Discussion:
An urn walks into a bar...
r***@gmail.com
2015-09-23 20:06:33 UTC
Raw Message
You're given an urn, full of red and black marbles, and
the following information:

i) If you draw a single marble, it will be red or black, 50-50
ii) If you draw a pair of marbles (from the full urn), and
they're matched colors, then draw another pair, the
second pair will match with 50-50 chance. (not
necessarily the same color as the first)

How many marbles in the urn?

--
Rich
Arturo Magidin
2015-09-23 20:34:02 UTC
Raw Message
Post by r***@gmail.com
You're given an urn, full of red and black marbles, and
i) If you draw a single marble, it will be red or black, 50-50
ii) If you draw a pair of marbles (from the full urn), and
they're matched colors, then draw another pair, the
second pair will match with 50-50 chance. (not
necessarily the same color as the first)
How many marbles in the urn?
Condition (i) tells you that the number r of red marbles and the number b of black marbles is the same: r=b.

So let 2n=r+b be the total number of marbles.

From condition (ii), the number of ways to pick a matched pair among the remaining (r-2)+b balls is the same as the number of ways of picking an unmatched pair.

The number of ways of picking a matching pair is
((r-2) choose 2) + (b choose 2)). The total number of ways of picking a pair is ((n-2) choose 2). So you want

((r-2)choose 2) + (b choose 2)) = ((n-2) choose 2) - ((r-2)choose 2) - (b choose 2).

Using b=r and n=2r, you get

2( [(r-2)(r-3)/2] + (r(r-1)/2)) = (2r-2)(2r-3)/2.

( r^2-5r + 6 + r^2 - r) = (4r^2 - 10r + 6)/2

2r^2 - 6r + 6 = 2r^2 - 5r + 3
r = 3.

So there are 6 marbles, three are red, and three are black.

Indeed, note that if that is the case, the odds of picking a red marble is 3/6 = 1/2, and the odds of picking a black marble are 3/6 = 1/2.

And if you pick two marbles of the same color, say red, then there remain in the urn 1 red marble and 3 black marbles. There are (4 choose 2) = 6 possible draws of a pair; three of these involve a red and a black marble, and three involve two black marbles, so the odds that the second pair will match is 1/2.
--
Arturo Magidin
r***@gmail.com
2015-09-24 21:45:21 UTC
Raw Message
Post by Arturo Magidin
Post by r***@gmail.com
You're given an urn, full of red and black marbles, and
i) If you draw a single marble, it will be red or black, 50-50
ii) If you draw a pair of marbles (from the full urn), and
they're matched colors, then draw another pair, the
second pair will match with 50-50 chance. (not
necessarily the same color as the first)
How many marbles in the urn?
Condition (i) tells you that the number r of red marbles and the number b of black marbles is the same: r=b.
So let 2n=r+b be the total number of marbles.
From condition (ii), the number of ways to pick a matched pair among the remaining (r-2)+b balls is the same as the number of ways of picking an unmatched pair.
The number of ways of picking a matching pair is
((r-2) choose 2) + (b choose 2)). The total number of ways of picking a pair is ((n-2) choose 2). So you want
((r-2)choose 2) + (b choose 2)) = ((n-2) choose 2) - ((r-2)choose 2) - (b choose 2)
So there are 6 marbles, three are red, and three are black.
And if you pick two marbles of the same color, say red, then there remain in the urn 1 red marble and 3 black marbles. There are (4 choose 2) = 6 possible draws of a pair; three of these involve a red and a black marble, and three involve two black marbles, so the odds that the second pair will match is 1/2.
Right. An easy problem, for any competent mathematician.
The reason I posted this, is that it's instructive,
as it contains a few concepts which frequently confuse students
of probability.

For instance, consider the commands:
a) "Draw a pair of marbles, AT THE SAME TIME"
b) "Draw a marble, then another one"
Is there any difference?

Also, the concept of combinations vs. permutations;
"hmmm, {B,R} looks the same as {R,B}, that counts as one
unit... no, they can be selected in different sequence...
does it matter?"

--
Rich