MJ: Question for Don or Norm regarding N0

[MJ]

I understand that N0 is the # hands required to overcome 1 standard deviation of negative variance. Another way to state it is # hands played such that EV = SD.

Now, I was messing around with CVCX and input N0 for my wonging game, which worked out to be 14,888 hands. I figured after I played this many hands, then the probability of my being down 1 or more units should be around 15.86%. The software confirmed my understanding.

Here is where I run into trouble. I figured if I double N0, then the probability of my being down 1 or more units should be around 2.5% b/c expectation is doubled which should overcome 2 SD of negative variance. HOWEVER, CVCX put it around 7%. In fact, according to CVCX it would require around 53,000 hands to have a 2.50% chance of being in the red. That is no where even close to 29,776 hands.

The area under the bell curve at two standard deviations to the left of the mean is about 2.50%. I don't understand why playing double N0 hands would not give the player a 2.50% chance of being behind 1 or more units. Thanks in advance for any answers.

[Don Schlesinger]

> Here is where I run into trouble. I figured if I
> double N0,

No, wrong!! :-)

> then the probability of my being down 1 or
> more units should be around 2.5% b/c expectation is
> doubled which should overcome 2 SD of negative
> variance. HOWEVER, CVCX put it around 7%. In fact,
> according to CVCX it would require around 53,000 hands
> to have a 2.50% chance of being in the red. That is no
> where even close to 29,776 hands.

2N0 needs 2^2 = 4 times the number of hands. 3NO needs 3^2 = 9 times the number of hands. This is an exponential function, not a linear one.

> The area under the bell curve at two standard
> deviations to the left of the mean is about 2.50%.

2.28%.

> I don't understand why playing double N0 hands would not
> give the player a 2.50% chance of being behind 1 or
> more units. Thanks in advance for any answers.

See above. Just work through a simple example. Suppose hourly e.v. = 1 unit and s.d. = 10 units. Then, if you play 100 hours, e.v. = 100 units and s.d. = 100 units, and 100 hands is your N0, since e.v. - s.d. = 0.

Now, how many hours do you need to play to overcome two s.d.s? If you play four times as many hours (400), you quadruple your e.v., but you only double (sqrt of 4) your s.d. So, e.v. is now 400 units, while s.d. is only 200 units. Therefore, 2.s.d.s would be 2 x 200 = 400 units, and 400 hours becomes your 2N0.

Clear?

Don

[MJ]

> Clear?

Crystal. Thanks for the great explanation!

Just one other quick question. How many standard deviations to the left of the mean would yield a 10% area under the bell curve? I know it is somewhere between 1 and 2 SDs, but not sure precisely how to figure this out. I think it has something to do with a z statistic. Thanks again.

MJ

[ES]

1.282

The standardized normal distribution or z distribution is symmetric about the y-axis. Therefore, the tables give areas to the left of values of z for z=0 to z=3.49, at which point the area is 1 to 4 decimal places. The area to the left of 0, or for z= minus infinity to 0, is .5, becaues of symmetry. The area to the left of -z equals the area to the right of z, for positive z, becaues of symmetry. This area equals 1 minus the value in the table becaue the area from minus infinity to infinity is one. The area to the left of 1.28 os .8997; the area to the left of 1.29 is .9015. Proportionately, the area to the left of 1.282 is .9. The area to the right of 1.282 and the area to the left of -1.282 is .1, or 10% of the total area under the curve.

[Don Schlesinger]

> Crystal. Thanks for the great explanation!

You're welcome.

> Just one other quick question. How many standard
> deviations to the left of the mean would yield a 10%
> area under the bell curve? I know it is somewhere
> between 1 and 2 SDs, but not sure precisely how to
> figure this out. I think it has something to do with a
> z statistic. Thanks again.

Thanks to ES for his contribution. Fact is, you don't figure these things out; rather, you look them up in a cumulative normal table, such as those to be found on pp. 147-148 of BJA3. On p. 147, go to the body of the table and look for 0.100 (10%). You will find it at -1.28 s.d.s.

