Number 64

August 1991

**EDITORIAL
Rick Rosner
5139 Balboa Blvd. #303
Encino, CA 91316-3430
(818) 986-9177
**

Jeff Ward, who publishes the Mega Society newsletter, and I have been talking with each other and with Ron Hoeflin and Chris Cole. We are going to attempt the merger of the Mega Society with the Noetic Society. A combined society could have a dozen or so more members than the unmerged Noetic Society. After some negotiation, we have decided on the following provisions (about which we want your feedback):

The name of the new, combined organization will be the Mega Society. Jeff Ward feels strongly about keeping the name, and I feel it's short and snappy. The name of this journal might change to something like the Megarian or the Mega Society Journal. Let us know your suggestions and preferences for the name of this publication.

All members who qualified through their score on the Mega or the Titan Test will be protected against future changes in qualifying scores based on renormings of those tests. That is, if the minimum one-in-a-million score would rise from 43 to 45, as it has in the past, people who score 43 or 44 would still be eligible for membership. Jeff Ward feels strongly about this, and I agree. Anyone who is currently a member on the basis of his or her score on a Hoeflin-written test can always continue to be a member.

Chris Cole would like to see membership limited to those who have qualified through a Hoeflin test. He feels the Mega, the Titan, the Ultra and their prototypes are different from any other IQ test. A few Mega members qualified through scores on other tests--the LAIT, childhood Stanford-Binets. We might ask those members to requalify by taking an abridged, non-time-consuming version of a Hoeflin test. People who qualified through non-Hoeflin tests would be welcome to remain contributing subscribers. Or the issue might not come up, if those people, most of whom aren't active, remain dormant. But hey, please don't remain dormant, dammit! The main reason we're merging is to revitalize the combined membership.

Let us know your thoughts on the merger.

Some other semi-related stuff:

We have two new members from Britain, Michael Clive Price (I think--I returned the score report containing his full name, and the rest of his mail says only M. C. Price.) and Peter A. Pomfrit. Both got whopping scores on the Titan. M. C. Price included these biographical details:

age 31, software engineer, physics B.Sc, quantum field theory M.Sc, atheist.

Interests: life extension/immortality, science especially QFT, history, money, mythology.

Dr. Pomfrit says: I am 48 in December, single and my Ph.D was in chemistry. Since an injury to my back I have been a private maths tutor. He also says: When I return from holiday I will send you some verbal analogies which might prove entertaining to the membership.

His letters also bear very attractive postage stamps.

Welcome aboard this somewhat silly vessel.

I think it was Jeff Ward who mentioned that in the past, second attempts were not accepted for Mega membership. I would find it reasonable to accept second attempts (though not third or subsequent attempts) submitted within a reasonable amount of time (and which are not tainted by evidence of cheating). Some people, myself included, would have worked harder on a Hoeflin test to make a good score better had they only known that they were in the right ball park. Of course, two, and soon, three different Hoeflin tests might reduce the need for second attempts. We might have to have a membership officer or committee to determine the eligibility of people who didn't qualify in their first attempt. Again, please let us know what you think!

TRIAL TEST #3

Ronald K. Hoeflin

P.O. Box 539

New York, NY 10101

1. Three-headed : Cerberus :: Many-headed : ?

2. Nemo : Nautilus :: Queeg : ?

3. Judaism : 6 :: Islam : ?

4. Gold miser : Midas :: Labyrinth monster : ?

5. One-eyed : Cyclops :: Two-faced : ?

6. Of ten : Factor :: Of magnitude : ?

7. Wisdom : Solomon :: Patience : ?

8. 100 : Are :: 10,000 : ?

9. 14 lines : Sonnet :: 17 syllables : ?

10. Jupiter : The planets :: Mozart : ?

11. Horse : Bucephalus :: Knot : ?

12. Sound : Phoneme :: Meaning : ?

13. Married : Alimony :: Living together : ?

14. Amphibian : Salamander :: Political district : ?

15. Cow : Moon :: Dish : ?

16. PD : Q :: ASA : ?

17. Them : Us :: Eskimo : ?

18. Riddle : Mystery :: Mystery : ?

19. Bait : Switch :: Tigress : ?

20. Qua Non : Sine :: Pro quo : ?

21. Plus ultra : Ne :: Ne sais quoi : ?

22. FY : I :: AK : ?

23. Crossing the : Rubicon :: Rich as : ?

24. sin x : -cos x :: e : ?

