MJ: Standard Deviation

[MJ]

Ok I changed the numbers to make things easier:

Unit = $25.00
Win Rate/Hr = $52.15
SD/Hand = 4.375 units = $110.00
SD/Hr = $43.75
100 Hands/Hr

if you play 100
> hands per hour, you'd have to multiply the
> per-hand SD by 10 (square root of 100) to
> get the hourly SD.

Ok now why would I have to calculate the Hourly SD if that is already computed by CVCX?
According to CVCX it is $43.75. However if I try and do it your way here is the result:

SD/Hand x 10(sq root of 100) = Hourly SD

$110.00 x 10 = $1100.00 Hourly SD

$1100.00 is way off from the SD/Hr of $43.75 given by CVCX(see above)!! The only way this could work out is if the Hourly SD is in Units/Hr and not $/Hr.

Here is what I mean:

SD/Hr = 4.375 units(SD/Hand) x 10 = 43.75 units(NOT dollars)
Now everything seems to match up.

Is this correct?

-MJ

[Don Schlesinger]

> Ok I changed the numbers to make things
> easier:

You can't just make them up; they have to make sense! :-)

> Unit = $25.00

Fine.

> Win Rate/Hr = $52.15

Fine.

> SD/Hand = 4.375 units = $110.00

Still OK.

> SD/Hr = $43.75
> 100 Hands/Hr

Bzzz. Quite impossible. If the per-hand SD is $110 and you play 100 hands per hour, the per-hour SD is $1,100. How would you want the per-hour SD to be less than half of the per-hand SD? Wouldn't make any sense, would it?

I think you may have meant SD/Hr = 43.75 units, without the dollar sign. In that case, the dollar amount would be $1,094, which makes sense now.

> Ok now why would I have to calculate the
> Hourly SD if that is already computed by
> CVCX?

It doesn't give the hourly SD in dollars -- just units. The chapter 10 charts in BJA3 give it in dollars, because I asked Norm to do it that way.

> According to CVCX it is $43.75.

See above.

> However if I
> try and do it your way here is the result:

> SD/Hand x 10(sq root of 100) = Hourly SD

> $110.00 x 10 = $1100.00 Hourly SD

There you go! Now, you're an SD maven! :-)

> $1100.00 is way off from the SD/Hr of $43.75
> given by CVCX(see above)!!

Check closely for the dollar sign in CVCX. It's not there, is it? :-)

> The only way this
> could work out is if the Hourly SD is in
> Units/Hr and not $/Hr.

See? You don't need me at all. :-)

> Here is what I mean:

> SD/Hr = 4.375 units(SD/Hand) x 10 = 43.75
> units(NOT dollars)
> Now everything seems to match up.

> Is this correct?

Right as rain!

Don

[MJ]

Thanks for clearing that up Don. Now let me see if I can finish the calculation. Here is the data once more:

Unit = $25.00
Win Rate/Hr = $52.15
SD/Hand = 4.375 units = $110.00
SD/Hr = 43.75 units = $1100.00
100 Hands/Hr

So the occurences are broken down as follows:

68% fall within -$1047.85 to $1152.15

95% fall within -$2147.85 to $2252.15

99.7% fall within -$3247.85 to $3352.15

Ok so let me see if I can make sense of the overall picture. I guess what this does is provide the counter with an EXTREMELY likely range where he can expect his earnings/losses to fall after playing a certain # of hands. It would be NEARLY impossible for earnings/losses to be outside of +/- 3SD right?

Now lets analyze the bell curve a bit. There is a 34% chance(+1 SD) earnings will be in the range of $52.15(Mu) to $1152.15. Now lets say I win $2000 at the tables. Would that occur 2SD - 1SD = 48% - 34% = 14% of the time?

$2000 is between 1 SD and 2 SD but I'm not certain if this logic is precise. Is it EXACTLY 14%? If not is there anyway I can calculate exactly how often I would be ahead $2000?

One last question. If my results are 1.5 SD to the right of the mean what range would my results fall into? After all you said SD is NOT linear but a square root function so how would compute this?

Thanks for any answers you can provide.

-MJ
> You can't just make them up; they have to
> make sense! :-)

> Fine.

> Fine.

> Still OK.

> Bzzz. Quite impossible. If the per-hand SD
> is $110 and you play 100 hands per hour, the
> per-hour SD is $1,100. How would you want
> the per-hour SD to be less than half of the
> per-hand SD? Wouldn't make any sense, would
> it?

> I think you may have meant SD/Hr = 43.75
> units , without the dollar sign. In that
> case, the dollar amount would be $1,094,
> which makes sense now.

