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Frequently Asked Questions (FAQS);faqs.458



For those who would prefer to see this worked out in detail:
Assume the smaller envelope is uniform on [$0,$M], for some value
of $M.  What is the expectation value of always switching?  A quarter of
the time $100 >= $M (i.e. 50% chance $X is in [$M/2,$M] and 50% chance
the larger envelope is chosen).  In this case the expected switching
gain is -$50 (a loss).  Thus overall the always switch policy has an
expected (relative to $100) gain of (3/4)*$50 + (1/4)*(-$50) = $25.
However the expected absolute gain (in terms of M) is:
  / M
  |    g f(g) dg,   [ where f(g) = (1/2)*Uniform[0,M)(g) +
  /-M                              (1/2)*Uniform(-M,0](g). ]

  = 0.  QED.

OK, so always switching is not the optimal switching strategy.  Surely
there must be some strategy that takes advantage of the fact that we
looked into the envelope and we know something we did not know before
we looked.

Well, if we know the maximum value $M that can be in the smaller envelope,
then the optimal decision criterion is to switch if $100 < $M, otherwise stick.
The reason for the stick case is straightforward. The reason for the
switch case is due to the pdf of the smaller envelope being twice as
high as that of the larger envelope over the range [0,$M). That is, the
expected gain in switching is (2/3)*$100 + (1/3)*(-$50) = $50.

What if we do not know the maximum value of the pdf?  You can exploit
the "test value" technique to improve your chances.  The trick here is
to pick a test value T.  If the amount in the envelope is less than the
test value, switch; if it is more, do not.  This works in that if T happens
to be in the range [M,2M] you will make the correct decision.  Therefore,
assuming the unknown pdf is uniform on [0,M], you are slightly better off
with this technique.

Of course, the pdf may not even be uniform, so the "test value" technique
may not offer much of an advantage.  If you are allowed to play the game
repeatedly, you can estimate the pdf, but that is another story...

==> decision/exchange.p <==
At one time, the Mexican and American dollars were devalued by 10 cents on each
side of the border (i.e. a Mexican dollar was 90 cents in the US, and a US
dollar was worth 90 cents in Mexico).  A man walks into a bar on the American
side of the border, orders 10 cents worth of beer, and tenders a Mexican dollar
in change.  He then walks across the border to Mexico, orders 10 cents worth of
beer and tenders a US dollar in change.  He continues this throughout the day,
and ends up dead drunk with the original dollar in his pocket.

Who pays for the drinks?

==> decision/exchange.s <==
The man paid for all the drinks.  But, you say, he ended up with the same
amount of money that he started with!  However, as he transported Mexican
dollars into Mexico and US dollars into the US, he performed "economic work"
by moving the currency to a location where it was in greater demand (and thus
valued higher).  The earnings from this work were spent on the drinks.

Note that he can only continue to do this until the Mexican bar runs out
of US dollars, or the US bar runs out of Mexican dollars, i.e., until
he runs out of "work" to do.

==> decision/newcomb.p <==
Newcomb's Problem

A being put one thousand dollars in box A and either zero or one million
dollars in box B and presents you with two choices:
    (1) Open box B only.
    (2) Open both box A and B.
The being put money in box B only if it predicted you will choose option (1).
The being put nothing in box B if it predicted you will do anything other than
choose option (1) (including choosing option (2), flipping a coin, etc.).

Assuming that you have never known the being to be wrong in predicting your
actions, which option should you choose to maximize the amount of money you
get?


==> decision/newcomb.s <==
This is "Newcomb's Paradox".

You are presented with two boxes: one certainly contains $1000 and the
other might contain $1 million.  You can either take one box or both.
You cannot change what is in the boxes.  Therefore, to maximize your
gain you should take both boxes.

However, it might be argued that you can change the probability that
the $1 million is there.  Since there is no way to change whether the
million is in the box or not, what does it mean that you can change
the probability that the million is in the box?  It means that your
choice is correlated with the state of the box.

Events which proceed from a common cause are correlated.  My mental
states lead to my choice and, very probably, to the state of the box.
Therefore my choice and the state of the box are highly correlated.
In this sense, my choice changes the "probability" that the money is
in the box.  However, since your choice cannot change the state of the
box, this correlation is irrelevant.

