A mischievous demon has captured you and a friend. He's always wanted a pet
human and now he has two. But before locking the cage he decides to play a
game. If you can beat him he'll set you free. If not, you're doomed to live out
your days counting hot and cold particles in a box. This is what he says.
I will be waiting for you in a room with a deck of 52 cards, no Jokers.
Only you may enter, your friend must wait outside. I will deal you five cards which
you can look at. You must choose one card and put it in your pocket. You must then
place each of the remaining four cards in envelopes marked one two three and
four and there is only room for one card in each envelope. You will exit the room
and your friend will enter. By looking at the cards in the envelopes he needs to
guess which card is in your pocket. If he guesses correctly you will both be set
free. And don't even try using other signals like folding or special placing
of the envelopes. He gives the two of you five minutes to
come up with a plan and he's bugged the room and can hear everything you say.
He's a very logical demon and he can choose which five cards you get. If there
is a flaw in your strategy, he will find it. Can you come up with a plan
guaranteed to set you free?
If the demon deals you five cards then at least one of the suits must repeat.
Remember you can choose which card to put in your pocket so you and your
friend agree that you will choose a card with a repeated suit. You then place the
other repeated card in the first envelope and use the remaining three
envelopes to communicate which is the card in your pocket. Let's try with this
hand. Diamonds is the repeated suit so you can choose either the queen or the
four of diamonds. But how do you decide which one? If we take any suit and choose
two cards there will always be a different amount of cards in between
them. This is easy to visualize if we place the cards in a ring. You and your
friend agree that you will always choose the card with the smallest count between
its and the other card, starting from that card and moving in a clockwise
direction. There are 1 2 3 4 cards when counting from the Queen to the 4 and 1 2
3 4 5 6 7 cards when counting from the 4 to the Queen, so you put the queen of
diamonds in your pocket. You can click on the information card in the right hand
corner of the screen to remind yourself which card is in your pocket at any
time throughout the video. We'll call this the mystery card and the 4
remaining cards from the demon's hand the messenger cards. The messenger cards
are the ace of spades, the 10 of Hearts, the 4 of diamonds and the 10 of clubs.
You both agree that the card in the first envelope will have the same suit
as the mystery card, so you place the 4 of diamonds in envelope number 1. Your
friend now knows that the mystery card is from the diamond suit and that one of
the repeated cards is the 4 of diamonds. The mystery card can't be any of these
cards because then the 4 of diamonds had the smallest count between it and
the other card when moving in a clockwise direction, so it would have
been selected as the mystery card. This narrows the search down to just six
cards. You both agree that envelope number two means the mystery card is in the
lower third of the remaining possible cards. Envelope three means it's in the
middle third and envelope four means it's in the upper third. The messenger
card with the highest value will be used to communicate this information, but
which messenger card has the highest value? If we go back card convention ten
is higher than ace but there at two tens. To get around this you assign
value to the suits. For example spades is greater than hearts, hearts is greater than
diamonds and diamonds is greater than clubs. This will safely break any ties
between card values. Because hearts is greater than clubs, the ten of Hearts has
the highest value. You place it in envelope number two to tell your friend
the mystery card is in the lower third, so now your friend knows the mystery
card is either the jack or the queen of diamonds. Continuing this pattern of
communication you both agree that if the card with the next highest value, the ten
of clubs, is placed in the envelope with the lowest value the mystery card is in
the lower half of the remaining two possible cards, and if placed in the
envelope with the highest value it's in the upper half. The mystery card is the
queen of diamonds. This strategy will work for any hand the demon
gives you. If he gives you a hand with two repeated suits, it doesn't matter which one you
choose, this strategy will work equally well with either one. It even works when
there are more than two repeated cards in a hand but I'll leave it up to you to
figure out how. This strategy works because once we've narrowed it down to
one suit we have three messenger cards and 12 or less possible mystery cards, as
there are 13 cards per suit and we know that the other repeated card is not it.
There are three possible envelopes we can place the second messenger card,
two envelopes we can place the third messenger card and one envelope we can
place the last messenger card. This makes a total of six possible arrangements, but
because we have 12 possibilities this isn't enough to get us down to just one
card. By choosing the repeated card with the smallest count we cut the number of
possible mystery cards in half, so there are always six or less possible mystery
cards and six ways to arrange the messenger cards so they will always be
able to point to the mystery card. This will work for a set of 52 of anything, it
doesn't have to be cards. As long as you devise an appropriate ordering system. So
that's it! You're free to go. Looks like this demon will have to spend eternity
counting gas particles alone.