Psychic Connection: Java program implementation
William Hale
ago that discussed a trick where four faceup cards encoded the
identity of a fifth facedown card. The discussion centered on
how to do such an encoding.
The original article is at:
http://mathforum.org/discuss/sci.math/m/436811/436811
One article with a solution is at:
http://mathforum.org/discuss/sci.math/m/436811/437190
I have implemented an online program in Java that allows you
to perform this trick for another person. The computer acts
as the "encoder" while you act as the "decoder". The url is:
http://www.geocities.com/billh04/poker-trick
The basic effect is as follows. The performer explains to
the spectator the premise of the effect (see below). The
web page with the Java program is called up. While the
performer looks away, the spectator selects five random
cards from a deck of 52 cards on the computer screen.
The computer shuffles the selections, removes the other cards,
and deals the five selections facedown. The spectator tells
the performer that he/she is done. The performer then
turns faceup four of the five facedown cards and,
with a little thought and with optional analysis,
reveals what the fifth facedown card is. This can be repeated.
The original premise was that the performer was psychic.
The premise that I propose is that the performer will
use new developments in mathematical probability theory
to determine what the fifth card is. The performer explains
that people cannot really select five random cards: they
usually leave several "tells" unconsciously. This is the
premise that I will base the explanation given below.
A third premise might be to use "psychology" analysis
of the spectator on what the four turned up cards reveals
about what the fifth remaining facedown card would be.
When you begin, you might also want to explain that your
guess might not be correct, but it should be close. This
will handle any arithmetic errors you might make.
If your "guess" is way off, just congratulate the
spectator's good random choice that time and repeat.
My Java program allows any of these premises to be used,
but I think the second one works best with minimal practice.
It also allows you to have note pages that you can consult
to "do the calculations". But, in fact, these notes
have what the decoding scheme is. Plus, you can
write on the notes if you can't do the calculations
in your head. Of course, there are other mathematical
equations and expressions on these notes that conceal
the actual decoding expressions that I give below.
Here is the method that you need to know, which is
essentially one of the encoding methods explained
in the original "Psychic Connection" thread.
First, you need to know what card to leave facedown
when you start turning four cards faceup. That card will
be the card you will reveal. I call that the hole card.
When you first start the trick, the computer will
place the hole card fourth from the left, counting
from 1. On repeats, the hole card will be placed
in position 1 through 5, depending on what the prior
hole card rank was, subtracting 5 as many times
as needed to get to the range 1-5. For example,
if the prior hole card was QH (rank 12), then
the next hole card will be places second from
the left.
Now, you randomly turn over the cards by clicking on
them, avoiding the hole card. The order that you
turn them over does not matter for what follows.
You want to leave the impression that any card
might have been left facedown as the fifth card
that you will then reveal.
Second, you need to know how to decode the suit.
The suit of the hole card will be the suit of the
first faceup card (from the left).
Third, you need to know how to decode the rank.
The rank of the hole card is determined by the order
of the remaining three faceup cards as follows (where
S denotes the smallest value, M denotes the
middle value, and L denotes the largest value):
-------------------
1 | SML | 10%
2 | SLM | 15%
3 | MSL | 20%
4 | MLS | 25%
5 | LSM | 15%
6 | LMS | 10%
------------------
Note that the third column is just to conceal
what this table is; it is not used at all.
The pattern of the remaining three faceup
cards indicate what row to choose. The
first number in that row is added to the
rank of the first faceup card that indicated
the suit of the hole card. Of course, if the
value you get is greater than 13, you need to
subtract 13 to get within the range of 1-13.
Here, 11 is a jack, 12 is a queen, and 13 is a king.
For example, if the value is 17 then you must
subtract 13 to get 4 as the rank of the hole card.
You might find it easier to take the unit digit 7
and substract 3.
In order to determine the smallest, middle, and
largest cards, you will sometimes have to use
the suit to break ties when the ranks are equal.
The suits are ordered as follows from the
smallest to the largest:
C < D < H < S
You might want to do some private test runs on your own
where you act as performer and spectator so that
you can see how things work. I also have a practice
mode where only the second phase is repeated with
five different random cards each time. To enter
the practice mode, hold down the shift key and
the control key simultaneously and then click
on any one of the 52 faceup cards in the first
phase of the trick where the spectator makes
his/her selections.
-- Bill Hale
Eric Osman
At first it looked like the solution failed, since the 3 cards used
for encoding the rank only allow for encoding 6 values, and not counting
the rank of the exposed card, there are still 12 values to choose from.
However, I soon realized that critical to your solution is that a
consistent scheme be chosen for which of the two cards of the chosen
suit be used for the hole card.
Since the encoding of the 3 cards gives a value of 1 through 6, for
example, if the two cards of the hole suit are an ace and an eight,
it's critical to use the eight in the hole and encode a 6 (which when
added to 8 gives 14, and then we subtract 13 to get 1) .
If instead, the ace were in the hole, we'd be in trouble, since to
get 8, we've have to encode 7 which we can't do.
I suggest the above be included in your revised write-up !
/Eric
Eric Osman
Ok, so with a 52 card deck, exposing 4 cards allow the second magician
to deduce the card in the hole.
I suspect 52 isn't the largest deck with which this is possible.
Is there an easy way to calculate the largest deck for which it is possible ?
/Eric os7...@attbi.com 10/2/02
William Hale
Look at this url:
http://people.brandeis.edu/~kleber/Papers/card.pdf
If you don't have the Acrobat reader, you can look at this url:
-- Bill Hale
Eric Osman
Thanks for mentioning that fascinating paper :
http://people.brandeis.edu/~kleber/Papers/card.pdf
For those of you not yet fascinated by this cool card trick,
here's a mini warmup puzzle:
You use a 3-card deck, with the cards labeled 1, 2, and 3.
The audience member picks 2 of the cards and hands them to your
"lovely" assistant, who then shows you ONE of the 2 cards.
You then announce the other of the 2 cards.
How is it done ?
A slightly more difficult puzzle, in two parts:
part I : What is the largest size deck of cards from which the
audience member can pick 3 cards, and then your
lovely assistant shows you 2 of them and you announce the
3rd ?
part II : How is it done ?
/Eric