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Peanut M&Ms (1 Viewer)

Despyzer

Despyzer

Lousy Attention Whore
I love peanut M&Ms. Ever since I was a little kid I always eat them three at a time (my own OCD hell). Oddly enough, no matter what size package, there is almost always one or two M&Ms left over (I usually give them away or throw them away). You would almost think that the probability that the number of M&Ms would be divisible by three would be roughly one in three, wouldn't you?

 
Despyzer said:
I love peanut M&Ms. Ever since I was a little kid I always eat them three at a time (my own OCD hell). Oddly enough, no matter what size package, there is almost always one or two M&Ms left over (I usually give them away or throw them away). You would almost think that the probability that the number of M&Ms would be divisible by three would be roughly one in three, wouldn't you?
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I don't know that I've ever seen a question with less practical importance.Please see mine. Thanks.

 
Despyzer said:
I love peanut M&Ms. Ever since I was a little kid I always eat them three at a time (my own OCD hell). Oddly enough, no matter what size package, there is almost always one or two M&Ms left over (I usually give them away or throw them away). You would almost think that the probability that the number of M&Ms would be divisible by three would be roughly one in three, wouldn't you?
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You would think that. But I have to ask you. Why wouldn't you just eat the last ones in the bag?
 
Mr. Pickles said:
I believe golddigger is the point man for all probability related questions.
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:shrug: :lmao:It may be that the number of M&M's in a bag is not random - the machines that fill the bags are probably calibrated to put X number of M&M's in the bag, where X is 1 or 2 mod 3. If the number of M&M's was truly random, then you're right, there would be a 1/3 chance that the number of M&M's would be a multiple of 3. Even then you'd expect to have some stragglers 2 out of 3 times, so it's not that surprising that you typically have 1 or 2 left over... not sure why you don't eat them though. :excited:
 
tomarken said:
Mr. Pickles said:
I believe golddigger is the point man for all probability related questions.
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:lmao: :lmao:It may be that the number of M&M's in a bag is not random - the machines that fill the bags are probably calibrated to put X number of M&M's in the bag, where X is 1 or 2 mod 3. If the number of M&M's was truly random, then you're right, there would be a 1/3 chance that the number of M&M's would be a multiple of 3. Even then you'd expect to have some stragglers 2 out of 3 times, so it's not that surprising that you typically have 1 or 2 left over... not sure why you don't eat them though. :shrug:
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I would think it was weight driven as to how many end up in the bag. Since each M&M is pretty standard you would expect each bag to contain essentially the same number of candies. Now since they make millions of bags a year the OPs personal experience is way to small a sample to say that there are never 3 left over in a bag if you eat them by 3s or to determine the possibility of such a happening.
 
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tomarken said:
It may be that the number of M&M's in a bag is not random - the machines that fill the bags are probably calibrated to put X number of M&M's in the bag, where X is 1 or 2 mod 3.
Click to expand...
No. When bulk fill packages like M&Ms (or pretty much any other bagged food item) is packaged, it is always by weight. Typically this is done using a multi-head scale. For example, if they are filling a 2oz bag, they'd probably use a 16 head scale, with each head having about .25oz of M&Ms. The scale would then calculate the combination of heads needed to dump to make weight, or just slightly over --- so the scale would only be dumping about half of it's charge per bag fill. Then it repeats that 80-90 times per minute.
 
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FatMax said:
tomarken said:
It may be that the number of M&M's in a bag is not random - the machines that fill the bags are probably calibrated to put X number of M&M's in the bag, where X is 1 or 2 mod 3.
Click to expand...
No. When bulk fill packages like M&Ms (or pretty much any other bagged food item) is packaged, it is always by weight. Typically this is done using a multi-head scale. For example, if they are filling a 2oz bag, they'd probably use a 16 head scale, with each head having about .25oz of M&Ms. The scale would then calculate the combination of heads needed to dump to make weight, or just slightly over --- so the scale would only be dumping about half of it's charge per bag fill. Then it repeats that 80-90 times per minute.
Click to expand...
Explain to me how this "multi-head scale" functions if ya don't mind.It would seem to me that the number of pieces in each bag would be distributed (guassian?) around a certain value (let's say 18.3 or something). If the spread of the distribution is very narrow (small sigma), it's quite possible that only the tail ends of that distro would include a multiple of 3.If the distribution were flat, then 1 out of 3 bags would contain a number of candies which is a multiple of 3; however, it is not a flat distro.
 
