[M T]

I'm not very learned about volleyball, but I wonder how many of our losses would have been wins if we had just eliminated (or greatly reduced) the service errors.  I've heard that a high number of service errors mean that you are being "aggressive" with your serves and that this will result in service aces and/or prevent your opponent from staying "in system."  But I don't see this happening with our serving.  We have low ace numbers and high service errors.  Am I missing something?

The Service Error to Ace ratio seemed high. I decided to work through it.

I just looked at the NCAA stats for Stanford WVB and compiled the service aces and errors for Stanford from 2012-13 to 2015-16 versus 2016-17 to 2021-22 (note that while both are 5 years, Hambly's number of sets is fewer because of COVID and the unfinished year).

		Stanford							Opponent
Coach/Years	Sets	Aces	SErrs	Aces/SErr	Aces/Set	SErrs/Set	z=0.4	z=0.35	Aces	SErrs	Aces/SErr	Aces/Set	SErrs/Set	z=0.4	z=0.35
Dunning 2012-2016	599	623	982	0.63	1.04	1.64	-0.03	0.10	541	948	0.57	0.90	1.58	-0.09	0.03
Hambly 2016-2022	488	670	1232	0.54	1.37	2.52	-0.19	0.01	416	888	0.47	0.85	1.82	-0.22	-0.08

Remember, most of the time, the receiving team wins the point.  Unfortunately, the NCAA stats don't have the data on side outs.  So, all we have is aces.  At about 30% more aces per set in the Hambly years, I suspect Stanford is winning more of its serves.

To be more precise, one should also take the service errors and consider the serving team may have won those points.  The Side-Out percentage is the percentage of service receptions that you win. I'll use something related, but not quite the same.  It is almost, but not quite 1 minus the Side Out percentage.

If Stanford typically wins X% of served points that are played (ignoring aces and service errors), and the opponent typically wins Y% of received points (ignoring A & SE), call Z = X+(1-Y)/2 as the expected number of played serves that Stanford wins.   Then Stanford should expect to win about Z% of each serve.  Let z = Z/100 (ie, the fraction, not the %).   A SE is a sure loss of a point, so it costs about z points over the expected value.   Likewise, an Ace is a sure win of a point, so it adds about 1-z points over the expected value.  So, when you figure in the A and SE, in a match, Stanford would get about  A*(1-z)-SE*z points for the serves that are either aces or service errors.  If that number is 0, then the aces and service errors balance out.

If a team were to win exactly half of its played serves (z=0.5), then it would want to have the same number of aces and service errors.  If a team is getting beat often on its played serves, then more aggressive serving is advantageous.  If z=0.33, then the numbers balance if you have twice as many service errors as aces.

However, there is a second order effect.  If the serves that are played cause the receiving team trouble, it will drive Z up.

Per set, Stanford averages about 4 serves that aren't played.  That means there are roughly 20 Stanford serves that are played.  If harder or riskier serves drive Z up by 1 percentage point (say, from 36% to 37%), that works out to a positive change of 20*0.01 = 0.20 points per set.  Obviously, this delta can't be known from stats.

A typical side out percentage is about 60%, (roughly) corresponding to z=0.4. The NCAA doesn't list the sideout percentage.  I found a site that had the stat for the UW game.  Washington's side out % was 65.5, so Stanford's z was about 0.345 (actually 0.382).  Stanford's side out % was 57.7, so Washington's z was about 0.423 (actually 0.411)

As we fans recognize, the team is serving more aggressively this year.  Before accounting for the second order effect, the aces and service errors are costing Stanford about 0.35 points per set if z=0.4 or 0.15 if z=0.35.  If the coach feels the more aggressive serves causes the team's z to increase a couple of percentage points, then he's going to be fine with it.

[M T]

I'm not very learned about volleyball, but I wonder how many of our losses would have been wins if we had just eliminated (or greatly reduced) the service errors.  I've heard that a high number of service errors mean that you are being "aggressive" with your serves and that this will result in service aces and/or prevent your opponent from staying "in system."  But I don't see this happening with our serving.  We have low ace numbers and high service errors.  Am I missing something?

The Service Error to Ace ratio seemed high. I decided to work through it.

I just looked at the NCAA stats for Stanford WVB and compiled the service aces and errors for Stanford from 2012-13 to 2015-16 versus 2016-17 to 2021-22 (note that while both are 5 years, Hambly's number of sets is fewer because of COVID and the unfinished year).

		Stanford							Opponent
Coach/Years	Sets	Aces	SErrs	Aces/SErr	Aces/Set	SErrs/Set	z=0.4	z=0.35	Aces	SErrs	Aces/SErr	Aces/Set	SErrs/Set	z=0.4	z=0.35
Dunning 2012-2016	599	623	982	0.63	1.04	1.64	-0.03	0.10	541	948	0.57	0.90	1.58	-0.09	0.03
Hambly 2016-2022	488	670	1232	0.54	1.37	2.52	-0.19	0.01	416	888	0.47	0.85	1.82	-0.22	-0.08

Remember, most of the time, the receiving team wins the point.  Unfortunately, the NCAA stats don't have the data on side outs.  So, all we have is aces.  At about 30% more aces per set in the Hambly years, I suspect Stanford is winning more of its serves.

To be more precise, one should also take the service errors and consider the serving team may have won those points.  The Side-Out percentage is the percentage of service receptions that you win. I'll use something related, but not quite the same.  It is almost, but not quite 1 minus the Side Out percentage.

If Stanford typically wins X% of served points that are played (ignoring aces and service errors), and the opponent typically wins Y% of received points (ignoring A & SE), call Z = X+(1-Y)/2 as the expected number of played serves that Stanford wins.   Then Stanford should expect to win about Z% of each serve.  Let z = Z/100 (ie, the fraction, not the %).   A SE is a sure loss of a point, so it costs about z points over the expected value.   Likewise, an Ace is a sure win of a point, so it adds about 1-z points over the expected value.  So, when you figure in the A and SE, in a match, Stanford would get about  A*(1-z)-SE*z points for the serves that are either aces or service errors.  If that number is 0, then the aces and service errors balance out.

If a team were to win exactly half of its played serves (z=0.5), then it would want to have the same number of aces and service errors.  If a team is getting beat often on its played serves, then more aggressive serving is advantageous.  If z=0.33, then the numbers balance if you have twice as many service errors as aces.

However, there is a second order effect.  If the serves that are played cause the receiving team trouble, it will drive Z up.

Per set, Stanford averages about 4 serves that aren't played.  That means there are roughly 20 Stanford serves that are played.  If harder or riskier serves drive Z up by 1 percentage point (say, from 36% to 37%), that works out to a positive change of 20*0.01 = 0.20 points per set.  Obviously, this delta can't be known from stats.

A typical side out percentage is about 60%, (roughly) corresponding to z=0.4. The NCAA doesn't list the sideout percentage.  I found a site that had the stat for the UW game.  Washington's side out % was 65.5, so Stanford's z was about 0.345 (actually 0.382).  Stanford's side out % was 57.7, so Washington's z was about 0.423 (actually 0.411)

As we fans recognize, the team is serving more aggressively this year.  Before accounting for the second order effect, the aces and service errors are costing Stanford about 0.35 points per set if z=0.4 or 0.15 if z=0.35.  If the coach feels the more aggressive serves causes the team's z to increase a couple of percentage points, then he's going to be fine with it.