Kellogg’s has issued Froot Loops fruit snacks in the shape of digits. (Side note: Cheez-Its need to get on board with this! There have been Scrabble tile Cheez-Its for years. We want numbers, operations and relational symbols!)
Naturally I bought some.
Tabitha (8 years old) asked—as she does in these scenarios which occur with great frequency—Are you just buying that because it’s mathy?
Yes, sweetie. Yes I am.
But how to put them to use?
After many rejected ideas, here’s my favorite.
The task
Here are the contents of one pack.
That’s 5, 2, 9, 1, 3, 2, 4, 3, 9. Their sum is 38.
I’m setting the over/under on the sum of the next pack at 41. Do you want the over or the under? Why?
Play along with your questions and answers in the comments.
I’ll open the pack on Wednesday, May 27.
Overthinking my teaching 
I’m choosing “over” 41. I don’t know if all packs have 9 pieces, but if they do, and there’s an equal chance of getting numbers 1-9, their sum is 45.
So, I was going to change my answer to “under” since there can be zeroes, and if each number could only occur once (which most likely won’t happen, but if it did…), then I think that out of those 10 combinations, under would occur 6 times and over would occur 4 times, for 9 pieces per pack. But if “sixes” can be considered “nines”, then I’m staying with “over” since under would now be 4 and over would be 6.
Yeah. All 9s. No 6s. It seems important to work this into your calculations.
I wonder what the margin of error will be for the number of fruit snacks in each bag? Watch the next one contain 8 fruit snacks, so I’ll play the under. Also, does each fruit snack weigh the same? Did they make all the numbers? Chris will have to keep us posted to let us know for sure that they make 5, 6, 7 & 8. Albeit, 8 is listed on the box, but you know how misleading advertisements can be.
Pete Kalenik
Orchard Park Middle School – Math Teacher
As far as I can tell, Pete, there are no 6s. Only 9s.
I’m over, for similar reasons as Monica. This pack was weighted low, with 4 of 8 numbers being 1, 2, or 3. If we assume equal production of the numbers, we’re going to have some packs with extra 6s, 7s, 8s. I’m curious, as you take data, if 45 does end up being the mean. I’m also curious if you’ll ever get a perfect pouch, with all 9 numbers and a sum of 45. The only thing I’m not curious about is the taste. Blech.
We don’t know if there’s a zero or not… we may never know 😉
I expect each pouch will contain the same fruit count, if only because kids will complain about this if they don’t get as many as they expect.
How many pouches will you need to open before seeing each digit once?
SUPER-LIKE!!!!!!!!!! 🙂
Yes, Bowen. There are zeroes.
You want the over or the under?
I was going to take the under anyway out of the possibility that it’s not a uniform distribution across 1-9, but with zeros, definitely under. You’ll need a lot of packages to have a convincing argument that the distribution is uniform, but you might not need so many if it turns out it isn’t.
I’m curious about this uniform distribution question, Bowen. Uniform distribution seems like a natural assumption to me. Non-uniform distribution seems to need a reason because someone would have had to make that decision. What reasons do you imagine the Kellogg corporation would have for skewing the distribution of Loops snacks?
Regarding the non-uniform distribution, this is suggested by there being two types of 3, while (possibly) only one type of 9. Say there were 16 different molds to use, you’d make them as 1122334455667890 or something like that. I’d want to see a second different color for some high number to believe it was uniform.
There’s a few things we don’t know. Will the next one have nine fruit loop numbers? Are there zeros. If there are no zeros the mean is 5. If there are zeros the mean is 4.5. Let’s call it 4.75 then. 9 x 4.75 is 42.75. That’s more than 41.
I’m going over.
(I like that the numbers can be more than one colour. I want to know if the colour and number are truly independent. Will need to open lots of packets for this…)
OK – so now I know there are zeroes…
That’s a mean of 4.5. 9 x 4.5 is 40.5.
Can I switch to under?
(If there are Cinderellas, or some other kind of fruit loop in there by some sort of production line mixup, I think they should count as zero too.)
Of course you may switch, Simon. We have time stamps on your comments so we’ll hold you to the last bet prior to package opening.
As someone who once worked scooping ice cream for a company who doesn’t even package candies—merely includes them in sundaes and such—yet still has official corporate policies about e.g. the number of Reese’s Pieces in a small child’s sundae (so that there are no inter-child fist fights, etc.), I’d be really surprised if the variance of fruit snack count were nonzero. Especially since nine candies are borderline subitizable, which could lead to almost instantaneous whining.