The following argument might be made: your expected gain if you take
both boxes is (nearly) $1000, whereas your expected gain if you take
one box is (nearly) $1 million, therefore you should take one box.
However, this argument is fallacious.  In order to compute the
expected gain, one would use the formulas:

    E(take one) =    $0 * P(predict take both | take one) +
                $1,000,000 * P(predict take one | take one)
    E(take both) =    $1,000 * P(predict take both | take both) +
                $1,001,000 * P(predict take one | take both)

While you are given that P(do X | predict X) is high, it is not given
that P(predict X | do X) is high.  Indeed, specifying that P(predict X
| do X) is high would be equivalent to specifying that the being could
use magic (or reverse causality) to fill the boxes.  Therefore, the
expected gain from either action cannot be determined from the
information given.


==> decision/prisoners.p <==
Three prisoners on death row are told that one of them has been chosen
at random for execution the next day, but the other two are to be
freed.  One privately begs the warden to at least tell him the name of
one other prisoner who will be freed.  The warden relents: 'Susie will
go free.'  Horrified, the first prisoner says that because he is now
one of only two remaining prisoners at risk, his chances of execution
have risen from one-third to one-half!  Should the warden have kept his
mouth shut?

==> decision/prisoners.s <==
Each prisoner had an equal chance of being the one chosen to be
executed.  So we have three cases:

Prisoner executed:         A    B    C
Probability of this case: 1/3  1/3  1/3

Now, if A is to be executed, the warden will randomly choose either B or C,
and tell A that name.  When B or C is the one to be executed, there is only
one prisoner other than A who will not be executed, and the warden will always
give that name.  So now we have:

Prisoner executed:  A    A    B    C
Name given to A:    B    C    C    B
Probability:       1/6  1/6  1/3  1/3

We can calculate all this without knowing the warden's answer.
When he tells us B will not be executed, we eliminate the middle two
choices above.  Now, among the two remaining cases, C is twice
as likely as A to be the one executed.  Thus, the probability that
A will be executed is still 1/3, and C's chances are 2/3.

Xref: bloom-picayune.mit.edu rec.puzzles:18140 news.answers:3072
Newsgroups: rec.puzzles,news.answers
Path: bloom-picayune.mit.edu!snorkelwacker.mit.edu!usc!sdd.hp.com!elroy.jpl.nasa.gov!ames!haven.umd.edu!uunet!questrel!chris
From: uunet!questrel!chris (Chris Cole)
Subject: rec.puzzles FAQ, part 5 of 15
Message-ID: <[e-mail]>
Followup-To: rec.puzzles
Summary: This posting contains a list of
     Frequently Asked Questions (and their answers).
     It should be read by anyone who wishes to
     post to the rec.puzzles newsgroup.
Sender: [e-mail] (Chris Cole)
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Archive-name: puzzles-faq/part05
Last-modified: 1992/09/20
Version: 3

==> decision/red.p <==
I show you a shuffled deck of standard playing cards, one card at a
time.  At any point before I run out of cards, you must say "RED!".
If the next card I show is red (i.e. diamonds or hearts), you win.  We
assume I the "dealer" don't have any control over what the order of
cards is.

The question is, what's the best strategy, and what is your
probability of winning ?

==> decision/red.s <==
If a deck has n cards, r red and b black, the best strategy wins
with a probability of r/n.  Thus, you can say "red" on the first card,
the last card, or any other card you wish.
Proof by induction on n.  The statement is clearly true for one-card decks.
Suppose it is true for n-card decks, and add a red card.
I will even allow a nondeterministic strategy, meaning you say "red"
on the first card with probability p.  With probability 1-p,
you watch the first card go by, and then apply the "optimal" strategy
to the remaining n-card deck, since you now know its composition.
The odds of winning are therefore: p * (r+1)/(n+1)  +
        (1-p) * ((r+1)/(n+1) * r/n  +  b/(n+1) * (r+1)/n).
After some algebra, this becomes (r+1)/(n+1) as expected.
Adding a black card yields: p * r/(n+1)  +
        (1-p) * (r/(n+1) * (r-1)/n  +  (b+1)/(n+1) * r/n).
This becomes r/(n+1) as expected.

==> decision/rotating.table.p <==
Four glasses are placed upside down in the four corners of a square
rotating table.  You wish to turn them all in the same direction,
either all up or all down.  You may do so by grasping any two glasses
and, optionally, turning either over.  There are two catches:  you are
blindfolded and the table is spun after each time you touch the
glasses.  How do you do it?
==> decision/rotating.table.s <==
1.  Turn two adjacent glasses up.
2.  Turn two diagonal glasses up.
3.  Pull out two diagonal glasses.  If one is down, turn it up and you're done.
    If not, turn one down and replace.
4.  Take two adjacent glasses.  Invert them both.
5.  Take two diagonal glasses.  Invert them both.

