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July 1

Tennis problem

In tennis, players have a "first serve" action, which is faster and more difficult for the opponent to return but more likely to go "out", and a "second serve" action, which is slower and easier to return but more likely to be "in". A player's statistics include percentages of first and second serves "in" (call these respectively p1 and p2), and percentages of points won on first and second serves that are "in" (call these respectively w1 and w2). For example, if p1 = 0.6 and w1 = 0.8, it means that the player hits 60% of his or her first serves "in", and of the 60% that are "in", 80% result in the server winning the point.

In order for the p's and w's to "make sense", the overall probability of winning a point using a "first serve" action followed (if necessary) by a "second serve" action must be greater than the probability using any other combination of "first serve" and "second serve" actions (otherwise the player would be more successful using that other combination). In other words:

p1*w1 + (1 - p1)*p2*w2 > p1*w1 + (1 - p1)*p1*w1
p1*w1 + (1 - p1)*p2*w2 > p2*w2 + (1 - p2)*p1*w1
p1*w1 + (1 - p1)*p2*w2 > p2*w2 + (1 - p2)*p2*w2

I believe (correct me if I'm wrong) that this is equivalent to

p2*w2 - p1*w1 > 0
p2*w2 - p1*w1 < p2*w2*(p2 - p1)

While the interpretation of the first of these conditions is straightforward (total probability of winning point on a second serve must be greater than total probability of winning on a first serve), I cannot formulate an interpretation of the second one. Can anyone see how to describe or interpret the second condition in a way that can be more easily visualised? 109.153.244.163 (talk) 17:34, 1 July 2015 (UTC)[reply]

The left-hand side is the marginal increase in winning on a second serve, compared to a first serve. The right hand side is the probability of winning on a second serve, times the marginal increase in the probability that the second serve is in:

dW2 < Prob(W2)*dp2

I believe this constrains the largest possible increase in winning on a second serve to a function of how much more likely your second serve is to actually be in. You can also divide both sides by p2*w2 and get:

1 - (Prob(W1) / Prob(W2) < dp2

So that 1 minus the ratio of the winning probabilities on first and second serve (which is less than one, from the first constraint) is less than the marginal probability of getting your second serve first in, compared to the first serve.

I don't think this quite gets you there, but hopefully it helps. OldTimeNESter (talk) 18:44, 1 July 2015 (UTC)[reply]