[Guest]

John put a number of colored balls in a box, with equal number of each color. After a while he added 20 balls of a new color to the box, and the addition of the new balls did not change probability of drawing two balls of the same color (of course without replacement).

Now the question is how many balls were in the box originally (before the addition of balls with the new color)?

[Guest]

20

Edited April 9, 2008 by Lost in space

[Guest]

20

Nice try LIS, but 20 is not the solution. I think it is very difficult to get it by guessing ...

[Guest]

hint please

[Guest]

does 0 class as an answer?

[Guest]

hint please ohmy.gif

I have seen Bonanova's Pair the marbles post about the expected number of drawing couples ... so I believe there are plenty of info there.

does 0 class as an answer?

No .... the box must be very large -- to hold all those balls

[Guest]

Let's say we begin with x colors and n of each color. This means that the probability of drawing two balls of the same color is (n-1)/(n*x). This is because it doesn't matter what the first ball we pick is, as long as the second ball is the same color.

Now to add 20 balls of a new color... am I on the right track?

Now that I think about it, I've probably already put too much work in to it.

[Guest]

Let's say we begin with x colors and n of each color. This means that the probability of drawing two balls of the same color is (n-1)/(n*x). This is because it doesn't matter what the first ball we pick is, as long as the second ball is the same color.

Now to add 20 balls of a new color... am I on the right track?

Now that I think about it, I've probably already put too much work in to it.

I don't think your formula is correct. I think it is good idea to start with the number of ways of drawing of two balls of any color; and then the number of ways of drawing of two balls with a particular color ... and then dividing the second by the first we have the probability.

[Guest]

40

[Guest]

None, One, Or trick question

That's unless you know the actual answer... No pi (ha ha) for this I'm thinking!

[Guest]

I wrote a little excel to do a trial & check method.

The way it was setup it would calculate the Probability Original (Po) and Probability New (Pn) (Anyone wants the equations, let me know)

I would enter a number for Number of balls per color, and number of colors. Once (Po-Pn) = Negative, I knew there was no sense if trying higher numbers in the color.

Ie; Lets look at 5;

Number of Balls per Color = 5

Number of colors = 1

Result, (Po - Pn) = .333

Number of Balls per Color = 5

Number of colors = 2

Result, (Po - Pn) = -.03831

And as the higher colors went the more negative the number went, so once Po-Pn = negative it was time to stop;

Here are a couple quick things I noticed;

When number of balls per color was 1-5, it changed to negative at 2

When number of balls per color was 11+, it never changed negative.

That leaves us with 6-10

The correct solution is;

10 balls per color. 19 different colors. Or 190 total balls were in there originally.

and the Proof:

(10/190)*(9/189) = .002506

Since there are 19 of these, 19*.002506 = 4.7619% of drawing the same color twice.

Add the new 20 balls,

(10/210)*(9/209) = .002501 * 19 = 3.8961% but we also have to add the new information

(20/210)*(19/209) = .008658 = .8658% + 3.8961% = 4.7619%

Edited April 9, 2008 by PolishNorbi

[Guest]

200

Edited April 9, 2008 by Athena

[Guest]

PolishNorbi has got it. Fantastic.

I would like to see if someone will come up with some explanation. Otherwise I will post my solution tomorrow.

Edited April 10, 2008 by brhan

[Guest]

PolishNorbi has got it. Fantastic.

I would like to see if someone will come up with some explanation. Otherwise I will post my solution tomorrow.

After re-using excel.. There is a distinct pattern;

Number of balls per color = 1/2 of balls added

Number of colors = Balls added - 1

Ie;

If you add 80 balls, then number of colors = 79, number of balls per color = 40.

It has to be some simple math reduction that I haven't been able to find.

[Guest]

Here is my solution.....

Before adding the new balls:

Let c be the number of colors, and n the number of balls of each color.

The number of ways drawing two balls: cn(cn − 1).

The number of ways of drawing two balls of the same color: n(n − 1).

Summing over all colors, the number of ways of drawing matching colors: cn(n − 1).

Therefore the probability of drawing matching colors = (cn(n - 1)) / (cn(cn - 1)) = (n - 1)/(cn - 1) .........(*)

After adding the new balls:

The number of ways of drawing two balls: (cn + 20)(cn + 19).

The number of ways of drawing matching colors: cn(n − 1) + 20*19.

Hence the probability of drawing matching colors: (cn(n - 1) + 20*19)) / ((cn + 20)(cn + 19)) .........(**)

Equating (*) and (**)

(cn − 1)(cn(n − 1) + 20*19)) = (n − 1)(cn + 20)(cn + 19)

After simplification, we get c(21 − 2n) = 19 ==> c = 19/(21-2n). Since 'c' and 'n' are natural numbers, 21-2n must be a factor of 19.

21-2n = 19 ==> n = 1 and c = 1

21-2n = 1 ==> n = 10 and c = 19

'n' can't be 1 because after adding 20 new balls of the same color, the probability will not be the same. Therefore the second one is the only solution.

The total number of balls before adding the new ones is 10*19=190