The Z Machine said:
FatMax said:
tomarken said:
It may be that the number of M&M's in a bag is not random - the machines that fill the bags are probably calibrated to put X number of M&M's in the bag, where X is 1 or 2 mod 3.
Click to expand...
No. When bulk fill packages like M&Ms (or pretty much any other bagged food item) is packaged, it is always by weight. Typically this is done using a multi-head scale. For example, if they are filling a 2oz bag, they'd probably use a 16 head scale, with each head having about .25oz of M&Ms. The scale would then calculate the combination of heads needed to dump to make weight, or just slightly over --- so the scale would only be dumping about half of it's charge per bag fill. Then it repeats that 80-90 times per minute.
Click to expand...
Explain to me how this "multi-head scale" functions if ya don't mind.It would seem to me that the number of pieces in each bag would be distributed (guassian?) around a certain value (let's say 18.3 or something). If the spread of the distribution is very narrow (small sigma), it's quite possible that only the tail ends of that distro would include a multiple of 3.

If the distribution were flat, then 1 out of 3 bags would contain a number of candies which is a multiple of 3; however, it is not a flat distro.
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Scale and photoThis one shows a 16 head scale with potato chips, or something that looks like it. I don't know the logic behind how they work, but the concept is simple. Each head (or pocket) is weighed and the combination closest to the package weight (but not under) is dumped. Of course, this is just another way for the companies to maximize profit. If there were 2.1 oz of M&Ms instead of 2oz, that is a 5% give away.

I'm sure the number of pieces is Gaussian in nature, but I have no idea what the deviation would be because it would be dependent upon consistency in production.

 
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The Z Machine said:
Explain to me how this "multi-head scale" functions if ya don't mind.

It would seem to me that the number of pieces in each bag would be distributed (guassian?) around a certain value (let's say 18.3 or something). If the spread of the distribution is very narrow (small sigma), it's quite possible that only the tail ends of that distro would include a multiple of 3.

If the distribution were flat, then 1 out of 3 bags would contain a number of candies which is a multiple of 3; however, it is not a flat distro.
Click to expand...
FatMax is the engineer, but I think I understand what he is talking about. Again as he said, it is weight dependent and not number dependent. Peanut M&M's are not uniform, as some are bigger than others. The multi-head scale would have 16 compartments of M&M's at around .25 ounces, give or take (some would be .23, others .28). For each bag, the scale would calculate which heads need to drop in order to get as close to 2 ounces as possible (assuming these numbers are correct). This site has some guy counting the peanut M&M's for 5 days (20, 21, 21, 21, 23).
 
Got it FM and KT.

Needed a picture to figure out what was up. Pretty smart way to maximize throughput while minimizing capital costs and excess package fill.

 
Despyzer said:
I love peanut M&Ms. Ever since I was a little kid I always eat them three at a time (my own OCD hell). Oddly enough, no matter what size package, there is almost always one or two M&Ms left over (I usually give them away or throw them away). You would almost think that the probability that the number of M&Ms would be divisible by three would be roughly one in three, wouldn't you?
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:goodposting:
(my own OCD hell).
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:thumbup:
 
Incidentally, I this is eerie timing as I just finished eating a handful of peanut M&M's literally in the last 10 minutes. :goodposting:

 
redman said:
Incidentally, I this is eerie timing as I just finished eating a handful of peanut M&M's literally in the last 10 minutes. :towelwave:
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Were they divisible by three? What size was the package?
 
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Despyzer said:
redman said:
Incidentally, I this is eerie timing as I just finished eating a handful of peanut M&M's literally in the last 10 minutes. :rolleyes:
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Were they divisible by three? What size was the package?
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Well, I'm not quite finished with it yet as it's a 56 oz. jumbo bag that I bought for Halloween mistakenly thinking that it had individually wrapped mini-bags in it. Give me another couple of weeks and a trip to my internist, and I'll have an answer for you.
 
Despyzer, I made fun of your OCD, but I have a confession. I always eat two at a time, and here's the thing. I pursue a balance of power approach, meaning that I don't want too many of any particular color to dominate, so if I've got a lot of one color, then I eat those. If a couple of colors predominate, then I eat both.

This doesn't make me crazy, right? :rolleyes:

 
Try adding a second variable of complexity by color coordinating your 3-nut job, such as only eating green and browns together, blues can only beaten with other blues, unless it is a yellow and blue (to make green) with a brown. Probablility of finishing with only 1 or 2 should decrease significantly...unless your OCD allows you to count mixing colors as a new single color. Who says math is not fun!!!