Don

[MJ]

Let us use your assumptions for my example...

"Suppose hourly e.v. = 1 unit and s.d. = 10 units. Then, if you play 100 hours, e.v. = 100 units and s.d. = 100 units, and 100 hands is your N0, since e.v. - s.d. = 0."

If I solve for how many hands it would take to have a 10% chance of being behind, then I would solve for 1.28N0.

1.28^2 = 1.6384 x number of hours (N0) = 1.6384 x 100 = 163.84 hours. After 164 hours our expectation is 164 units. Sq root of 164 hours x 10 equals 128 units for ONE std deviation. Now, 128 units x 1.28 stds = 164 units. So, after playing 164 hours we have accumulated sufficient EV such that there is only a 10% chance of being behind.

Is this correct? :-) Thanks.

MJ

[Don Schlesinger]

> Let us use your assumptions for my example...

> "Suppose hourly e.v. = 1 unit and s.d. = 10
> units. Then, if you play 100 hours, e.v. = 100 units
> and s.d. = 100 units, and 100 hands is your N0, since
> e.v. - s.d. = 0."

OK.

> If I solve for how many hands it would take to have a
> 10% chance of being behind, then I would solve for
> 1.28N0.

No, you would solve for 1.28 s.d.s, which is 1.28^2N0. It isn't N0 that is multiplied by 1.28, because it isn't a linear function. Remember that for TWO s.d.s, you wound up with FOUR N0.

> 1.28^2 = 1.6384 x number of hours (N0) = 1.6384 x 100
> = 163.84 hours. After 164 hours our expectation is 164
> units. Sq root of 164 hours x 10 equals 128 units for
> ONE std deviation. Now, 128 units x 1.28 stds = 164
> units. So, after playing 164 hours we have accumulated
> sufficient EV such that there is only a 10% chance of
> being behind.

Exactly right.

> Is this correct? :-)

A+. :-)

>Thanks.

Any time.

Don

[Don Schlesinger]

By the way, for much more on this, see BJA3, p. 21, especially Table 2.2.

Don

[MJ]

You have shown how to use N0 to figure out how much time is required to overcome "x" std devs.

But now I would like to figure out a variation on this theme. How many hours would it require to be 90% certain of being ahead 400 units? The assumptions are the same as above.

I have been messing around with it and came up with an equation.

If x = number of hours played, then:

x - 10(sqrt(x)) = 400
-10(sqrt(x) = 400 - x
[-10sqrt(x)]^2 = (400 - x)^2
100x = 160,000 - 800x + x^2
0 = x^2 - 900x + 160,000

I don't think this can be factored. BUT, not to worry we can employ the quadratic formula.

Quadratic Formula:
x = -b +/- [Sq Rt(b^2 - 4AC)]/2A

A = 1
b = -900
C = 160,000

If you plug in the variables, you get the following:

x = 900 +/- [sq rt ((-900)^2 - 4(1)(160,000))]/(2x1)
= (900 +/- 412.31)/2

= (900 + 412.31)/2 = 656.15 hours Solution #1
-OR-
= (900 - 412.31)/2 = 243.845 hours Solution #2

Now, let us check our solutions in the original equation:
x - 10(sqrt(x)) = 400

Solution #1
656.15 - 10(sqrt (656.15) = 400 CORRECT SOLUTION

Solution #2
243.845 - 10(sqrt (243.845) = 84 INCORRECT SOLUTION

We can conclude that after playing 656 hands there will be an 84% chance of being ahead 400 or more units. But, that is not was asked! I want 90% certainty of being ahead 400 units. What adjustment do I need to make to my original equation to overcome 1.28 std dev and be ahead 400 units?

Also, would Solution #2 be considered an extraneous solution? Usually with extraneous solutions one solution is positive and the other is negative. Clearly, that is not the case here.

I'll bet there must be a simpler way to solve this without using all this algebra. Don?

MJ