25. For example : e.g. :: Which was to be demonstrated : ?

(Instructions same as for above test--Ed.)

2. April : Showers :: May : ?

3. Fuller : Buckminster :: Dome : ?

4. Of salt : Pillar :: 's wife : ?

5. Horseman : Headless :: Crane : ?

6. A ___ in time saves nine : Stitch :: ___ lips sink ships : ?

7. Low : High :: Woofer : ?

8. 2.54 : Inch :: 454 : ?

9. 2.54 : Inch :: 39.37 : ?

10. 2.54 : Inch :: 3.26 : ?

11. A, AB, B, BO, O : BO :: A, C, G, T, U : ?

12. Pocus : Hocus :: Pokery : ?

13. Thingama : Bob :: Do : ?

14. Head : Stagehand :: Key : ?

15. Errand boy : Gofer :: Lighting electrician : ?

16. C, D, I, K, L, M, V, X : K :: B, E, M, N, R, S, T, U : ?

17. Far : Near :: Stratosphere : ?

18. - 1 : Birdie :: - 2 : ?

19. Eggs : Grading :: Wounded : ?

20. Wedding assistant : Best man :: Movie production assistant : ?

21. Mock : Mach :: Oiler : ?

22. 1951 : Wittgenstein :: 1970 : ?

23. Mom : Dad :: Sire : ?

24. Bird : Cat :: Tweetie : ?

25. 1953 : Stalin :: 1940 : ?

August 23, 1991

Ronald K. Hoeflin

P.O. Box 539

New York, NY 10101

Dear Rick,

Your comments on "Evolution" in *Noesis*
#61 parallel some thoughts I have encountered by others that have impressed me.
Specifically, you remark that the "unfittest" are often obliged to
"blaze new evolutionary trails," particularly by achieving "some
slightly higher level of awareness"--i.e., intelligence.

The alleged racist Nobel Prize winner William Shockley argued, for example,
that studies of the visual acuity of blacks and whites show that blacks have,
as a group, systematically higher visual acuity than whites. Shockley's view was that the systematically
higher intelligence level of whites had enabled them to compensate for their
visual weakness.

Studies of mathematically precocious youth show that the percentage of those
with myopia increases with increased precocity, which seems to support
Shockley's view that there is a connection between intelligence and
compensation for visual weakness.

Rick's remark that "I want narrative evolution--evolution with heroes and
a story line" recalls to mind the following passage from a book on
aesthetics by the American philosopher Stephen C. Pepper (1891-1972), who
taught at the University of California at Berkeley (1919-1958), where he served
as chairman of its Art Department (1938-1953) and of its Philosophy Department
(1953-1958), and who was the author of nearly a dozen books. The following passage occurs on page 109 of
his 1945 book, *The Basis of Criticism in
the Arts*:

The representation of men as they are seen with the normal eye, penetrating to
the traits that count, dwelling seriously on what is serious in life, laughing
at what is silly, and altogether showing through the representation a balanced
view of human values, that is the representation of the norm. That is what we sense in Shakespeare and
Moliere, and Rembrandt, and Breughel, and, in another but not ultimately so
different a way, in Bach and Beethoven.
The representation of the norm is not necessarily in the manner of the
direct, naive conception of the process--namely, a depiction of heroic
man. A caricature can strikingly represent
the norm if the artist's comment is implicit in the picture and tells just how
abnormal the depiction is. And strangely
enough a tragedy can represent the norm best of all. It is entirely natural that the tragic flaw
theory of tragedy had its origin in Aristotelian thought. For how can you most effectively depict the
power of the ideal or normal man? Not by
depicting a thoroughly well adjusted man.
He makes no mistakes, and the full potentialities of his nature are not made
apparent. A weak man, of course, will
not do. But depict a strong man who is
almost normal but has some flaw. He is
over ambitious, he is jealous, he is over reflective and acts too late. In all other respects he has heroic
proportions, but this flaw throws him out of adjustment with his environment
and precipitates a struggle. Then we see
what man can do, and what he is like when he exerts himself to the utmost, and
we become aware of man's complete potentialities by perceiving what he would
have been without the flaw.

