I pick two numbers, randomly, and tell you one of them.

Additional material

[ Editor: Barry Merriman's summary of the replies to his posting of
this game, with an intuitive explanation of why the strategy works. ]

Thanks to everyone who sent comments on my weird game.  With your help,
I have pieced together a simple, intuitive understanding of why the
strange strategy works.  I think even the people who _can't_ do the
three card Monte problem can understand this :-)

Follow me through my explanation; once you understand the principle,
its fun and easy to create your own probability "paradoxes".  As an
example, I create one that is simpler than my original "game".

Recall the game: the dealer has put two real numbers, (U, L with L < U)
in a hat.  You draw one number from the hat---call it x.  You then have
to guess whether x = L or x = U.

[Reminder: You know _absolutely nothing_ about how U, L were chosen: you
don't know their magnitudes, signs, or whether they were chosen
randomly, or whether they were just the ages of the dealer's parents in
minutes, etc.  U, L simply exist, with no assumption about how they were
chosen. ]

The winning strategy was to pick a random number y according to some
probability density p(y) (e.g. Gaussian), and pretend that the number
left in the hat was y, and guess accordingly.  (see original post for
precise details...its a little more complicated, or see the simplified
version later on in this post.)  The mathematics shows that this does
increase your chances of guessing correctly to beyond 50%.  In fact,
your odds are (1/2)*( 1 + P[ L < y < U ])

The amazing thing is that it seems like your odds in the game as stated
should be limited to 50-50.  Since the mathematics is correct, we need
to beef up our intuition.  The key questions:

Q1 "Why can my odds exceed 50%?"

Q2 "What am I really doing by using this prob. density mumbo-jumbo,
    and why does my advantage depend on P[L < y < U]?"

Q3 "How can I see simply what my odds are?"

Q4 "What have we learned from this "paradox"?" <---most important

The Answers:

To get us in a good position to see whats going on, consider this
simpler strategy first:

Simple Strategy:

idea: Pretend that zero lies between L and U, i.e. pretend the picture
is:

                <-------*----0--------*---->
                        L             U

and guess accordingly.

As a strategy, this means

   (a) If I draw x > 0, I guess x = U
   (b) If I draw x < 0, I guess x = L

This is a well defined strategy (even though the idea that led to it
may be false.)

result:

   (1) If 0 is between U, L I guess correctly 100% of the time.
   (2) If 0 is not really between U and L, my guess is as good as
       random, and I am right 50% of the time.

Note: we have here a strategy that (depending on U,L) can increase our
odds of guessing correctly.  This is essentially a model for what
happens in my complex strategy.

We can see clearly why this simple strategy "works": If I do have a
number y that lies between U, L (y=0 in this strategy) then I can figure
out which number I have drawn by comparing it to y: if x < y, I must
have x = L, etc.---I guess right 100% of the time when I have such a
in-between point y.

Because we know nothing about U,L, we don't _know_ such a y for sure,
so we can't have a guaranteed advantage.

Moral: having a y between U,L guarantees you can guess with 100%
accuracy.  Without such y, your accuracy drops to 50%.

Since we need such a y to get an edge, but don't know one, we may as
well pick one randomly---there is some chance that we get a "good" y,
and that will boost our accuracy.  If we get a "bogus" y, our accuracy
stays at 50%.

This leads to the following strategy (essentially the same as the one I
gave originally, but more simply stated):

  (1) Try and guess a y lying between L, U by choosing y randomly (from
      your favorite prob. density, eg Gaussian).

  (2) assume y lies between U and L, and guess about x accordingly; ie
      assume the picture:

          <-------*------y----*------->
                  L           U

      so we guess x = L if x < y, or guess x = U if x > y

The probability of a correct guess using this scheme is easy to find;
the result is:

  (a) If y is between U, L we will guess x correctly
  (b) If y is not, our guess will have 50-50 accuracy:

which leads to the probability of a correct guess being:

P[correct guess about x] = (prob. of y between U, L)*1
                         + (prob of y not between U,L)*(1/2)

                         = P[ L<y<U ]*1
                         + ( 1 - P[ L<y<U ] )*(1/2)

                         = (1/2)*( 1 + P[ L<y<U ] )

Note this is the exact same probability we computed in my previous
post, although there it took a more complicated argument, and it was
not clear why it worked.

So, we can summarize the "magic strategy", and its effects very
intuitively as follows:

Game: L < U are in a hat.  We draw a number from the hat, x, and must
guess if x = U or x = L.

Strategy: we try to guess a y lying between L and U, and determine our
x guess based on the assumption L < y < U.

Result: This assumption is right with probability P[ L < y < U], and in
that fraction of the time, we guess x correctly.  Otherwise, our x guess
has 50% accuracy.

Note:

The Simple Strategy discussed initially (take y = 0) corresponds to the
special case of the general Strategy, in which y is a random variable
concentrated at the origin (---i.e. when you draw y, it is always 0).
So the motivating example I considered fits nicely into the general
scheme.

The answer to the four original questions:

Q1 "Why can my odds exceed 50%?"
A1 "If you know a number between U,L your odds jump to 100%.  You can't
    be sure you have such a number, but you can guess one and have some
    chance of it being so.  This _chance_ of knowing such an in-between
    number increases your odds."

Q2 "What am I really doing by using this prob. density mumbo-jumbo,
    and why does my advantage depend on P[L < y < U]?"
A2 "You are trying to guess a number that lies between L, U.  Drawing a
    number according to some probability distribution is the formal
    method of "guessing".  Since your odds are only enhanced when you
    find y with L < y < U, your odds depend on the probability of this
    being correct.

Q3 "How can I see simply what my odds are?"
A3 "You get 100% accurate x guess if y is between U,L---this occurs
    P[ L < y < U ] of the time.  You otherwise get 50% accuracy in your
    x guess."

Q4 "What have we learned from this "paradox"?" <---most important
A4 "If you can have an advantage by knowing a number y that has
    property X, then you can also have a (lesser) advantage by trying to
    guess such a number.  As long as your guess has some probability of
    having property X, you are be better off (unless mistakenly thinking
    y has X hurts you).

To illustrate our new found insight, here is a much simpler game and
strategy with the same paradoxical features.  You can all go out and
make up your own too---just make use of A4.

Game: the dealer has picked a real number N, and you know nothing about
      it or how he selected it.  You have to guess whether N is
      positive or negative.

"Paradox": You can have a better than 50% chance of guessing the sign
           of N correctly.

Strategy: If we knew a number y between N and 0, we could guess the
          sign of N accurately.  We don't have such a y, but we "guess
          y", by randomly selecting y (from, e.g. a Gaussian
          distribution.)  By our standard argument,

      P[guessing sign of N] = (1/2)* ( 1 + P[ y between N,0])

As long as y has nonzero probability of lying between 0 and N (which it
would if it was chosen with Gaussian density) we have increased our
odds above 50%.

Well, probability is fixed and I can sleep at night again... :-)

-- Barry Merriman