return.
Points = (1-P) x C x 0.6 + P x C x 0.8 + (1-P) x (1-C) x .85 + P x (1-C) x 0.4 =
-.25 C + .65 CP - .45 P + .85
If P = 0, servers expected point gain is -.25 C + .856. To minimize our win probability, the receiver should play the strategy, C = 1, and in consequence, our point expectation is 0.6 point. This is as expected.
If P = 0.2, we win -.25 C + .13 C - .09 + .85. To minimize our win, C should be 1, and our point expectation is 0.64.
If P = 0.4, we win -.25 C + .26 C - .18 + .85. To minimize our win, C should be 0 (but it hardly matters) and our point expectation is 0.66.
If P = 0.6, we win -.25 C + .39 C - .27 + .85. C should be 0 and our point expectation is 0.63. Now we see that our expectation value is declined from a maximum near 0.66.
In fact, the optimum crossing probability for the numbers I've picked is 0.3846, for which the probability of winning the point is its maximum of .6769. At this poaching value, it doesn't matter if the receiver hits crosscourt all the time or down the alley all the time, or mixes it up. At this poaching value, and only at this poaching value, it doesn't matter what the receiver does.