[Making Ice]

should be in a tray to prevent the tray above it from ruining the cubes below?

Assuming 4, then I would make 1 tray of cubes and then use it to make 3 trays with 4 cubes each and fill those with water with another tray with only water on top.

This will then give us 36 cubes. Again I put in 24 cubes in 6 trays to prop them all up and one tray of water on top.

This gives me 60 cubes in this cycle. Rinse and repeat for 60 cubes in each cycle...

[Line Up]

Firstly, switching entire columns to the left and right does not make any difference to the result and neither does switching entire rows up or down.

Assume x is the guy who is the TOSC (tallest person of the shortest in each column)

Imagine the table looks like below (after I sorted the rows and columns to get x in the bottom right):
A B
x C

A has a bunch of people in the same column as x
C has a bunch of people in the same row as x
B has a bunch of people who are not in the same row or column as x

The SOTR (shortest of tallest person from all the rows) can be someone from A, B, C or x
If SOTR is from A, then that person is taller than x (because x is the shortest from his column)
If SOTR is from B, then that person is taller than the person in the same row from A and everyone in A is taller than x
If SOTR is from C, then that person is taller than x
If SOTR is x, then that person is same height as TOSC.

Therefore the TOSC is always shorter or same height than SOTR.

[Simple classic puzzle: The inheritance]

The standard solution of switching camels I've heard before, so I put myself in the shoes of the old man.

If I were the old man, I would ask them to play tug of war with their camels (if 'crossing' the finish line is a problem, then I would install (by foot) and use a pulley on the other side of the finish line for the tug of war

This does not have anything to do with the question, but I found it interesting that even VERY old questions can be entertaining!

[Many More Discursive Judges]

If we assume P and Q are independent (I declare that they are!), then the subjective likelighood of P being true is 6/9 and for Q, it is 7/9 which implies overall, likelihood of P&Q is 42/81 which makes Jacques guilty in A.

In B and C, Gilles would be guilty by this method. Anyone agree?

[Many More Discursive Judges]

- Two suspects: Jacques & Gilles. It's been proven that one of them is guilty and deserves decapitation, but we do not know which one.

- Jacques is guilty if and only if both statements P and Q are true. Otherwise, Gilles gets the guillotine...

- There are 9 independent judges and an arbiter (you) who interprets their collective decision with no information about the case.

Situation A

- 4 judges think P&Q are true

- 3 believe only P is true and that Q is false

- 2 believe only Q is true and that P is false

Situation B

- 5 judges think both P&Q are true

- 4 judges think P is false and Q is false

Situation C

- 5 judges think P&Q are true

- 2 believe only P is true and that Q is false

- 1 believes only Q is true and that P is false

- 1 believes that P is false and Q is false

What is your verdict for each of these situations? Perhaps this thread needs to move to philosophical banter?

Cheers!

[Discursive Judges]

Actually I feel even there mathematically, the judges are 2/3 sure of P and 2/3 of Q which means they are probabilistically only 4/9 certain of the man's guilt. I think a not guilty decision is fair, but this got me thinking (in another thread...)

[A paradox with derivatives]

Derivative measures the rate at which a given quantity changes with a change in x. When you write x*x as a sum of x x's, one of them becomes a constant. for instance, if x = 5, we start with 25. But if you increase it to 6, x^2 becomes 36, but (x+x+x+x+x) becomes just 30 (with the plus notation, what you differentiate does not mathematically capture the 'x times' part because you do not magically add a d/dx(x) term when you increase x.

[Initial reactions - Lyrics]

But I figured AIK TIA IA IATL is " I know that I am I am I am"

The Luckiest:

I don't get many things right the first time
In fact, I am told that a lot
Now I know all the wrong turns
The stumbles and falls brought me here

And where was I before the day
That I first saw your lovely face?
Now I see it every day

And I know
That I am
I am
I am
The luckiest

[Whether to switch]

Let us assume we knew before drawing that one envelope has $1000 and the other could be either 500 or 2000.

In that case, what is the expected value of the draw? The expected value is then 1125.

In our problem we have now drawn the $1000 envelope. This is disappointing since the expected value is higher. Given the option to switch, it now makes sense to do so!

[Can Infinity Be Less Than One?]

1/n is basically "one divided into n parts"... using this definition, "one divided into an uncountable number of parts" will be naturally close to zero. I think the jump you made was

- infinity is uncountable

- 1/2, 1/3, 1/4.... 1/infinity

- 1/infinity is uncountable (Why? 1/infinity is clearly defined as zero..but let's say I grant you this even if I do not agree)

- infinity is less than zero (Why? just because it is uncountable, it is not infinite )

Cheers!

[Berry's Paradox]

Back here after almost 6 years of absence I used to be known as kingofpain, but I lost the password... I still remember discussions with bonanova, so it only made sense that start off replying to him

An assumption you make is that all numbers are specifiable. This is not proven.

- The list of specifiable numbers with less than 23 syllables is finite.

- "The smallest number not specifiable using fewer than 23 syllables" does not exist

The conclusion here is that if a number is specifiable, it can be specified with less than 23 syllables. In other words, the list of specifiable numbers is finite. PAradox resolved!