Ron Hoeflin

P.S. Regarding recruitment of new
members, the Mega Society has a more widespread reputation than the Noetic
Society due to its mention for many years in the _*Guinness Book of World Records*_ and other reference books as well
as its mention in the issue of _*Omni*_
in which my Mega Test first appeared. As
a result, Jeff Ward, who serves as the membership officer for the Mega Society,
often gets membership inquiries that the Noetic Society does not. For example, Jane V. Clifton, who was the
highest female scorer on the Mega Test when the summary report concerning the
test was published in _*Omni*_ in its
January 1986 issue, recently inquired about membership in the Mega Society
after a lapse of six years. Since Jeff
is a subscriber to _*Noesis*_, it might
be reasonable to ask him to share with us the addresses of new members or
would-be members of the Mega Society. He
did write to me to confirm Ms. Clifton's score on my Mega Test, but he did not
mention her current address to me. I do
include, incidentally, the names and addresses of ten high-IQ societies in my
leaflet for the Top One Percent Society, which I send to everyone who enquires
about or takes the Mega Test or who seeks membership in the Top One Percent
Society. This list includes both the
Mega and Noetic Societies.

P.P.S. Regarding the remark on the last
page of issue 61 that "Feynman issued especially strong cautions against
trying to construct mental pictures of quantum events," I assume most members are aware that this
injunction is now somewhat obsolete, since it is now possible to reduce fairly
large globs of matter to temperatures just millionths of a degree above
absolute zero, at which temperature one can see on the macroscopic level
various quantum phenomena that previously were hidden from view.

Another P.S. I don't see the Editor's
address in issue 61. I think the Editor
should include his address in each issue.
The envelope only gives Chris Cole's address.

Another P.S. Regarding my tests, I have
nearly completed the fourth in a planned series of ten 25-problem trial tests
leading up to a new Mega-like 48-problem test, half verbal and half non-verbal. But my larger aim is to produce a purely
spatial test somewhere down the line that might be more "culture
fair" than my usual tests, or at least fairer for those whose native
language is not English.

Editor's comment:

With tangential regard to William Shockley's theorizing on race and visual
acuity, I've tended to suspect that racial oppression could be a form of false
unfitness which might force some of its victims to higher levels of adaptation. But this could be mistaken for the kind of
thinking that got Jimmy "the Greek" Snyder fired from his sportscasting job. I don't mean to suggest that slavery is
good for people.

What do other members think of the role of oppression in the history of humans
as a whole and races in particular? Are
there analogues to racial strife in the rest of the animal kingdom?

**STEVE SWEENEY'S MOBIUS LETTER
**Dear Rick,

I thought I should finally get around to writing at least a few words on the Mobius strip problem. It's not going to be more than a few words, however, since I'm just about completely burnt out on this problem and have virtually no interest in it anymore. The source of my interest for so long was that I kept thinking, "If I just do this one little thing, I can get a few more pieces, and that'll be the maximum." I now admit defeat, and am quite apathetic about it. First, let me restate the problem, verbatim:

Consider the torus, a doughnut-shaped solid that is perfectly circular at each perpendicular cross section, and a Mobius strip which has a single 180 degree twist and a uniform curvature throughout its length. Suppose a torus is sliced three times by a knife that each time precisely follows the path of such a Mobius strip. What is the maximum number of pieces that can result if the pieces are never moved from their original positions?

I've discussed the problem with, or at least heard from, about twenty different people. Of these approximately twenty people (including One-in-a-Million members and people who solved the cones/cylinder problem), almost everyone had a different interpretation. A few shared the same numerical answer, but had different (often erroneous) ways of getting it.

The above, as stated, is problematic for several reasons. By far the greatest source of controversy is the interpretation of the Mobius strips. Several people searched for a mathematical definition of "curvature" and argued that "uniform curvature" was impossible. Others, such as myself, were content with an intuitive understanding of "uniform curvature." With this interpretation, the Mobius strip is traced by a line segment as it rotates 180 degrees at a constant rate, with its midpoint tracing out a circle, also at a constant rate. This is clearly what Ron Hoeflin had in mind. Even if we accept this interpretation, however, we still do not arrive at a definitive solution.

