MJ: Question for Don: Team standard deviation

[MJ]

Don,

A while back, I asked you a question about how to determine the cumulative standard deviation for different BJ games and then figure out how likely the event is to occur.

You wrote, "...e.v.s (or SCOREs) are simply additive. As you know, s.d.s are not, but their squares, variances, are. So, if you know the individual s.d.s, which you do, you square them, giving you the individual variances. Then you weight those variances by number of hours, add, giving you the combined variance, then you take the square root. The result is the desired s.d. for the combination of play".

This seems like an important concept, so I propose the following problem just to see it put into practice.

Problem:

There is a team of 3 counters that play under a team manager. Collectively, the team is down 400 units. The manager suspects that there might be skimming so he would like to know how likely this event is to occur.

Here is a summary of the team's play.

Bob:
1-deck
EV/Hr: 1 unit
SD/Hr: 10 units
#Hrs played: 25

Joe:
6-deck
EV/Hr: 2 units
SD/Hr: 26 units
#Hrs played: 64

Bill:
8-deck
EV/Hr: 1.5 units
SD/Hr: 30 units
#Hrs played: 36

Solution:

First we calculate the cumulative EV for the team given their 125 hours of play:

25 units + 128 units + 54 units = 207 units

Next, we figure out the Variance for the combination of play:

(10^2 units x 25) + (26^2 units x 64) + (30^2 units x 36) = 78164 units

SD = sq rt (78164) units ~ 280 units

Finally, the mean here is 207 units, less the unfortunate loss of 400 units results in underperforming expectation by 607 units. 607 units divided by the SD of 280 units gives a Z-statistic of -2.17. Ergo, the probability of the team being down 400 units is 1.54%. Yup, the manager was correct, somebody is SKIMMING! :-)

Is the solution correct?

Thanks for any assistance!
MJ

[Don Schlesinger]

> Don,

> A while back, I asked you a question about how to
> determine the cumulative standard deviation for
> different BJ games and then figure out how likely the
> event is to occur.

> You wrote, "...e.v.s (or SCOREs) are simply
> additive. As you know, s.d.s are not, but their
> squares, variances, are. So, if you know the
> individual s.d.s, which you do, you square them,
> giving you the individual variances. Then you weight
> those variances by number of hours, add, giving you
> the combined variance, then you take the square root.
> The result is the desired s.d. for the combination of
> play".

Sounds like a pretty good answer to me! ;-)

> This seems like an important concept, so I propose the
> following problem just to see it put into practice.

> Problem:

> There is a team of 3 counters that play under a team
> manager. Collectively, the team is down 400 units. The
> manager suspects that there might be skimming so he
> would like to know how likely this event is to occur.

> Here is a summary of the team's play.

> Bob:
> 1-deck
> EV/Hr: 1 unit
> SD/Hr: 10 units
> #Hrs played: 25

> Joe:
> 6-deck
> EV/Hr: 2 units
> SD/Hr: 26 units
> #Hrs played: 64

> Bill:
> 8-deck
> EV/Hr: 1.5 units
> SD/Hr: 30 units
> #Hrs played: 36

> Solution:

> First we calculate the cumulative EV for the team
> given their 125 hours of play:

> 25 units + 128 units + 54 units = 207 units

> Next, we figure out the Variance for the combination
> of play:

> (10^2 units x 25) + (26^2 units x 64) + (30^2 units x
> 36) = 78164 units

> SD = sq rt (78164) units ~ 280 units

> Finally, the mean here is 207 units, less the
> unfortunate loss of 400 units results in
> underperforming expectation by 607 units. 607 units
> divided by the SD of 280 units gives a Z-statistic of
> -2.17. Ergo, the probability of the team being down
> 400 units is 1.54%.

Up to this point, you did a perfect job!

> Yup, the manager was correct, somebody is SKIMMING! :-)

> Is the solution correct?

The math is right, but if you're going to conclude that something that happens 1.54% of the time is "impossible," by chance, or bad luck alone, then I've seen an awful lot of "skimming" myself out of money! :-)

Something that happens once every 65 attempts is an "unusual," or "somewhat rare," event; it isn't an "unthinkable" event. So, when you conclude that there has been stealing, you may very well be absolutely correct, but it isn't the math that is telling you that beyond a shadow of a doubt, and that's a very, very important distinction to be made.

> Thanks for any assistance!

One final comment: In my rather considerable experience, it seems that, whenever a team experiences a 2-s.d. event, there's about a 90% chance that it's to the negative side! :-)

Don

[MJ]

Thanks for the quick response!

Just one other unrelated question.

How many ways can the letters in the state Mississippi be arranged without any repetition?

Is it 11!/(4!4!2!) ?

MJ

[Don Schlesinger]

> Thanks for the quick response!

> Just one other unrelated question.

> How many ways can the letters in the state Mississippi
> be arranged without any repetition?

> Is it 11!/(4!4!2!) ?

Yes.

Don