References
    Probing the Rotating Table"
    W. T. Laaser and L. Ramshaw
    _The Mathematical Gardner_,
    Wadsworth International, Belmont CA 1981.

    ... we will see that such a procedure exists if and
    only if the parameters k and n satisfy the inequality
    k >= (1-1/p)n, where p is the largest prime factor
    of n.

The paper mentions (without discussing) two other generalizations:
more than two orientations of the glasses (Graham and Diaconis)
and more symmetries in the table, e.g. those of a cube (Kim).

==> decision/stpetersburg.p <==
What should you be willing to pay to play a game in which the payoff is
calculated as follows:  a coin is flipped until in comes up heads on the
nth toss and the payoff is set at 2^n dollars?

==> decision/stpetersburg.s <==
Classical decison theory says that you should be willing to pay any
amount up to the expected value of the wager.  Let's calculate the
expected value:  The probability of winning at step n is 2^-n, and the
payoff at step n is 2^n, so the sum of the products of the
probabilities and the payoffs is:

    E = sum over n (2^-n * 2^n) = sum over n (1) = infinity

So you should be willing to pay any amount to play this game.  This is
called the "St. Petersburg Paradox."

The classical solution to this problem was given by Bernoulli.  He
noted that people's desire for money is not linear in the amount of
money involved.  In other words, people do not desire $2 million twice
as much as they desire $1 million.  Suppose, for example, that people's
desire for money is a logarithmic function of the amount of money.
Then the expected VALUE of the game is:

    E = sum over n (2^-n * C * log(2^n)) = sum over n (2^-n * C' * n) =  C''

Here the C's are constants that depend upon the risk aversion of the
player, but at least the expected value is finite.  However, it turns
out that these constants are usually much higher than people are really
willing to pay to play, and in fact it can be shown that any
non-bounded utility function (map from amount of money to value of
money) is prey to a generalization of the St. Petersburg paradox.  So
the classical solution of Bernoulli is only part of the story.

The rest of the story lies in the observation that bankrolls are always
finite, and this dramatically reduces the amount you are willing to bet
in the St. Petersburg game.

To figure out what would be a fair value to charge for playing the game
we must know the bank's resources.  Assume that the bank has 1 million
dollars (1*K*K = 2^20).  I cannot possibly win more than $1 million
whether I toss 20 tails in a row or 2000.

Therefore my expected amount of winning is

    E = sum n up to 20 (2^-n * 2^n) = sum n up to 20 (1) = $20

and my expected value of winning is

    E = sum n up to 20 (2^-n * C * log(2^n)) = some small number

This is much more in keeping with what people would really pay to
play the game.

Incidentally, T.C. Fry suggested this change to the problem in 1928
(see W.W.R. Ball, Mathematical Recreations and Essays, N.Y.: Macmillan,
1960, pp.  44-45).

The problem remains interesting when modified in this way,
for the following reason. For a particular value of the bank's
resources, let

      e denote the expected value of the player's winnings; and let
      p denote the probability that the player profits from the game, assuming
        the price of getting into the game is 0.8e (20% discount).

Note that the expected value of the player's profit is 0.2e.  Now
let's vary the bank's resources and observe how e and p change.  It
will be seen that as e (and hence the expected value of the profit)
increases, p diminishes.  The more the game is to the player's
advantage in terms of expected value of profit, the less likely it is
that the player will come away with any profit at all.  This
is mildly counterintuitive.

==> decision/switch.p <==
Switch? (The Monty Hall Problem)

Two black marbles and a red marble are in a bag. You choose one marble from the
bag without looking at it. Another person chooses a marble from the bag and it
is black. You are given a chance to keep the marble you have or switch it with
the one in the bag. If you want to end up with the red marble, is there an
advantage to switching? What if the other person looked at the marbles remaining
in the bag and purposefully selected a black one?

==> decision/switch.s <==
Generalize the problem from three marbles to n marbles.

If there are n marbles, your odds of having selected the red one are 1/n. After
the other person selected a black one at random, your odds go up to 1/(n-1).
There are n-2 marbles left in the bag, so your odds of selecting the red one
by switching are 1/(n-2) times the odds that you did not already select it
(n-2)/(n-1) or 1/(n-1), the same as the odds of already selecting it. Therefore
there is no advantage to switching.