> It doesn't give the hourly SD in dollars --
> just units. The chapter 10 charts in BJA3
> give it in dollars, because I asked Norm to
> do it that way.

> See above.

> There you go! Now, you're an SD maven! :-)

> Check closely for the dollar sign in CVCX.
> It's not there, is it? :-)

> See? You don't need me at all. :-)

> Right as rain!

> Don

[Wolverine]

Don and MJ,
Sounds like were looking for a two tailed t-test to determine the probability of an event occuring given the event, the known mean, and a known s.d. Two tails because you have chances to be on either side of the mean, even though you are only testing for one specific event.

If that is true (wait for Don to confirm this), then there are tables (published in statistics books) which will allow you to get that information OR your MS Excel spreadsheet should have a t-test function on it as well.

A t-test basically takes the event and determines how many S.d. from the mean it occured. The resulting factor of "how many standard deviations from the mean" (I think you quoted 1.5 above) can then be translated into the "probability of likelihood" of said event occuring given the mean and s.d. you have calculated through the tables or function in Excel.

Hope this helps. And I really hope Don, Cac, Zen and the other statistical experts agree with my thoughts too! :-)

[Don Schlesinger]

> Thanks for clearing that up Don. Now let me
> see if I can finish the calculation. Here is
> the data once more:

> Unit = $25.00
> Win Rate/Hr = $52.15
> SD/Hand = 4.375 units = $110.00
> SD/Hr = 43.75 units = $1100.00
> 100 Hands/Hr

Much better.

> So the occurrences are broken down as
> follows:

> 68.3% fall within -$1047.85 to $1152.15

> 95.4% fall within -$2147.85 to $2252.15

> 99.7% fall within -$3247.85 to $3352.15

I added a couple of decimals for you, for the ranges. The dollar amounts look right.

> Ok so let me see if I can make sense of the
> overall picture. I guess what this does is
> provide the counter with an EXTREMELY likely
> range where he can expect his
> earnings/losses to fall after playing a
> certain # of hands.

Right.

> It would be NEARLY
> impossible for earnings/losses to be outside
> of +/- 3SD right?

Except if you play against MY dealers! :-)

> Now let's analyze the bell curve a bit. There
> is a 34% chance(+1 SD) earnings will be in
> the range of $52.15(Mu) to $1152.15.

After a single hour.

> Now
> let's say I win $2000 at the tables. Would
> that occur 2SD - 1SD = 48% - 34% = 14% of
> the time?

> $2000 is between 1 SD and 2 SD but I'm not
> certain if this logic is precise. Is it
> EXACTLY 14%? If not is there anyway I can
> calculate exactly how often I would be ahead
> $2000?

This is called a z-statistic. Basically, what you do is take your win ($2,000) and subtract what you should have won from it, namely, the mean, or $52.15. This gives your outperformance, above the mean, or $1,947.85. Next, you express this value in terms of standard deviations. So, you divide 1948 by 1100, getting 1.77 (SDs).

Finally, you need a table of the cumulative normal distribution (handily provided in BJA3, pp. 147-48!) to see how often you win 1.77 SDs or more. There are two ways to find this value. If you go to the positive SDs, you look up 1.77 and find 0.9616. This means that you will win $2,000 or LESS 96.16% of the time. To find out how often you will win $2,000 or MORE, do one of two things: 1) Either subtract the above value from 100%, yielding 3.84% of the time, or 2) Do what I do, which is to look at the negative value of -1.77, instead, in the page 147 chart, and get 0.0384, or 3.84% directly (since the curve is symmetrical around the mean).

> One last question. If my results are 1.5 SD
> to the right of the mean what range would my
> results fall into? After all you said SD is
> NOT linear but a square root function so how
> would compute this?

Again you need the table. We read, for +1.5 SD, 93.32%. So, you're above that 1.5-SD value 6.68% of the time. Note that, if you want to express the probability of being within a 1.5-SD bandwidth around the mean, you need to subtract 6.68% (to the LEFT of -1.50) from 93.32%, yielding 86.64% as the area under the curve, between -1.5 SD and +1.5 SD.

> Thanks for any answers you can provide.

Hope all is clear now.

Don

[Don Schlesinger]

[Don Schlesinger]

Can you tell me why you double-post your questions both here and then again at bj21.com? Are you somehow dissatisfied with either the clarity or the speed of the responses you get here?

Don

[MJ]

Thanks much Don. I understand everything you explained. I found a Z chart and all the numbers you provided seem to match up.

Here is a link for those who do not have one:
http://www.statsoft.com/textbook/sttable.html#z

So basically a Z chart is just the area under a normal distribution curve on the interval of 0 to Z. As I understand it Z is just the # of standard deviations from the mean and the area under the curve represents the probability of being within a given SD. Thanks again for the assistance.

