A monastry is filled with lots of monks who meet only once a day at
breakfast all of whom have a vow of silence, which includes no writing
or signing. A cardinal tell them (no vows there) that at least one of
them has been infected by a disease that gives them a red mark on
their head where they can't see (but others can). As the disease is so
awful he tells them that if they have it they must kill themselves the
day they discover they have it.

On the fifth day a number of monks kill themselves. How many and why?

Hi m00se,
This may interest you:
http://physicsforums.com/showthread.php?t=6845&page=5&pp=15

and this :
http://www.creativepuzzels.nl/spel/speel1/puzzel61-2.htm#q11
http://www.creativepuzzels.nl/spel/speel1/370-2.htm#q11

Best regards,
Rainbow

Hi mOOse,

I like this kinda stuff. Without looking at Rainbow's references, (you
will have to trust me here), I think the answer is either 5 or all.
That is, one a day or all of them on the 5th day. Then again, it can't
be one a day, as all kill themselves ON the 5th day. Still not sure
what breakfast has to do with it though. The once a day reference
might lead to the answer.

If they can't see it, I presume they have a mark on the back of their
collective shaved heads. So all others can see, but them. The vow of
silence suggests that no other monk can tell the afflicted that they
are so afflicted. That leaves the Cardinal. But the vow prohibits the
monks from even asking the Cardinal. Then, if the disease is
contagious, they might not have to ask, as the Cardinal should retreat
from them so as not to contract the disease. There is no mention of
how many monks you start with, yet the question asks how many choose
to kill themselves and why. Is the number you start with irrelevant to
the answer?

Ahaa .. another thought. Suicide turns to murder, as all others can
see, excepting those afflicted. So the non-afflicted monks murder the
diseased. Still can't work out how many though.

Are we getting near the answer? I know you want the answer, but what's
the chance of drawing this out for a bit, to see if we can get to it
through a group effort?

Phil

The explanation that was at the physicsforums site was great.
Essentially 5 means 5.

We know that at least one has a mark (given).

If there was a monk (A) that saw no other marks, that monk should kill
himself the first night, regardless of how many other monks exist.

All the other monks will know that monk (A) has a mark, and wait till
night 1 passes to see if monk (A) kills himself. If he does, it is
because monk (A) only sees unmarked people.

Now, if there are two marked monks, monk (A) sees exactly one marked
monk (B), and marked monk (B) sees exactly one marked monk (A). All
the rest of the monks see two marked monks (A) and (B). Monk (A) waits
the first night to see if monk (B) kills himself. Essentially, every
monk waits to see if the two kill themselves. Outside of A and B,
every other monk sees exactly two marked monks. Monk A returns to find
Monk B still alive and marked, which means Monk B still sees at least
one marked monk. Since Monk A doesn't see any other marked monks, Monk
A realizes he is marked. Monk B has the same realization. Monk B
wonders why Monk A didn't kill himself, and realizes that Monk B still
sees at least one marked monk. Since Monk B doesn't see any OTHER
marked monks, Monk B realizes Monk A sees Monk B with a mark. All
other observer monks see only 2 marked monks. Night 2 comes and Monk B
and Monk A kill themselves.

That much is trivial and can be extrapolated.

3 markings:
A sees 2.
B sees 2
C sees 2
observer sees 3

1st night:
no suicides.
A thinks B and C see each other only.
B, C each think the same of the other two.
observers still see 3

2nd night: 
no suicides.
A wonders why, because A only sees 2 and thought that B and C only saw
each other. Why didn't B and C kill themselves? Answer: they saw at
least one other mark. A only sees two marks, and realizes that he must
be the other.
Observers see three marks, and wait.

3rd night:
3 suicides.
Observers breathe a sigh of relief.

Skipping ahead to 5 nights:

A-E see 4 apiece.
1st night:
no suicides. 
A says to himself, B-E saw three. Each of course wishes to exclude
himself from the list. If A saw 4, and assumed that B saw 3, B would
naturally exclude himself looking at C, D, E. A is thinking B would
say, "I see 3. If I am to be excluded, then each of C, D, and E must
see 2. If C excludes himself, C must think that D and E see each
other."
In other words:
A knows there are 4. A thinks B sees 3 only. If A's assumption is
correct, B has the same assumption and thinks C sees 2. If what A
thinks B thinks C says is correct, D must see only E and will wait for
E to kill himself.
 
A sees B  A thinks B sees C  If A is correct, B says 
   ""  C                  D            C must see D
   ""  D                  E                       E
   ""  E          but not A         but not B or A

If B only sees C, D, E then C must only see D and E. 
If C only sees D and E, then D must only see E.

2nd night:
no suicides.
D notices E didn't kill himself, so E must see at least one more mark.
D must realize that there are at least 2.

3rd night.
no suicides
A knows there are 4... D and E didn't kill themselves, so D and E must
notice that there is at least one more (3).

4th night.
no suicides.
C didn't kill himself. A thinks B only sees 3. If A thinks B only sees
3, then why doesn't C kill himself? B must realize that C sees at
least one more than D and E. If B sees 3 and realizes that C sees 3,
then B must realize that B is marked, if A isn't, and A still assumes
that A isn't.

5th night.
A noticed that B (who A thinks only sees 3) didn't kill himself over
the 4th night (A thinks B noticed that C saw at least one more than 2.
If B saw at least one more than 3 (C plus D & E), that means B saw 4.)
Wait a second. A sees 4. In order for B to see at least one more than
3, A must be marked. Each B, C, D, E are also following the same
logic. By this night, each realizes that those they observe must be
seeing 4. At that point, they kill themselves.

Observers are waiting to see if this happens. It doesn't really matter
how many non-marked monks there are. At the point that the 5 didn't
kill themselves on the 5th night, the monk(s) that saw 5 marked monks
would kill himself on the 6th night.

I hope this is intelligible.
Regards.