If the person looked into the bag and selected a black one on purpose, then
your odds of having selected the red one are not improved, so the odds of
selecting the red one by switching are 1/(n-2) times (n-1)/n or (n-1)/n(n-2).
This is (n-1)/(n-2) times better than the odds without switching, so you
should switch.

This is a clarified version of the Monty Hall "paradox":

You are a participant on "Let's Make a Deal." Monty Hall shows you
three closed doors.  He tells you that two of the closed doors have a
goat behind them and that one of the doors has a new car behind it.
You pick one door, but before you open it, Monty opens one of the two
remaining doors and shows that it hides a goat.  He then offers you a
chance to switch doors with the remaining closed door.  Is it to your
advantage to do so?

The original Monty Hall problem (and solution) appears to be due to
Steve Selvin, and appears in American Statistician, Feb 1975, V. 29,
No. 1, p. 67 under the title ``A Problem in Probability.''  It should
be of no surprise to readers of this group that he received several
letters contesting the accuracy of his solution, so he responded two
issues later (American Statistician, Aug 1975, V. 29, No. 3, p. 134).
I extract a few words of interest, including a response from Monty
Hall himself:

   ...  The basis to my solution is that Monty Hall knows which box
   contains the prize and when he can open either of two boxes without
   exposing the prize, he chooses between them at random ...

   Benjamin King pointed out the critical assumptions about Monty
   Hall's behavior that are necessary to solve the problem, and
   emphasized that ``the prior distribution is not the only part of
   the probabilistic side of a decision problem that is subjective.''

   Monty Hall wrote and expressed that he was not ``a student of
   statistics problems'' but ``the big hole in your argument is that
   once the first box is seen to be empty, the contestant cannot
   exchange his box.''  He continues to say, ``Oh, and incidentally,
   after one [box] is seen to be empty, his chances are not 50/50 but
   remain what they were in the first place, one out of three.  It
   just seems to the contestant that one box having been eliminated,
   he stands a better chance.  Not so.''  I could not have said it
   better myself.

The basic idea is that the Monty Hall problem is confusing for two
reasons:  first,  there are hidden assumptions about Monty's motivation
that cloud the issue in some peoples' minds; and second, novice probability
students do not see that the opening of the door gave them any new
information.

Monty can have one of three basic motives:
1.  He randomly opens doors.
2.  He always opens the door he knows contains nothing.
3.  He only opens a door when the contestant has picked the grand prize.

These result in very different strategies:
1.  No improvement when switching.
2.  Double your odds by switching.
3.  Don't switch!

Most people, myself included, think that (2) is the intended
interpretation of Monty's motive.

A good way to see that Monty is giving you information by opening doors is to
increase the number of doors from three to 100.  If there are 100 doors,
and Monty shows that 98 of them are empty, isn't it pretty clear that
the chance the prize is behind the remaining door is 99/100?

Reference (too numerous to mention, but this one should do):
    Leonard Gillman
    "The Car and the Goats"
    The American Mathematical Monthly, 99:1 (Jan 1992), pp. 3-7.

==> decision/truel.p <==
A, B, and C are to fight a three-cornered pistol duel.  All know that
A's chance of hitting his target is 0.3, C's is 0.5, and B never misses.
They are to fire at their choice of target in succession in the order
A, B, C, cyclically (but a hit man loses further turns and is no longer
shot at) until only one man is left.  What should A's strategy be?

==> decision/truel.s <==
This is problem 20 in Mosteller _Fifty Challenging Problems in Probability_
and it also appears (with an almost identical solution) on page 82 in
Larsen & Marx _An Introduction to Probability and Its Applications_.

Here's Mosteller's solution:

  A is naturally not feeling cheery about this enterprise.  Having the
first shot he sees that, if he hits C, B will then surely hit him, and
so he is not going to shoot at C.  If he shoots at B and misses him,
then B clearly {I disagree; this is not at all clear!} shoots the more
dangerous C first, and A gets one shot at B with probability 0.3 of
succeeding.  If he misses this time, the less said the better.  On the
other hand, suppose A hits B.  Then C and A shoot alternately until one
hits.  A's chance of winning is (.5)(.3) + (.5)^2(.7)(.3) +
(.5)^3(.7)^2(.3) + ... .  Each term cooresponds to a sequence of misses
by both C and A ending with a final hit by A.  Summing the geometric
series we get ... 3/13 < 3/10.  Thus hitting B and finishing off with
C has less probability of winning for A than just missing the first shot.
So A fires his first shot into the ground and then tries to hit B with
his next shot.  C is out of luck.

