[xquery-talk] Izzit Bcos I is functional?

Ihe Onwuka ihe.onwuka at gmail.com
Sun Jul 5 00:12:19 PDT 2015

On Thu, Jun 18, 2015 at 12:23 PM, daniela florescu <dflorescu at me.com> wrote:

> Now contrast that with the little experiment that I was running recently:
> here is the simplest statistical problem you can imagine.
> You have 3 boxes and 6 balls: 3 white and 3 back. You put the balls in the
> boxes, 2 in each. I take one ball out of one box and it is black.
> What is the probability that the secod ball in the same box is also black ?
> I asked this question to CTOs of data science companies, Phd in math, Phd
> In CS, the head of machine learning at Google, hordes of “data scientists”.
> Out of tens of people I asked, I got only TWO correct answers: both where
> Phd in physics. (not the CTOs, not the head of machine learning..)
I now remember arguing the equivalent problem with a Math PhD student  on a
long drive back from a rugby tournament while I was on study abroad. He got
it wrong, easy for me to say, at the time  I had just taken a probability
class. This is a camouflaged version of the Monty Hall problem typically
used  to introduce Bayes Rule.

Here it's camouflaged because what is really being asked is what is the
(conditional) probability of selecting the box with 2 black balls (the door
with the prize).

A priori it is 1/3 but after you take out 1 black ball do your
probabilities change. In Bayesian terms, have you been given  information
that should update your probabilistic beliefs? Alot of people go wrong by
thinking that they stay the same but you have been given information that
eliminates one of the possibilities (one of the doors without the prize).
The next pitfall is intuiting that because you now only have 2 boxes to
chose from then the probability is 1/2 or worse that the answer is 3/4
because 3 of the remaining 4 possible balls are black.

But the question is really asking for a conditional probability. What is
the probability that the 2 black ball box was chosen given that I have
shown you that it contains at least one black ball, so you have to divide
by the conditioning event -  the probability of drawing a black ball. This
is usually calculated  using the law of total probability but we it can be
intuited here as being 1/2 (3 of the 6 balls in the boxes are black).

So the answer is obtained by dividing the 1/3 (the original chance of
selecting the box with 2 black balls) by the conditioning event 1/2 giving
1/6...........DOH!!! .... 2/3.
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