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ALERT: THIS IS A MATH PROBLEM

Here's how it works, each day I dump out all the skittles and whatever side faces no 'S' then I eat it. Next day I do the same till there's none left.

And no I can't actually buy a big bag of Skittles and try it. So how would I explain this in a paragraph?

2007-11-12 09:01:36 · 24 answers · asked by ? 6 in Science & Mathematics Mathematics

That was the only info I was given

2007-11-12 09:25:16 · update #1

24 answers

Assuming average distribution, roughly half of the Skittles will be face down, so you'd be eating half of them each day. Under strictly mathematical analysis, you'd never finish the bag because there's always half the remainder left. In practice, the number of Skittles is finite and sooner or later there would be only 1 left, and it would come out face down eventually.

What you need to do is figure out how many Skittles are in a bag. For example, if Skittles weigh about 1 gram each and there's 454 grams in a pound, you'd start with 454 Skittles. The consumption would look something like this, assuming 50% chance of any Skittle coming our face up or face down:

Day 1 - 454 to start, eat 227, 227 left
Day 2 - 227 to start, eat 114, 113 left
Day 3 - 113 to start, eat 56, 57 left
Day 4 - 57 to start, eat 29, 28 left
Day 5 - 28 to start, eat 14, 14 left
Day 6 - 14 to start, eat 7, 7 left
Day 7 - 7 to start, eat 3, 4 left
Day 8 - 4 to start, eat 2, 2 left
Day 9 - 2 to start, eat 1, 1 left
Day 10 - 1 to start, eat 1, none left

So, about 10 days. However, the exact numbers depend on chance, and the actual distribution becomes more uncertain as the number of Skittles decreases.

2007-11-12 09:13:25 · answer #1 · answered by dukefenton 7 · 1 0

Day 1: 1lb
Day 2: 1/2 lb
Day 3: 1/4 lb
Day 4: 1/8 lb
Day 5: 1/16 lb
Day 6: 1/32 lb
Day 7: 1/64 lb
Day 8: 1/128 lb
Day 9: 1/256 lb
Day 10: 1/512 lb
Day 11: 1/1024 lb
Day 12: 1/2048 lb
(the weights are how much is left)

You would have to find the weight of 1 skittle and then determine where it falls on the chart. If you only have a gram scale, then use the conversion 1lb=452.6g.

2007-11-12 09:10:22 · answer #2 · answered by HT-5 2 · 1 0

There's not really an answer to that in an exact number. Every Skittle has exactly a 50% chance of being on one side or the other. It's safe to say that the numbers would be approximately halved every day, until you get down to the last few. You might get them all at once, or you might have several stragglers that keep landing 'S' side up. Then, of course, there's always an infinitesimal chance of having them all in one direction, and that increases the fewer you have left. It would be entirely possible that you have half a dozen or so that last for several rolls after the rest are finished off.

2007-11-12 09:06:39 · answer #3 · answered by fishtrembleatmyname 5 · 1 0

It would be based on probability, like flipping a coin. You have a 50/50 chance of eating a Skittle.

I think, in order to get a number, you'd need to figure out approximately how many Skittles are in a big bag.

2007-11-12 09:06:14 · answer #4 · answered by xK 7 · 0 0

every time you dump the bag out, assume half are S up and half are S down.

that means you get rid of half each day

You would find out how many skittles are in a one pound bag by weighing one skittle and converting the pound into an approximate number of skittles.

than divide that number in half until you are come to a number less than 1

count how many times you divided the bag in half and that is how many days it took you

2007-11-12 09:10:13 · answer #5 · answered by KEYNARDO 5 · 2 0

2 days

2007-11-13 21:34:21 · answer #6 · answered by Anonymous · 0 1

Each day you eat half the skittles. so after a day you have 8 oz after two days you have 4 oz. Eventually you get to one skittle and odds are in a day or two it will be an "S" and you're done.

2007-11-12 09:48:21 · answer #7 · answered by tfloto 6 · 1 0

assuming this is probability, you'd have 1/2 on the 1st day, 1/4 on the secon day, and carrying on like that and assuming it went along those lines, you'd be there untill you ended up with only 2 left and then you'd have one and still have a 50/50 chance of it not landing on the S

2007-11-12 09:07:06 · answer #8 · answered by claire 3 · 1 0

2 days

2007-11-12 09:03:50 · answer #9 · answered by Steve B 1 · 1 2

well it depends how many there are just say there is 50..

there is a 50/50 chance of it being on the "s" side so if u poured 50 out there might be around 25 that are "s" sided.

2007-11-12 09:06:58 · answer #10 · answered by Alex S 2 · 0 0

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