As much as I respect Mosteller, I have some serious problems with this
solution.  If we allow the option of firing into the ground, then if
all fire into the ground with every shot, each will survive with
probability 1.  Now, the argument could be made that a certain
strategy for X that both allows them to survive with probability 1
*and* gives less than a probability of survival of less than 1 for
at least one of their foes would be preferred by X.  However, if
X pulls the trigger and actually hits someone what would the remaining
person, say Y, do?  If P(X hits)=1, clearly Y must try to hit X, since
X firing at Y with intent to hit dominates any other strategy for X.
If P(X hits)<1 and X fires at Y with intent to hit, then
P(Y survives)<1 (since X could have hit Y).  Thus, Y must insure that
X can not follow this strategy by shooting back at X (thus insuring
that P(X survives)<1).  Therefore, I would conclude that the ideal
strategy for all three players, assuming that they are rational and
value survival above killing their enemies, would be to keep firing
into the ground.  If they don't value survival above killing their
enemies (which is the only a priori assumption that I feel can be
safely made in the absence of more information), then the problem
can't be solved unless the function each player is trying to maximize
is explicitly given.
--
    -- [e-mail] (Chris Long)

OK - I'll have a go at this.

How about the payoff function being 1 if you win the "duel" (i.e. if at some
point you are still standing and both the others have been shot) and 0
otherwise? This should ensure that an infinite sequence of deliberate misses
is not to anyone's advantage. Furthermore, I don't think simple survival
makes a realistic payoff function, since people with such a payoff function
would not get involved in the fight in the first place!

[ I.e. I am presupposing a form of irrationality on the part of the
  fighters: they're only interested in survival if they win the duel. Come
  to think of it, this may be quite rational - spending the rest of my life
  firing a gun into the ground would be a very unattractive proposition to
  me :-)
]

Now, denote each position in the game by the list of people left standing,
in the order in which they get their turns (so the initial position is
(A,B,C), and the position after A misses the first shot (B,C,A)). We need to
know the value of each possible position for each person.

By definition:

    valA(A) = 1            valB(A) = 0            valC(A) = 0
    valA(B) = 0            valB(B) = 1            valC(B) = 0
    valA(C) = 0            valB(C) = 0            valC(C) = 1

Consider the two player position (X,Y). An infinite sequence of misses has
value zero to both players, and each player can ensure a positive payoff by
trying to shoot the other player. So both players deliberately missing is a
sub-optimal result for both players. The question is then whether both
players should try to shoot the other first, or whether one should let the
other take the first shot. Since having the first shot is always an
advantage, given that some real shots are going to be fired, both players
should try to shoot the other first. It is then easy to establish that:

    valA(A,B) = 3/10       valB(A,B) = 7/10       valC(A,B) = 0
    valA(B,A) = 0          valB(B,A) = 1          valC(B,A) = 0
    valA(B,C) = 0          valB(B,C) = 1          valC(B,C) = 0
    valA(C,B) = 0          valB(C,B) = 5/10       valC(C,B) = 5/10
    valA(C,A) = 3/13       valB(C,A) = 0          valC(C,A) = 10/13
    valA(A,C) = 6/13       valB(A,C) = 0          valC(A,C) = 7/13

Now for the three player positions (A,B,C), (B,C,A) and (C,A,B). Again, the
fact that an infinite sequence of misses is sub-optimal for all three
players means that at least one player is going to decide to fire. However,
it is less clear than in the 2 player case that any particular player is
going to fire. In the 2 player case, each player knew that *if* it was
sub-optimal for him to fire, then it was optimal for the other player to
fire *at him* and that he would be at a disadvantage in the ensuing duel
because of not having got the first shot. This is not necessarily true in
the 3 player case.

Consider the payoff to A in the position (A,B,C). If he shoots at B, his
expected payoff is:

    0.3*valA(C,A) + 0.7*valA(B,C,A) = 9/130 + 0.7*valA(B,C,A)

If he shoots at C, his expected payoff is:

    0.3*valA(B,A) + 0.7*valA(B,C,A) = 0.7*valA(B,C,A)

And if he deliberately misses, his expected payoff is:

    valA(B,C,A)

Since he tries to maximise his payoff, we can immediately eliminate shooting
at C as a strategy - it is strictly dominated by shooting at B. So A's
expected payoff is:

    valA(A,B,C) = MAX(valA(B,C,A), 9/130 + 0.7*valA(B,C,A))

A similar argument shows that C's expected payoffs in the (C,A,B) position are:

    For shooting at A: 0.5*valC(A,B,C)
    For shooting at B: 35/130 + 0.5*valC(A,B,C)
    For missing:       valC(A,B,C)

So C either shoots at B or deliberately misses, and:

    valC(C,A,B) = MAX(valC(A,B,C), 35/130 + 0.5*valC(A,B,C))

Each player can obtain a positive expected payoff by shooting at one of the
other players, and it is known that an infinite sequence of misses will
result in a zero payoff for all players. So it is known that some player's
strategy must involve shooting at another player rather than deliberately
missing.

