
The Slippery Eel of Probability - retupmoc01
https://www.quantamagazine.org/20150730-the-slippery-eel-of-probability/
======
Nadya
_> “You can paint a single sea creature of your choice on each of these
canvases, with one condition. I would like to see a picture of an eel when I
move my gaze vertically or horizontally. So there must be at least one eel in
the vertical column and at least one in the horizontal row.”_

It does not mention each canvas must be a unique sea creature - merely that it
must be single (one sea creature per canvas). I'd paint an eel on every single
canvas, out of laziness.

Introducing a bet to not paint in the corner means I would paint on 5
canvases, not including the corner piece. This changes what I am predicting
though by skewing my decision. It's asking a completely different question. As
catbird pointed out, this is like asking the probability of flipping a coin.
Then asking what if you knew the coin was biased to always land on heads and
you placed a bet with a friend about the result being heads or tails, what
would you vote?

The probability is 1 in 6. Their are 6 canvases and only a single corner
piece. When introducing the bet, do we assume you want to win $2 over paying
$1 100% of the time? If so the probability turns from 1/6 to 0/6 - as the
artist, in their best interest, would never paint an eel in the corner.

The probabilities of 1/4 and 1/3 are asking different questions. It's no
longer asking if the piece is a corner piece but whether it is at a specific
end of a row or column.

It's extremely naive because it doesn't say which end of the row or column
forms the L with the column or row. Therefore both ends are valid and it would
be 2/4 and 2/3 because the eel could be at either end and still form an L.

So it's 0/1, 1/6, 1/3, 1/4, 2/4, and 2/3 depending on the actual question
being asked.

------
hervature
To those saying that the question is essentially asking "predict a coin flip"
without knowing if the coin is biased or not ascribe to the tenants of
Frequentist probability. That is, conclusions cannot be made until trials have
been done. In this specific case, you'd have to ask the artist to paint an
infinitely many L shaped canvases before the true probability is revealed. To
those that ascribe to Bayesian probability, our first assumption is to assume
all valid orientations are part of a uniform distribution. By my counting, I
get 32 valid combinations with an eel in the corner and 21 combinations with
no eel in the corner. This means that I get 32/53 ~= 0.604 as the probability
that the eel is in the corner.

~~~
Nadya
Do you mind explaining how you got 32?

If all 6 paintings have an eel then all 53 combinations of the 6 paintings
would have an eel in the corner, as it is impossible to not have an eel in the
corner if every painting is of an eel.

The solution, if an eel is not in the corner, requires 2 paintings of eels.

It seems you made an assumption that there is only a single eel painting?

~~~
hervature
I believe the correct way to do the counting for this problem is to list the
valid orientations with an eel in the corner and divide by the total amount of
valid orientations. This is correct if you assume that the artist is choosing
uniformly from the set of valid orientations. Let me list a couple of valid
orientations (X marks no eel and O marks an eel). Pretend the first position
is the top of the L and you go down then to the right as you move through my
list (corner is in the 3rd position if 0-indexed):

XXXOXX OXXOXX XOXOXX XXOOXX XXXOOX XXXOXO OOXOXX OXOOXX ... OOOOOO

These are some that are not valid as it doesn't satisfy the owner's
conditions:

XXXXXX OXXXXX OOOXXX XXXXOO

If you wish me to list all the valid combinations I will, but it a simple
matter of combinations.

EDIT: Reformated my answers as HN clobbered my newlines

~~~
Nadya
/Begin Edit

Ignore me I got confused. :D I see what you did. I'll leave the below so that
others might see my confusion. I was confused by your emphasis of "eel in the
corner" because there are valid combinations that lack that.

/End Edit

The issue is that all combinations are valid given the conditions.

 _> You can paint a single sea creature of your choice on each of these
canvases, with one condition. I would like to see a picture of an eel when I
move my gaze vertically or horizontally. So there must be at least one eel in
the vertical column and at least one in the horizontal row._

"At least one" means there can be multiple - and, in fact, without having an
eel in the corner it is necessary to have at least 2 eels. So "a single"
simply means a painting cannot be of an eel _and_ a sea turtle.

Given X = No eel and O = Eel

    
    
      X
      O
      X
      XXO
    

The above is a valid combination. As is the below:

    
    
      O
      O
      O
      XOO
    

Now if you consider:

    
    
      O
      O
      O
      OOO
    

All combinations will result with an eel in the corner and all combinations
are equally valid. All paintings are of a single sea creature (an eel) and
meet the requirement of vertical and horizontal inclusion.

------
catbird
Wow that puzzle is complete nonsense! Either the artist paints an eel in the
corner, or not, is entirely at their own discretion. That's like asking you to
predict a coin flip when you have no info on whether it is biased or not.

~~~
o_nate
The point of the puzzle is to show that there is not a unique way of applying
the Principle of Indifference when there are different ways of enumerating the
possibility space.