 
Despyzer said:
I love peanut M&Ms. Ever since I was a little kid I always eat them three at a time (my own OCD hell). Oddly enough, no matter what size package, there is almost always one or two M&Ms left over (I usually give them away or throw them away). You would almost think that the probability that the number of M&Ms would be divisible by three would be roughly one in three, wouldn't you?
Click to expand...
The answer is obvious: 50/50 chance. It's either going to happen...or it's not
 
Mr. Pickles said:
Norman Einstein said:
Mr. Pickles said:
Norman Einstein said:
opened one of our Halloween leftover M&M fun sized packages and found 8 inside....all blue...what are the odds?
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10^42,855Basically, it's not possible.
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hmmm...i must be colorblind then or maybe I ate a grand prize they were giving away...either way the kids thought it was cool
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Would you like to see the odds of DNA spontaneously forming?
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i'll believe my eyes vs. fancy statsthat DNA thingy would be pretty interesting though
 
More M&M eccentricities

Whenever I get a package of plain M&Ms, I make it my duty to continue the strength and robustness of the candy as a species. To this end, I hold M&M duels.

Taking two candies between my thumb and forefinger, I apply pressure, squeezing them together until one of them cracks and splinters. That is the "loser," and I eat the inferior one immediately. The winner gets to go another round.

I have found that, in general, the brown and red M&Ms are tougher, and the newer blue ones are genetically inferior. I have hypothesized that the blue M&Ms as a race cannot survive long in the intense theater of competition that is the modern candy and snack-food world.

Occasionally I will get a mutation, a candy that is misshapen, or pointier, or flatter than the rest. Almost invariably this proves to be a weakness, but on very rare occasions it gives the candy extra strength. In this way, the species continues to adapt to its environment.

When I reach the end of the pack, I am left with one M&M, the strongest of the herd. Since it would make no sense to eat this one as well, I pack it neatly in an envelope and send it to M&M Mars, A Division of Mars, Inc., Hackettstown, NJ 17840-1503 U.S.A., along with a 3x5 card reading, "Please use this M&M for breeding purposes."

This week they wrote back to thank me, and sent me a coupon for a free 1/2 pound bag of plain M&Ms. I consider this "grant money." I have set aside the weekend for a grand tournament. From a field of hundreds, we will discover the True Champion.

There can be only one.
Click to expand...
 
Despyzer said:
More M&M eccentricities

Whenever I get a package of plain M&Ms, I make it my duty to continue the strength and robustness of the candy as a species. To this end, I hold M&M duels.

Taking two candies between my thumb and forefinger, I apply pressure, squeezing them together until one of them cracks and splinters. That is the "loser," and I eat the inferior one immediately. The winner gets to go another round.

I have found that, in general, the brown and red M&Ms are tougher, and the newer blue ones are genetically inferior. I have hypothesized that the blue M&Ms as a race cannot survive long in the intense theater of competition that is the modern candy and snack-food world.

Occasionally I will get a mutation, a candy that is misshapen, or pointier, or flatter than the rest. Almost invariably this proves to be a weakness, but on very rare occasions it gives the candy extra strength. In this way, the species continues to adapt to its environment.

When I reach the end of the pack, I am left with one M&M, the strongest of the herd. Since it would make no sense to eat this one as well, I pack it neatly in an envelope and send it to M&M Mars, A Division of Mars, Inc., Hackettstown, NJ 17840-1503 U.S.A., along with a 3x5 card reading, "Please use this M&M for breeding purposes."

This week they wrote back to thank me, and sent me a coupon for a free 1/2 pound bag of plain M&Ms. I consider this "grant money." I have set aside the weekend for a grand tournament. From a field of hundreds, we will discover the True Champion.

There can be only one.
Click to expand...
Click to expand...
:lmao:
 
Despyzer said:
More M&M eccentricities

Whenever I get a package of plain M&Ms, I make it my duty to continue the strength and robustness of the candy as a species. To this end, I hold M&M duels.

Taking two candies between my thumb and forefinger, I apply pressure, squeezing them together until one of them cracks and splinters. That is the "loser," and I eat the inferior one immediately. The winner gets to go another round.

I have found that, in general, the brown and red M&Ms are tougher, and the newer blue ones are genetically inferior. I have hypothesized that the blue M&Ms as a race cannot survive long in the intense theater of competition that is the modern candy and snack-food world.

Occasionally I will get a mutation, a candy that is misshapen, or pointier, or flatter than the rest. Almost invariably this proves to be a weakness, but on very rare occasions it gives the candy extra strength. In this way, the species continues to adapt to its environment.

When I reach the end of the pack, I am left with one M&M, the strongest of the herd. Since it would make no sense to eat this one as well, I pack it neatly in an envelope and send it to M&M Mars, A Division of Mars, Inc., Hackettstown, NJ 17840-1503 U.S.A., along with a 3x5 card reading, "Please use this M&M for breeding purposes."

This week they wrote back to thank me, and sent me a coupon for a free 1/2 pound bag of plain M&Ms. I consider this "grant money." I have set aside the weekend for a grand tournament. From a field of hundreds, we will discover the True Champion.

There can be only one.
Click to expand...
Click to expand...
:tumbleweed: :tumbleweed: :tumbleweed:
 
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