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Topic: Colored Balls (Read 7313 times) |
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Bassem
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I got this riddle today over a phone interview for MS. it goes like this: You have a single box full of balls of four different colors. 45 Black, 38 White, 25 Blue and 9 Green balls. What is the maximum number of balls that you need to pull from the box to be able to get two balls of the same color(which can be any color).
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Neelesh
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Re: Colored Balls
« Reply #1 on: Nov 17th, 2005, 11:04am » |
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on Nov 17th, 2005, 10:53am, Bassem wrote:| I got this riddle today over a phone interview for MS. it goes like this: You have a single box full of balls of four different colors. 45 Black, 38 White, 25 Blue and 9 Green balls. What is the maximum number of balls that you need to pull from the box to be able to get two balls of the same color(which can be any color). |
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Maximum number of balls : 45 + 38 + 25 + 9 = 117
(I am so dumb that I am unable to understand that I got the two balls of the same color unless I pull out all of them ....)
 Ooops probably you meant "the number of balls in the worst case" or something like that
So in the worst case, 4 + 1 = 5 balls.
Or may be I have not understood the question properly.
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Bassem
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You understood it correctly Neelesh. 5 is the correct answer.
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krishnaiah narukulla
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(I got this riddle today over a phone interview for MS. it goes like this: You have a single box full of balls of four different colors. 45 Black, 38 White, 25 Blue and 9 Green balls. What is the maximum number of balls that you need to pull from the box to be able to get two balls of the same color(which can be any color).
Total number of balls in worst case is=45+38+25+1=109
Best case: 1+1=2
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SMQ
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Re: Colored Balls
« Reply #4 on: Dec 27th, 2005, 7:32am » |
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You might want to rethink that worst-case answer...
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Grimbal
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Re: Colored Balls
« Reply #5 on: Dec 27th, 2005, 10:49am » |
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Can we choose the balls as we take them from the box? 
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deoBRAT
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Re: Colored Balls
« Reply #6 on: Feb 6th, 2006, 5:17am » |
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Hey, the answer is very simple.
Consider this -
You have balls of 4 colors. In the worst case, if you pick up 4 balls, they will all be of different colors. The fifth ball you pick up, has to match with at least one of the balls picked up earlier.
So the answer is - 5 balls.
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koojealion
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Re: Colored Balls
« Reply #7 on: Mar 27th, 2006, 7:13pm » |
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I think the question was lost in translation. The answer 5 is right, but the question should be, "What is the minimum number of balls you must pull out to be sure to get two of a color?"
(not "what is the maximum?")
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towr
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Re: Colored Balls
« Reply #8 on: Mar 27th, 2006, 11:48pm » |
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No, maximum is right. Because the minimum is when you take two balls of the same color at the start. Once you have two of the same color, you're sure you have them.
But you never need to take more then 5 to be sure there's at least two of the same color. So that's the maximum you need.
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Grimbal
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Re: Colored Balls
« Reply #9 on: Mar 28th, 2006, 8:27am » |
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It all depends whether you can look at the ball you have taken out or not.
If you need to take them out in the dark and you only see later what are the colors, you need to take minimum 5. If you just pull balls until you have a matching pair, it is maximum 5.
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NumBeast
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Re: Colored Balls
« Reply #10 on: May 12th, 2006, 1:04pm » |
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For the porposes of this or similar problems, the number of balls(or other objects) doesn't matter, it is only the number of types of balls(or other objects) that makes any difference.
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algo_wrencher
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Re: Colored Balls
« Reply #11 on: Jul 24th, 2006, 6:37am » |
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This was asked from me and my friend too Seems lot of ppl at MSFT failed to answer this and hence it became a test bench.
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| « Last Edit: Jul 24th, 2006, 6:38am by algo_wrencher » |
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TheNumberScott
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Re: Colored Balls
« Reply #12 on: Oct 29th, 2006, 5:40pm » |
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hmmm, what would it be if you pulled 2 balls at the same time, then discarded them if they were not the same color? I don't know the answer yet, but it seems interesting, and a little harder than the other way.
Also, what color would they be?
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| « Last Edit: Oct 29th, 2006, 6:04pm by TheNumberScott » |
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TimMann
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Re: Colored Balls
« Reply #13 on: Oct 29th, 2006, 10:01pm » |
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on Oct 29th, 2006, 5:40pm, TheNumberScott wrote:hmmm, what would it be if you pulled 2 balls at the same time, then discarded them if they were not the same color? I don't know the answer yet, but it seems interesting, and a little harder than the other way.
Also, what color would they be? |
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In the worst case you can pull them all out 2 at a time and never get a pair that are the same color. For example, it's possible that the first 34 times you get black/white, the next 4 times blue/white, then 11 times blue/black, then 9 times blue/green, and finally have just one blue left.
It might be more challenging to ask the following:
1) Suppose you pull out balls 2 at a time and discard them if they aren't the same color? What's the expected (average) number of tries until you either get two of the same color or run out of balls?
2) Suppose instead that you pull out balls 2 at a time, and if they aren't the same color, you drop them back in the box. Then what's the expected number of tries until you get two the same color?
I didn't work these out.
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shrek43
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Re: Colored Balls
« Reply #14 on: Jan 4th, 2007, 4:38pm » |
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its 5 because there are only 4 different colors, 5 would gaurantee that u get 2 of the same color.
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A
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Re: Colored Balls
« Reply #15 on: Mar 9th, 2007, 12:23am » |
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on Oct 29th, 2006, 10:01pm, TimMann wrote:
2) Suppose instead that you pull out balls 2 at a time, and if they aren't the same color, you drop them back in the box. Then what's the expected number of tries until you get two the same color?
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We might end with picking same 2 balls of different color rite ?
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jollytall
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Re: Colored Balls
« Reply #16 on: Mar 10th, 2007, 2:31am » |
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2. Is easy:
I count rounds (pulling two, throwing them back if different colour).
The probability that we get two of the same colour in any draw (and so already at the first) is:
P = 45/117*46/116+38/117*37/116+25/117*24/116+9/117*8/116 = cca 30%.
So the expected number of rounds is:
E = P*1 + (1-P) * (E+1), or E = 1/P, cca. 3.34 rounds.
1. Is much more complicated, at the moment I do not see an easy way, other than a brute force calculation.
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