Given: an opaque jar with a large number of white & black marbles, same
number of each. If I pick 2 marbles randomly, it seems that the
probabilities depend upon how I do the drawing.

If I pick one marble, then pick another, the probability of drawing one
black and one white marble is 50%. (4 possibilities, 2 of which give 1
white & 1 black.)

If I reach in and pick 2 marbles AT ONCE, the probability is 33%. (3
possibilities, 1 of which gives 1 of each color.)

It just doesn't SEEM right that the probabilities could be different.
Why is it different? Or is it not different? Is there really 4
possibilities when drawing 2-at-once?

Drawing 3 marbles gives even worse results. One-at-a-time gives:
1/8 probability of all black,
1/8 all white,
3/8 2 black + 1 white, &
3/8 2 white + 1 black.

3 at once gives:
1/4 all black,
1/4 all white,
1/4 2 black + 1 white, &
1/4 2 white + 1 black.
The probability of drawing all same color 3-at-once is twice that of
one-at-a-time!

Is there a statistician in the house?

Bob

On 2011-11-06, Bob Engelhardt wrote:
Given: an opaque jar with a large number of white & black marbles, same
number of each. If I pick 2 marbles randomly, it seems that the
probabilities depend upon how I do the drawing.

If I pick one marble, then pick another, the probability of drawing one
black and one white marble is 50%. (4 possibilities, 2 of which give 1
white & 1 black.)

If I reach in and pick 2 marbles AT ONCE, the probability is 33%. (3
possibilities, 1 of which gives 1 of each color.)

3 possibilities, unequal probabilities.

Your reasoning is wrong.

i


It just doesn't SEEM right that the probabilities could be different.
Why is it different? Or is it not different? Is there really 4
possibilities when drawing 2-at-once?

Drawing 3 marbles gives even worse results. One-at-a-time gives:
1/8 probability of all black,
1/8 all white,
3/8 2 black + 1 white, &
3/8 2 white + 1 black.

3 at once gives:
1/4 all black,
1/4 all white,
1/4 2 black + 1 white, &
1/4 2 white + 1 black.
The probability of drawing all same color 3-at-once is twice that of
one-at-a-time!

Is there a statistician in the house?

Bob

"Ignoramus27678" wrote in message
...

On 2011-11-06, Bob Engelhardt wrote:
Given: an opaque jar with a large number of white & black marbles, same
number of each. If I pick 2 marbles randomly, it seems that the
probabilities depend upon how I do the drawing.

If I pick one marble, then pick another, the probability of drawing one
black and one white marble is 50%. (4 possibilities, 2 of which give 1
white & 1 black.)

If I reach in and pick 2 marbles AT ONCE, the probability is 33%. (3
possibilities, 1 of which gives 1 of each color.)

3 possibilities, unequal probabilities.

Your reasoning is wrong.

i\

Jeez, you're harsh. d8-) Why don't you explain that the two possibilities of
one each black and white, in the first example, actually give the same
result and are really just two different orders for the *same*
possibility -- one white, one black?

Inelegant, but maybe helpful...

--
Ed Huntress




It just doesn't SEEM right that the probabilities could be different.
Why is it different? Or is it not different? Is there really 4
possibilities when drawing 2-at-once?

Drawing 3 marbles gives even worse results. One-at-a-time gives:
1/8 probability of all black,
1/8 all white,
3/8 2 black + 1 white, &
3/8 2 white + 1 black.

3 at once gives:
1/4 all black,
1/4 all white,
1/4 2 black + 1 white, &
1/4 2 white + 1 black.
The probability of drawing all same color 3-at-once is twice that of
one-at-a-time!

Is there a statistician in the house?

Bob

On 2011-11-06, Ed Huntress wrote:


"Ignoramus27678" wrote in message
...

On 2011-11-06, Bob Engelhardt wrote:
Given: an opaque jar with a large number of white & black marbles, same
number of each. If I pick 2 marbles randomly, it seems that the
probabilities depend upon how I do the drawing.

If I pick one marble, then pick another, the probability of drawing one
black and one white marble is 50%. (4 possibilities, 2 of which give 1
white & 1 black.)

If I reach in and pick 2 marbles AT ONCE, the probability is 33%. (3
possibilities, 1 of which gives 1 of each color.)

3 possibilities, unequal probabilities.

Your reasoning is wrong.

i\

Jeez, you're harsh. d8-) Why don't you explain that the two possibilities of
one each black and white, in the first example, actually give the same
result and are really just two different orders for the *same*
possibility -- one white, one black?

Inelegant, but maybe helpful...


I am sligfhtly drunk, I did not mean to be harsh or rude, sorry.

i

On Sat, 05 Nov 2011 22:36:27 -0400, Bob Engelhardt wrote:

Given: an opaque jar with a large number of white & black marbles, same
number of each. If I pick 2 marbles randomly, it seems that the
probabilities depend upon how I do the drawing.

If I pick one marble, then pick another, the probability of drawing one
black and one white marble is 50%. (4 possibilities, 2 of which give 1
white & 1 black.)

If I reach in and pick 2 marbles AT ONCE, the probability is 33%. (3
possibilities, 1 of which gives 1 of each color.)

When you hold those marbles in your hand, you know that there are two
marbles in two positions. The marble in _one_ position can be either
white or black: that's two possibilities. The marble in the _other_
position can also be white or black; those two possibilities times the
first two make four.

(This is easier if you just break down and enumerate the positions: 1, 2,
etc.. I was trying to avoid that for pedagogical reasons).

