```This came up at the Webster Christmas-Eve party.  An old chestnut that I'd
explained to my satisfaction ages ago, but I'd forgotten the gist of it and
had to come up with it again.

Three men rent a hotel room for the night for \$30.  The desk clerk finds
that he should have charged the men only \$25.  The desk clerk gives a
bellboy \$5 to give back to the men.  The bellboy, in an effort to make
things easy for people, keeps \$2 and gives the men \$3 which is easily
divided among them.

So, the men have spent \$9 each, which makes \$27.  The bell boy has \$2, so
that adds up to \$29.  Where is the other \$1?

The problem here is not showing that something is wrong somewhere.  The
problem is explaining it so that everyone agrees you've found the error.

My explanation is that you can't add the \$2 the bellboy has to the amount
the men spent because that \$2 is part of what the men spent already.  You're
trying to add it in twice.  What you really ought to say is that the men
spent \$27, the hotel has \$25 of that and the bellboy has \$2.

Thinking about this further, I found the method of reducing things to an
absurd point makes things clearer as well.  I pointed out that since the
method was suspect, you could try out other prices for the room and other
amounts kept by the bellboy (so long as what he returns to the men is
divisible by 3 you understand) and get other weird answers.  The best case
is where the hotel decides to give the room away for free.

Three men rent a hotel room for the night for \$30.  The desk clerk finds
that he should have charged the men only \$0.  The desk clerk gives a bellboy
\$30 to give back to the men.  The bellboy, in an effort to make things easy
for people, keeps \$3 and gives the men \$27 which is easily divided among
them.

So, the men have spent \$1 each, which makes \$3.  The bellboy has \$3, so that
adds up to \$6.  Where is the other \$24?

Stated with those numbers, most everyone is likely to say, "What do you mean
where's the other \$24?  They only spent \$3 and the bellboy has that."  So I
ask then, why did you think you could add it in before?

I've found this method is useful in other situations.  At another family
gathering, our annual trip to the lake, someone brought up an interesting
card "trick" someone had shown them.  You take a deck arranged as it is when
you first take it out of the box and cut it multiple times.  After a certain
number of times you find that the deck is back in it's original order.
People asked why this was.

After a bit of thought, I decided to look at the worst case cut.  If you cut
the cards with just one card you don't get the same result as you do when
you cut roughly in half each time.  From this I could say that there must
not be anything deterministic in the process, and, Aunt B I think it was,
said that yes the number of cuts was not the same all the time.

>From that, looking at the mechanics of the thing closely, you notice that
merely cutting the deck does not change how the cards are arranged in
relation to each other within the halves and enough tries will eventually
bring you back to the start.

While I was in Los Angeles, Lucy and I stopped in at a magic shop outside
the Paramount Studios, I think it's the Paramount Studios right?  The slight
of hand man on call that day had never seen this phenomenon, but he could
create the same effect in a deterministic manner with either a trick deck or
by nudging a card in a regular deck to make a place marker.

I then bought a book on dealing three card Monte which I've read through a
few times but still haven't gotten around to trying out.  I even have the
suggested style of cards, still unwrapped.

```