Another problem involves the interpretation of Hoeflin's torus. Most people did not have any problem with it, but a couple notable people did. The torus is "perfectly circular at each perpendicular cross section." "Perpendicular to what?" asks Chris Langan. This question was not commonly asked, since most of us have ready-made idea of what a torus is. Chris Langan does point out, however, that this is an unsatisfactory definition. He also points out that nothing stated prevents the circular cross sections from varying in diameter [one guy did give an answer of infinity, but I don't know how he got it]. A satisfactory, albeit long-winded definition would be: "a solid whose surface is formed by rotating a circle about an axis such that the circle and the axis are in the same plane, but such that the axis does not intersect the circle."

One or two people even complained about the "knife." It doesn't bother me much, but they do have a point, I suppose. The "knife" would have to be a line segment, or a piece of the Mobius strip. One guy let the knife be a line and got some interesting results. The same result can be achieved if we just use a really long line segment. In this case the Mobius strip intersects itself, which was also an issue of controversy. Is such a figure a Mobius strip? My argument was that it isn't, because it seems that such a figure is four-sided, whereas a Mobius strip should be one-sided. But it is not even clear that such a figure is four-sided. The self-intersection creates a kind of "fence" in the area of intersection. I think both the one-sided and four-sided views are reasonable. [What do members think?]

A plot of submitted answers would surely be multi-modal. The mode was certainly 8, which was your answer and would be correct if we employ the intuitive sense of "uniform curvature," as described above, but with the Mobius strips confined to the interior of the torus. Quite a few people had 12 as an answer. Several people were clustered around 17 (my answer) or 18. From then on we have a decreasingly dense scattergram, with several answers in the twenties, some in the thirties, and a sprinkling above 40. the highest finite answer that I have seen submitted was 117. I should note that most of the solutions I have seen contained errors, even given a particular interpretation of the surfaces. Even erroneous solutions, however, raised legitimate issues, some of which are detailed above.

I guess I finally stopped caring about this problem when it became clear that the solution is almost certainly extremely unaesthetic, that is, really ugly. Since diagrams are not possible in the current format of Noesis, I won't submit any. [I previously said we couldn't do diagrams, but Chris Cole says we can. So send in any diagrams y'all have lying around.--Ed.] I probably wouldn't anyway, since it would take too much time. Let me just say that I'm certain that the answer is 29 or greater, and possibly much more (this is under the popular interpretations of the surfaces). This is achieved by letting each Mobius strip be tangent to the surface of the torus twice. I also let each Mobius strip be wide enough so that it is tangent to its own edge at one point. [If we let it intersect itself, we get a lot more pieces.]

But I don't necessarily think I'm onto the solution here. One guy gave an answer in the thirties with an intriguing approach, but I haven't checked it out thoroughly due to lack of interest; ie. burn-out. This was the only guy who described his surfaces using geometry (he uses cylindrical coordinates). He might have the right idea. Maybe not. But to illustrate a point, he now claims his original answer was too low--now he can get 76 pieces. It just goes on and on.

TANGENTIAL CHALLENGE: If we accept the most popular interpretation of the Mobius strip, but we trace it out with a line (as opposed to a line segment), the set of points of self-intersection trace out something that looks very much like a normal, or Gaussian curve. Can anyone determine whether this curve is normal? What is it?

Future considerations: from our correspondence, it sounds like Ron Hoeflin will scrap this problem entirely, and go back to his original Trial Test "A" version, where the three Mobius strips are confined to the interior of the torus. He will still have to come up with satisfactory definitions of the surfaces, however. My suggestion to him was to state the problem something like this:

Consider a torus, a surface formed by rotating a circle about an axis such that the circle and the axis are in the same plane, but such that the axis does not intersect the circle. An consider a Mobius strip with a single 180 degree twist, formed as follows: take a line segment and let its midpoint trace out a circle at a constant rate, and at the same time, let the line segment rotate 180 degrees, also at a constant rate. Now let three such Mobius strips intersect inside a torus. What is the maximum number of pieces thus formed?

This is the best I could do. I believe that this yields the desired single-digit answer of "8." Do One-in-a-Million members agree? I can foresee a potential problem with the rotation of the line segment, but I don't know how to avoid it without adding several more sentences. The bottom line should be that intelligent people share the same interpretation, and I think they would under my definition, while they do not under Ron Hoeflin's definition.