It just doesn't SEEM right that the probabilities could be different.
Why is it different? Or is it not different? Is there really 4
possibilities when drawing 2-at-once?

Drawing 3 marbles gives even worse results. One-at-a-time gives: 1/8
probability of all black,
1/8 all white,
3/8 2 black + 1 white, &
3/8 2 white + 1 black.

3 at once gives:
1/4 all black,
1/4 all white,
1/4 2 black + 1 white, &
1/4 2 white + 1 black.
The probability of drawing all same color 3-at-once is twice that of
one-at-a-time!

Is there a statistician in the house?

The same argument holds no matter how many marbles you draw out, or how:
they are still N distinct marbles, so you still have to treat them as
distinct even if you grab them all at once. Consequently, if the jar has
an infinite number of marbles then there are 2^N, a great many of which
are repeats of other possibilities once you jumble the marbles.

Note that the probabilities only go to exactly 2^N as the number of
marbles in the jar approaches infinity: If there are 10 marbles of each
color in the jar and the first one out is black, then the probability
that the second one will be black is 9/19. So the overall probability of
black/black is 10/20 * 9/19, or 90/380, or 9/38 after you reduce. So is
the probability of white/white. The probability of drawing a black/white
is 10/20 * 10/19, or 10/38, as is the probability of drawing a white/
black.

The more draws compared to the total number of marbles the more skewed
the probabilities get: given a jar with five marbles of each color, the
probability of drawing six white marbles ain't 1:64, it's just plain old
0 (unless you have a felt-tip marker in your back pocket).

--
www.wescottdesign.com

"Ignoramus27678" wrote in message
...

On 2011-11-06, Ed Huntress wrote:


"Ignoramus27678" wrote in message
...

On 2011-11-06, Bob Engelhardt wrote:
Given: an opaque jar with a large number of white & black marbles, same
number of each. If I pick 2 marbles randomly, it seems that the
probabilities depend upon how I do the drawing.

If I pick one marble, then pick another, the probability of drawing one
black and one white marble is 50%. (4 possibilities, 2 of which give 1
white & 1 black.)

If I reach in and pick 2 marbles AT ONCE, the probability is 33%. (3
possibilities, 1 of which gives 1 of each color.)

3 possibilities, unequal probabilities.

Your reasoning is wrong.

i\

Jeez, you're harsh. d8-) Why don't you explain that the two possibilities
of
one each black and white, in the first example, actually give the same
result and are really just two different orders for the *same*
possibility -- one white, one black?

Inelegant, but maybe helpful...


I am sligfhtly drunk, I did not mean to be harsh or rude, sorry.

I think that drinking often makes our personality appear the opposite of our
normal one. For example, it makes me very mellow. d8-)

--
Ed Huntress

On 11/5/2011 10:36 PM, Bob Engelhardt wrote:
Given: an opaque jar with a large number of white & black marbles, same
number of each. If I pick 2 marbles randomly, it seems that the
probabilities depend upon how I do the drawing.

If I pick one marble, then pick another, the probability of drawing one
black and one white marble is 50%. (4 possibilities, 2 of which give 1
white & 1 black.)

That is incorrect.

Suppose there are 100 marbles in the jar, 50 black and 50 white. Your
probability of drawing a white marble first is 50/100 = 0.50, and the
probability that the second marble will be black is 50/99 = 0.505. The
combined probability of these events is 0.50 * 0.505 = 0.2525. It is
equally likely (P = 0.2525) that you will draw a black marble, followed
by a white one. The probability that one or the other of these will
happen is 0.2525 + 0.2525 = 0.5050, or a little more than half, that you
will draw one marble of each color.

If you pick one marble, *put it back*, then pick another, your
probability of picking one of each color is in fact exactly 50%.

If I reach in and pick 2 marbles AT ONCE, the probability is 33%. (3
possibilities, 1 of which gives 1 of each color.)

That also is incorrect. Just because there are only three possibilities
does not mean they are equally likely. In fact, they are not: there is
only one way you can draw two white marbles, and only one way you can
draw two black ones, but there are *two* ways you can draw one of each
color: white-black, and black-white. It is exactly twice as likely that
you will have one of each color, than that you will have two white ones.

It just doesn't SEEM right that the probabilities could be different.

It doesn't seem right because it isn't right.

Why is it different? Or is it not different?

It is different, but not in the way you suppose (as I have attempted to
explain).

Is there really 4
possibilities when drawing 2-at-once?

Yes.

Drawing 3 marbles gives even worse results. One-at-a-time gives:
1/8 probability of all black,
1/8 all white,
3/8 2 black + 1 white, &
3/8 2 white + 1 black.

3 at once gives:
1/4 all black,
1/4 all white,
1/4 2 black + 1 white, &
1/4 2 white + 1 black.
The probability of drawing all same color 3-at-once is twice that of
one-at-a-time!

No, it isn't, actually (although I can easily understand how it seems
like it). Probability is full of pitfalls for the unwary. You found a
few of them. :-)

Is there a statistician in the house?

I'm a math professor. Will that do?

"Bob Engelhardt" wrote in message
...
...
If I pick one marble, then pick another, the probability of drawing one
black and one white marble is 50%. (4 possibilities, 2 of which give 1
white & 1 black.)

If I reach in and pick 2 marbles AT ONCE, the probability is 33%. (3
possibilities, 1 of which gives 1 of each color.)

Bob

There are still 4 possibilities, two of which are degenerate and thus give
equivalent results.

jsw