-MJ

-MJ

> Much better.

> I added a couple of decimals for you, for
> the ranges. The dollar amounts look right.

> Right.

> Except if you play against MY dealers! :-)

> After a single hour.

> This is called a z-statistic. Basically,
> what you do is take your win ($2,000) and
> subtract what you should have won from it,
> namely, the mean, or $52.15. This gives your
> outperformance, above the mean, or
> $1,947.85. Next, you express this value in
> terms of standard deviations. So, you divide
> 1948 by 1100, getting 1.77 (SDs).

> Finally, you need a table of the cumulative
> normal distribution (handily provided in
> BJA3, pp. 147-48!) to see how often you win
> 1.77 SDs or more. There are two ways to find
> this value. If you go to the positive SDs,
> you look up 1.77 and find 0.9616. This means
> that you will win $2,000 or LESS 96.16% of
> the time. To find out how often you will win
> $2,000 or MORE, do one of two things: 1)
> Either subtract the above value from 100%,
> yielding 3.84% of the time, or 2) Do what I
> do, which is to look at the negative value
> of -1.77, instead, in the page 147 chart,
> and get 0.0384, or 3.84% directly (since the
> curve is symmetrical around the mean).

> Again you need the table. We read, for +1.5
> SD, 93.32%. So, you're above that 1.5-SD
> value 6.68% of the time. Note that, if you
> want to express the probability of being
> within a 1.5-SD bandwidth around the mean,
> you need to subtract 6.68% (to the LEFT of
> -1.50) from 93.32%, yielding 86.64% as the
> area under the curve, between -1.5 SD and
> +1.5 SD.

> Hope all is clear now.

> Don

[MJ]

When I was a student in college my Calculus Professor once asked me "Why do you attend my lecture in the afternoon and then sit in on Professor Sydney's class in the evening? Is there something wrong with the way I teach?"

I then explained I like to hear the same material being explained by different people. Everybody has their unique way of explaining things, even the uniform concepts of Mathematics. This practice served me well too...I got an A in the course! :-)

With that said, I trust 150% of your information Don. I peruse your answers to my questions MANY times until I get the concept. Don't read too much into my double posting. Your answers are meticulous, well explained, and arrive in a swift manner. However, if you want me to cease the practice of double posting then I will.

-MJ

> Can you tell me why you double-post your
> questions both here and then again at
> bj21.com? Are you somehow dissatisfied with
> either the clarity or the speed of the
> responses you get here?

> Don

[Magician]

> Finally, you need a table of the cumulative
> normal distribution ...

No you don't. Using a normal approximation, the probability of being ahead $2000 is exactly zero. Bonus marks for answering the question he meant to ask though. ;-)

[Don Schlesinger]

> No you don't. Using a normal approximation,
> the probability of being ahead $2000 is
> exactly zero. Bonus marks for answering the
> question he meant to ask though. ;-)

Right. I automatically ignored the question he asked, preferring, in fact, to answer the question he meant to ask.

I'll try not to do that too often in the future! :-)

Don

[Don Schlesinger]

> When I was a student in college my Calculus
> Professor once asked me "Why do you
> attend my lecture in the afternoon and then
> sit in on Professor Sydney's class in the
> evening? Is there something wrong with the
> way I teach?"

> I then explained I like to hear the same
> material being explained by different
> people. Everybody has their unique way of
> explaining things, even the uniform concepts
> of Mathematics. This practice served me well
> too...I got an A in the course! :-)

I admire that. Congratulations.

> With that said, I trust 150% of your
> information Don. I peruse your answers to my
> questions MANY times until I get the
> concept. Don't read too much into my double
> posting. Your answers are meticulous, well
> explained, and arrive in a swift manner.

We aim to please! :-)

> However, if you want me to cease the
> practice of double posting then I will.

No, not at all. Feel free to do whatever makes you happy. It's just that, after receiving the answer here, I was hoping you'd be satisfied and find the response adequate enough to not have to ask the same question elsewhere -- especially since you won't always get the right answer! :-)

But, I understand your motivation, and you should continue to do whatever you're comfortable with.

Don

[MJ]

I think I get it. So the probability of winning more or less then $2000 can be calculated. However, the probability of winning EXACTLY $2000 is 0%?

I guess that makes sense intuitively as there are MANY different amounts one can win/lose. Graphically this also makes sense as there would be no area under the curve for an exact win amount. It would be like integrating a function on an interval from a to a. There is no area to compute hence the probability is 0%.

-MJ

> No you don't. Using a normal approximation,
> the probability of being ahead $2000 is
> exactly zero. Bonus marks for answering the
> question he meant to ask though. ;-)