Of course the problem with this wording is that it is very long and will produce a lot of blank stares. Not ideal for an IQ test. So can any members come up with a shorter and more comprehensible wording for this problem? I'm sure Ron Hoeflin would be interested in seeing it.

Note to members: If you have comments on any of the above, please send them to Rick Rosner for publication in

Take it easy Rick.

Sincerely,

Steve Sweeney

1075 Shadow Hill Way

Beverly Hills, CA 90210

by Chris Cole

I maintain a USENET rec.puzzles Frequently Asked Questions List (FAQL). Rec.puzzles is the USENET news group for recreational puzzles. As might be expected, puzzles come in waves, with certain topics becoming fashionable. For example, the Monty Hall problem was big on the net at the same time Marilyn was getting all her publicity with the problem.

Recently, the classic "two envelope" problem has been hot:

I show you two sealed envelopes and say that one envelope has x dollars and the other has 2x but don't say what x is. It is not possible to determine which envelope has the larger amount without opening them. I hand you one envelope (at random - whatever that means) and say that you have two options:

(1) Open the envelope that you have and keep whatever you find.

(2) Exchange envelopes with me, open the envelope that you receive from the exchange, and keep the amount you find. End of game.

You analyze: I have an unknown amount y in my hand. If I exchange I get y/2 or 2y with equal probability. The expected value of the exchange is (1/2)((y/2) + (2y)) or 1.25y. Expected value from not exchanging is y. Therefore it is better to exchange.

This is clearly wrong. What is wrong? It is not sufficient to answer that the right analysis is something else. The question is what is wrong with the above.

Let's follow the argument carefully, substituting real numbers for variables, to see where we went wrong. In the following, we will assume the envelopes contain $100 and $200. We will consider the two equally likely cases separately, then average the results.

First, take the case that y=$100.

"I have $100 in my hand. If I exchange I get $200. The value of the exchange is $200. The value from not exchanging is $100. Therefore, I gain $100 by exchanging."

Second, take the case that y=$200.

"I have $200 in my hand. If I exchange I get $100. The value of the exchange is $100. The value from not exchanging is $200. Therefore, I lose $100 by exchanging."

Now, averaging the two cases, I see that the expected gain is zero.

So where is the slip up? In one case, switching gets y/2 ($100), in the other case, switching gets 2y ($200), but y is different in the two cases, and I can't simply average the two different y's to get 1.25y. I can average the two numbers ($100 and $200) to get $150, the expected value of switching, which is also the expected value of not switching, but I cannot under any circumstances average y/2 and 2y.

However, a lot of people simply do not relate to this answer. They want an answer involving the probability density function (pdf) of the amount in the envelope. Consider the following solution submitted by Richard Harter:

------------------------------------------------------------------------------------------------------

Let us suppose, per the problem, that we draw a sample L using a given pdf. We place in one envelope the amount $L and in a second $2L. One of the envelopes is picked at random (with selection probabilities of .5) and is opened to reveal that it contains $x. What is the expected value of the other envelope?

Now, given the above information, the other envelope contains either $x/2 or $2x. In the first case L was x/2; in the second it was L. Let d(L) be the pdf in question. Then the probabilities are:

Pr(L=x/2) = d(x/2)/(d(x) + d(x/2)) Pr(L=x) = d(x)/(d(x) + d(x/2))

According the expectation of the contents of the other envelope is

E(other) = Pr(L=x/2)*($x/2) + Pr(L=x)*($x)

And the expected gain from switching is

E(switch) = Pr(L=x)*x - Pr(L=x/2)*(x/2)

Now comes the switcheroo. We do not know what these probabilities are because we do not know what the pdf is. However we do know that the mean of these probabilities is each 1/2. (Follows from condition of the problem.) So we argue that we can replace the unknown probabilities by their means to get

E(switch) = .5*$x - .5*$(x/2) = .25*$x.

Conclusion 1: In any particular instance the claim is not valid because the probabilities for L, although unknown, are not 50%. The observation that the chosen envelope contains $x gives us information, even though we do not know what the information was.

But why is it not correct, on average, to replace the unknown probabilities by their mean values? To do so amounts to the claim that

Mean(Eswitch) = Mean (Pr(L=x)*x) - Mean(Pr(L=x/2)*(x/2))

= Mean(Pr(L=x))*Mean(x) - Mean(Pr(L=x/2)*Mean(x/2)

= .5*Mean(x) -.5*Mean(x/2)

= .25*Mean(x)

The error lies in the second step where the mean of a product is replaced by the product of means.

