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# Lining up my skittles

Best Answer superprismatic, 07 November 2013 - 10:30 PM

Sheesh people. 17 then

Spoiler for in that case

Go to the full post

9 replies to this topic

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Posted 05 November 2013 - 12:23 PM

My pack of skittles comes in five flavors, grape, lemon, green apple, orange, and strawberry (odd that they don't have the cherry blue flavor found in the UK but alas).

Assume that my pack had the following proportion of flavors 2.5 lemon, 2 orange, 1.5 grape, 1.5 green, 1 strawberry.  If i were to pour my skittles on the table and line them up in an offset two row pattern (so that at most, a single skittle touches four skittles [see image].  What are the chances that two skittles of the same color won't touch?

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### #2 jamieg

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Posted 05 November 2013 - 04:28 PM

Is the number of skittles in the image correct? You don't otherwise provide how many are being poured. Or do you want a general solution for arbitrary N?

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Posted 06 November 2013 - 01:24 AM

Is the number of skittles in the image correct? You don't otherwise provide how many are being poured. Or do you want a general solution for arbitrary N?

The image was just to show what I meant by how to line them up. Calculate for any N.
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### #4 dgreening

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Posted 06 November 2013 - 05:12 PM

Yes, the total number of candies makes a difference [if very large, the answer approaches 0]

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Posted 06 November 2013 - 11:51 PM

hmmm... okay assume there are 20 skittles.

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### #6 superprismatic

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Posted 07 November 2013 - 09:34 PM

hmmm... okay assume there are 20 skittles.

In order for a pack to contain flavors with those exact proportions, the pack must contain a multiple of 17 Skittles.  So, a pack of 20 wouldn't satisfy the hypotheses of the problem!

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Posted 07 November 2013 - 09:53 PM

Sheesh people. 17 then
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### #8 superprismatic

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Posted 07 November 2013 - 10:30 PM   Best Answer

Sheesh people. 17 then

Spoiler for in that case

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### #9 antel0pe

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Posted 10 November 2013 - 07:36 PM

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### #10 superprismatic

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Posted 10 November 2013 - 09:33 PM

I worked on it for a while until I decided it was just to complex for me to do without writing a program.

So, I generated all possible pairs of 9 & 8 long strings which together had 5 lemon, 4 orange, 3 grape,

3 green, and 2 strawberry colors.  There are 17!/(5!*4!*3!*3!*2!) = 1,715,313,600 of these.  Then, I

checked each one of them and counted those which had no same-color skittles touching.  There were

1,232,600 like that.  That gave me the probability I posted.  The program was pretty sloppy but it

ran in 69 seconds on one core of my desktop machine.

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