Re: Another probability "paradox".

On 2 Sep, 20:01, Virgil <Vir...@xxxxxxxxx> wrote:
In article
<022240ee-4490-4e16-a3a1-4e8ad077c...@xxxxxxxxxxxxxxxxxxxxxxxxxx>,
 ju...@xxxxxxxxxxxxx wrote:
On 2 Sep, 08:24, Ray Koopman <koop...@xxxxxx> wrote:
On Sep 1, 11:34 pm, ju...@xxxxxxxxxxxxx wrote:
On 2 Sep, 07:23, Ray Koopman <koop...@xxxxxx> wrote:
On Sep 1, 10:28 pm, Bill Taylor <w.tay...@xxxxxxxxxxxxxxxxxxxxx>
wrote:

It seems to be the season for Monty Hall, etc, again.  So:

This time, you roll two balanced dice, and you're
hoping to get a "double".  The probability is  1/6.
No problem so far.

Now, you roll the dice, and they roll out of your sight,
but a friend tells you "Hey, you got a six on the blue die!"
(The dice are red and blue respectively, and your friend
 is very grammatical.)

The conditional probability of having got a double, given this
information, is still  1/6 , easily obtained.    Still no problem;
after all, to get a double, it doesn't really matter what you
got on the blue die.  He could have told you that you'd
got a one, or a two, or anything, and it wouldn't have
changed your  1/6  probability.        Still no problem.

HOWEVER, next time, the dice still roll out of your sight,
but within your friend's sight, and he tells you,
   "Hey, you got at least one six this time!"

Now, if you work out the conditional probabilities
based on this new information, it turns out that
the probability of having got a double has dropped
sharply, to  1/11 (!)

Why should this be so different!?  How can it make
so much difference whether or not you are told
the colour of the di(c)e that got a six?

But even that isn't the real paradox!  Because the thing is,
it's still the same whether he said "six" or any other number!

P( double | at least one "1") = 1/11 ;
:
:
P( double | at least one "6") = 1/11 .

That is, your friend can reduce your odds by almost 1/2
merely by telling you you got at least one of something,
which you knew you had to get anyway!!

What the *#&%!   How can this be?

-- Odd Odds Bill

               B l u e

        1   2   3   4   5   6

    1  11  12  13  14  15  16

    2  21  22  23  24  25  26
 R
    3  31  32  33  34  35  36
 e
    4  41  42  43  44  45  46
 d
    5  51  52  53  54  55  56

    6  61  62  63  64  65  66

If I know only that I have at least one 6 then I'm in row or column 6,
which means 1 double in 11 chances. But if I know that I have a blue 6
then I'm in column 6, where my chances are 1 in 6.

Aren't you in row 6 XOR column 6?

Wouldn't XOR exclude the diagonal?

Ok, debatable, it depends on how one formalizes yours and then my
assertion.

Anyway, if my fried tells me: "Hey, you got at least one six this
time!", that to me means:

  "Either the blue die is 6 or the red die is 6, or both."

Then, I'd have these cases (keep in mind I have no specific notions of
conditional probabilities):

-- The BLUE die is 6, and the RED is not: 5/11
-- The RED die is 6, and the BLUE is not: 5/11
-- The BLUE die is 6, and the RED is 6 too: 1/11

This amounts to a total of 11 possibilities, where only one of them is
the winning one. But, my overall chance to have a couple is still 1/6,
because:

   [ chance_of_couple / solution_space ] / asserted_initial_chance =

 = [ (1/11) / (11/36) ] / (1/6) =

 = 1/6

I am guessing with the math, but, from a logical stand-point, it
doesn't seem to make very much sense that "you have at least one of
something" and "you have at least a 6" can make a difference as to the
chance of having a couple of them.

After the dice have been rolled, the "probability" of two sixes is
either 0 or 1, even if one does not know which.

Is this conditional probability? You are just saying that either one
wins or s/hr does not. Not very informative.

On the other hand, if the experiment of rolling the dice is modified so
as to reroll the pair of dice until until at least one 6 appears, then
the a priori probability the the experiment will produce double 6's is
indeed 1/11.

So you are saying that 1 out of 11 tries I get a double six? Ok, no,
you are rather saying that 1 out of 11 times I get a six, what I get
is a double six.

Still, each time I *do* get *a* six, the (conditional) probability
that the *other* die is six is 1 out of 6.

Isn't it?

-LV
.

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