BrainDen.com - Brain Teasers
• 0

# Lining up my skittles

## Question

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?

## Recommended Posts

• 0

Sheesh people. 17 then

The probability that no two touching Skittles have the same color is precisely 6163/8576568 which is approximately 0.00071858580.

##### Share on other sites

• 0

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?

##### Share on other sites

• 0

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.

##### Share on other sites

• 0

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

##### Share on other sites

• 0

hmmm... okay assume there are 20 skittles.

##### Share on other sites

• 0

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!

##### Share on other sites

• 0

Sheesh people. 17 then

• 0

##### Share on other sites

• 0

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.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.