Now look at this from the point of view of player B. He knows that *if* it
is sub-optimal for him to shoot at another player, then it is optimal for at
least one of the other players to shoot. He also knows that if the other
players choose to shoot, they will shoot *at him*. If he deliberately
misses, therefore, the best that he can hope for is that they miss him and
he is presented with the same situation again. This is clearly less good for
him than getting his shot in first. So in position (B,C,A), he must shoot at
another player rather than deliberately miss.

B's expected payoffs are:

    For shooting at A: valB(C,B) = 5/10
    For shooting at C: valB(A,B) = 7/10

So in position (B,C,A), B shoots at C for an expected payoff of 7/10. This
gives us:

    valA(B,C,A) = 3/10     valB(B,C,A) = 7/10     valC(B,C,A) = 0

So valA(A,B,C) = MAX(3/10, 9/130 + 21/100) = 3/10, and A's best strategy is
position (A,B,C) is to deliberately miss, giving us:

    valA(A,B,C) = 3/10     valB(A,B,C) = 7/10     valC(A,B,C) = 0

And finally, valC(C,A,B) = MAX(0, 35/130 + 0) = 7/26, and C's best strategy
in position (C,A,B) is to shoot at B, giving us:

    valA(C,A,B) = 57/260   valB(C,A,B) = 133/260  valC(C,A,B) = 7/26

I suspect that, with this payoff function, all positions with 3 players can
be resolved. For each player, we can establish that if their correct
strategy is to fire at another player, then it is to fire at whichever of
the other players is more dangerous. The most dangerous of the three players
then finds that he has nothing to lose by firing at the second most
dangerous.

Questions:

(a) In the general case, what are the optimal strategies for the other two
    players, possibly as functions of the hit probabilities and the cyclic
    order of the three players?

(b) What happens in the 4 or more player case?

    -- David Seal <[e-mail]>

==> english/acronym.p <==
What acronyms have become common words?

==> english/acronym.s <==
The following is the list of acronyms which have become common nouns.
An acronym is "a word formed from the initial letter or letters of each
of the successive parts or major parts of a compound term" (Webster's Ninth).
A common noun will occur uncapitalized in Webster's Ninth.

Entries in the following table include the year in which they first
entered the language (according to the Ninth), and the Merriam-Webster
dictionary that first contains them.  The following symbols are used:

NI1    New International (1909)
NI1+    New Words section of the New International (1931)
NI2    New International Second Edition (1934)
NI2+    Addendum section of the Second (1959, same as 1954)
NI3    Third New International (1961)
9C    Ninth New Collegiate (1983)
12W    12,000 Words (separately published addendum to the Third, 1986)

asdic    Anti-Submarine Detection Investigation Committee (1940, NI2+)
dew    Distant Early Warning (1953, 9C)
dopa    DihydrOxyPhenylAlanine (1917, NI3)
fido    Freaks + Irregulars + Defects + Oddities (1966, 9C)
jato    Jet-Assisted TakeOff (1947, NI2+)
laser    Light Amplification by Stimulated Emission of Radiation (1957, NI3)
lidar    LIght Detection And Ranging (1963, 9C)
maser    Microwave Amplification by Stimulated Emission of Radiation (1955, NI3)
nitinol    NIckel + TIn + Naval Ordinance Laboratory (1968, 9C)
rad    Radiation Absorbed Dose (1918, NI3)
radar    RAdio Detection And Ranging (ca. 1941, NI2+)
rem    Roentgen Equivalent Man (1947, NI3)
rep    Roentgen Equivalent Physical (1947, NI3)
scuba    Self-Contained Underwater Breathing Apparatus (1952, NI3)
snafu    Situation Normal -- All Fucked (Fouled) Up (ca. 1940, NI2+)
sofar    SOund Fixing And Ranging (1946, NI2+)
sonar    SOund NAvigation Ranging (1945, NI2+)
tepa    Tri-Ethylene Phosphor-Amide (1953, 9C)
zip    Zone Improvement Plan (1963, 9C)