Conclusion 2: The claim in error on average because it makes the implicit assumption that the mean of a product is the product of means.

Remark: There are a number of interesting results associated with this problem. For example, for a given pdf, there is a number x0 such that it pays to switch if the sample x is less than x0 and not switch if it is greater. For extra credit determine whether x0 is always less than the mean of the pdf, greater than the mean, or dependent on the particular pdf.

------------------------------------------------------------------------------------------------------

As a result of these discussions, I have come to the conclusion that the envelope problem is really four problems rolled into one. Let's consider the cases:

1. You do not know the probability density function (pdf) of the envelope values nor your particular envelope value. Clearly in this case it is irrational to switch, because you have not sampled the envelope and have nothing to go on to decide if the other envelope is likely to be more.

2. You do know the envelope pdf, but not the current contents. Without sampling the envelope, any knowledge about the pdf is useless. After all, what can the pdf tell you about whether you chose the larger or smaller value?

3. You do get to look into the envelope, but you do not know the pdf of the envelope values. You cannot decide whether to switch, because some pdf's would tell you to switch and some would not, and you don't know which kind of pdf you are looking at. You could perhaps perform a lot of experiments to try to estimate the pdf, but on the first try you don't know enough to conclude anything.

4. You do get to look into the envelope and you do know the pdf of the envelope values. Then you can compute an optimal strategy for switching. For example, suppose the envelopes contain $100 and $200. If yours contains $100, you should switch. Otherwise, not.

by Dean Inada

We are on usenet, are a fully registered domain, and can be reached via our internet address:

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postmaster@Peregrine.COM

Or, if you want to use a usenet ! path:

uunet!peregrine

ames!elroy!peregrine

ucsd!peregrine

It is indeed a unix system at Peregrine, Inc.

From our usenet map file:

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# peregrine .peregrine.com peregrine = peregrine.com peregrine altnet(DEAD), ccicpg(HOURLY*2+HIGH), dhw68k(DAILY+HIGH), elroy(DEMAND*3+HIGH), felix(DAILY+HIGH), gordian(DEAD), imt3b2(DEAD), infoexp(DAILY+HIGH), irt(DAILY/2+HIGH), lawnet(DAILY), laxsqnt(DEAD), mbf(DAILY), sceard(EVENING+HIGH), sequent(WEEKLY), silogic(DAILY), skywest(DEAD), spsd(EVENING+HIGH), ucsd(DAILY/4+HIGH), uiucuxc(DAILY+LOW), uunet(EVENING+HIGH), xenix38(WEEKLY+LOW), zardoz(DAILY+HIGH),

(GNU or the Free Software Foundation has little to do with our email system, except that they wrote the editor you've been using to write your mail)

by Rick Rosner

I much prefer the statistics generated by catching people with fake ID's. Having set my 1991 goal at 1,000, I've caught 887 1/2 so far. Of those, about 20 percent have been false ID's obtained through various state departments of motor vehicles by people brazen enough to go to the DMV with fraudulent documents. These are interesting statistically because they contain a date of issue, so that I can tell how long they've been in use. It's hard to tell how long someone has been using a borrowed or altered ID, but it's easy to refer to the issue date on an official-but-fake ID. Given my lazy desperation to publish something in this issue of Noesis, I hereby present "months of use" statistics for a sample of 71 official-but-fake ID's caught from late May through early September of this year.

Numbers on the left represent the number of months an ID has been in circulation when I catch it. Dots represent the number of ID's in each category. Over half of the ID's of this type are returned to their users, so most of these remain in circulation.

A low percentage of this type of ID is detected and confiscated. Probably a much greater proportion are lost, stolen or accidentally destroyed. Most are probably simply outgrown as the user turns 21. Occasionally, someone who is of age loses their real ID (or is afraid to go to the DMV and get a real license) and continues to use their fake ID though they are of age. Sometimes we catch very old fakes used by people who are careless because they are about to turn 21.

My overall rate for misidentifying a real ID as fake is about 50 fake ID's accurately identified for every real ID I misidentify. Another error I could make is misidentifying a borrowed ID as an official-but-fake ID. I am confident that no more than three members of this sample of 71 are included because of misidentification errors.