[Brandonb]

I misinterpreted brhan's "colored ball" post and accidentally came up with this. I believe it is much simpler but still entertaining.

Billy put a number of different colored balls (ex. 2 blue, 5 red, 1 purple, 7 green...) in a box. After a while he added 20 more balls (all of the same color as one color already in the box) to the box, and the addition of the new balls did not change probability of drawing two balls of the same color (with replacement).

How many different colors of balls are in the box? And how many of each color were in the box originally?

[Guest]

I'll take snapshot of work up later, because I hate typing math solutions (exponents/fractions/etc), I get lost when typing. There are 19 colors in the box, 10 of each.

[Brandonb]

I'll take snapshot of work up later, because I hate typing math solutions (exponents/fractions/etc), I get lost when typing. There are 19 colors in the box, 10 of each.

-Sorry, that answer worked for brhan's question, but not for this one.

Edited April 10, 2008 by Brandonb

[Guest]

-Sorry, that answer worked for brhan's question, but not for this one.

ohhhhhhhh yeah, same as "one color already in box." Oops! Will try again later...

[Brandonb]

The example in the question is merely an example nothing more

The # of different colors is less than 20

Also, please post your answers in a "Spoiler"

Edited April 10, 2008 by Brandonb

[EventHorizon]

of course, it'll work if there is only one color in the box (any number of them).

Will now work on possibilities for more than one color....

[Brandonb]

of course, it'll work if there is only one color in the box (any number of them).

Will now work on possibilities for more than one color....

Well of course, but that's why I said "Billy put a number of different colored balls... in a box."

Edited April 11, 2008 by Brandonb

[EventHorizon]

Well of course, but that's why I said "Billy put a number of different colored balls... in a box."

ok...here's another solution....

2 reds

2 blues

2 greens

2 yellows

12 blacks

and either 20 reds,blues,greens, or yellows can be added in without changing the probability of getting two of the same (with replacement)

I can find more....but this was the first I tried that fit my equation.

[Brandonb]

ok...here's another solution....

2 reds

2 blues

2 greens

2 yellows

12 blacks

and either 20 reds,blues,greens, or yellows can be added in without changing the probability of getting two of the same (with replacement)

I can find more....but this was the first I tried that fit my equation.

Hmm, maybe I wasn't clear in my original post... The object is to achieve a probability after the 20 balls were added, that is identical to the probability before the 20 balls were added. Which I believe there is only one answer.

Your response allows for 3 comparable probabilities for only after the balls were added, regardless of the original probability before the 20 were added.

[Guest]

Hmm, maybe I wasn't clear in my original post... The object is to achieve a probability after the 20 balls were added, that is identical to the probability before the 20 balls were added. Which I believe there is only one answer.

Your response allows for 3 comparable probabilities for only after the balls were added, regardless of the original probability before the 20 were added.

Actually;

If you have 4 different colors at 2, and 1 color at 12 here is the original and new;

(2/20)*(2/20) = 1% for any of the 4 2s

(12/20)*(12/20) = 36%

Total of 1% + 1% + 1% + 1% + 36% = 40%

New probability;

(22/40)*(22/40) = 30.25%

(12/40)*(12/40) = 9%

(2/40)*(2/40) = .25% (3 of them)

Total = 30.25 + 9 + .25 + .25 + .25 = 40% Identical

9 colors of 4, and 1 color 24 will yield 20% average but still fits the answer

Edited April 11, 2008 by PolishNorbi

[Brandonb]

Actually;

If you have 4 different colors at 2, and 1 color at 12 here is the original and new;

(2/20)*(2/20) = 1% for any of the 4 2s

(12/20)*(12/20) = 36%

Total of 1% + 1% + 1% + 1% + 36% = 40%

New probability;

(22/40)*(22/40) = 30.25%

(12/40)*(12/40) = 9%

(2/40)*(2/40) = .25% (3 of them)

Total = 30.25 + 9 + .25 + .25 + .25 = 40% Identical

9 colors of 4, and 1 color 24 will yield 20% average but still fits the answer

Sorry,

using that reasoning, in the origin each of the 2s would have a 10% chance of being drawn, and the 12 would have a 60% chance of being drawn.

Once the 20 are added to one of the 2s then there are 22 of that color in the box (obviously). This means that the remaining 2s have a 5% chance of being drawn, the 12 now have a 30% chance of being drawn, and the 22 have a 55% chance of being drawn.

It's not even close. Not by a long shot.

[EventHorizon]

Sorry,

using that reasoning, in the origin each of the 2s would have a 10% chance of being drawn, and the 12 would have a 60% chance of being drawn.

Once the 20 are added to one of the 2s then there are 22 of that color in the box (obviously). This means that the remaining 2s have a 5% chance of being drawn, the 12 now have a 30% chance of being drawn, and the 22 have a 55% chance of being drawn.

It's not even close. Not by a long shot.

The probability we are trying to keep identical is the probability you draw (with replacement) the same color twice. This is essentially the sum of the squares of the probability to draw a given color.

Yes, the probability to draw one of the 2s is 10%....but the probability you draw it twice is 1%....just as PolishNorbi calculated.

Similarly with 60% = .6, .6^2 = .36 = 36%.

If {2,2,2,2,12} isn't an answer to your puzzle, then it looks like you didn't explain your question well....but as explained...it was an interesting exercise.

[Brandonb]

The probability we are trying to keep identical is the probability you draw (with replacement) the same color twice. This is essentially the sum of the squares of the probability to draw a given color.

Yes, the probability to draw one of the 2s is 10%....but the probability you draw it twice is 1%....just as PolishNorbi calculated.

Similarly with 60% = .6, .6^2 = .36 = 36%.

If {2,2,2,2,12} isn't an answer to your puzzle, then it looks like you didn't explain your question well....but as explained...it was an interesting exercise.

Ah, I see the problem, thank you.

I'll make another one that is phrased correctly.

well done!

[EventHorizon]

ok...here's another solution....

2 reds

2 blues

2 greens

2 yellows

12 blacks

and either 20 reds,blues,greens, or yellows can be added in without changing the probability of getting two of the same (with replacement)

I can find more....but this was the first I tried that fit my equation.

here's a few more. The ones that can have 20 balls added will be marked with asterixes

{1*,1*,9,4}

{4*,12,1,1}

{2*,2*,2*,2*,2*,5,15}