Here is definitive proof that variance is non-zero, Lusto. Ten packs, 92 snacks.
Christopher – of those ten packs, did you determine the average sum?
No, Monica. Hopefully someone will be so inspired and report back. The photo provides all we the info we need in order to do that.
I love that there’s 9 – I think that’s intentional. So I’ll do as above, call 5 the average, and say the next pack is over.
I want to know why you set the over at 41 instead of 38. I also want to know the probability of 1,2,3,4,5,6,7,8,9.
P(0) doesn’t interest you, goldenjo?
I actually meant the probability of a straight. So now I’ll include 0-8 as well as 1-9. Numerically, didn’t know about 0, or 6=9. I’ll assume 0-9 digits, count 6s as 9s, so 4.8 average. 41 makes more sense now, so I’ll take under. 2 tenths is two tenths, right?
Game proposal: high low. Take your pack and make the biggest sum of two digit numbers you can and the smallest, no repeats. 1 point for each, no points for ties. Play until someone wins. Has some interesting strategy given the digits in your pack. Would the play for your open be: 12+23, 98+95? Strong pack! (9/6 can be used as you wish.)
Game proposal 2: patterns. You play a number, they play a number. Last to play continuing the pattern mod 10 gets a point and starts the next pattern.
Doh! Can’t multiply. Did 4×9+.8×6. So I say EV= 43.2, I’ll take over.
Random additional thought Chris – Can you differentiate between 6 versus 9?!?!?
How will we decide whether a given digit is a 6 or a 9?
Christopher, how did you decide that those were 9s and not 6s?
Also: as every package opened so far has had a digit sum of under 41, my working wager is to go with the “under” bet. Data is king! Long live data!
As we discussed on Twitter, Justin, I don’t think my ability to tell a 9 from a 6 is unique among math teachers.
Under. For kids snacks, I would think that the more 0-5 numbers would be made.
Say more about why this should be, lukewalsh.
Do some digits have more calories than others? I notice there are exactly 70 calories per pack.
I have no data to offer on this front, Max. I have tried in vain to apply my patented Oreo-Component-Caloric-Analysis Techniques (OCCAT(TM)).
I’ve got my eye on you.
Makes me want to go buy a pack and see what I get. I think over, hoping for those 6s that can be 9s.
Unders. Data skewed right because of the 9’s.
Wait! How is this not an argument for the over?
Because the 38 is being pulled UP to 38 by the 9s. If the next one isn’t pulled up by 9s, it will be lower.
Ah. I see. The 9s in THIS sample make its sum larger than we’d expect. Got you down for the under then.
I’m thinking if 0-9 are all equally likely, which seems reasonable and not blatantly contradicted by the pack we see, then the peak & middle of the distribution for a packet of *10* would be the sum of 0 to 9, aka 5*4 + 5*5, or 45. If you then took one away to make a packet of 9, on average, you’d be taking away 4.5. So for the packet of 9, I’m thinking 40 & 41 would be the most likely, with the probabilities falling away symmetrically on either side (with 0 and 81 being the lowest, while still possible).
So you’d think I’d pick under 41 as being more likely than over 41.
BUT. It is clear Christopher has hexaphobia (fear or hatred of the number 6), since he has not yet given a satisfying explanation for why he claims those are 2 9’s instead of a 9 and a 6 or 2 6’s. If there is a 6, he’s going to call it a 9, and Christopher is in control of the counting. On average, there will be close to 1 6 per package whose true nature Christopher will refuse to see. So Christopher will add almost 3 to many totals, making the most likely outcome more like 43-44. So I’m picking over.
My love of hexagons is well documented. How does this interact with your psychiatric theorizing, Monica?
Hmmm. The documented love of hexagons did stump me, until I got to your 12/15/11 comment about your “never-ending quest to problematize the routine”.
NEW THEORY: You don’t actually hate or fear 6’s, you’re just pretending to for the sake of making the problem more mathematically interesting.
Pseudohexaphobia!
I don’t know you well enough to know whether you will consistently maintain there are no 6’s, or whether you will suddenly abandon the pseudohexaphobia like Lucy yanking the football.
The battle of wits has begun. It ends when you decide and we both open a packet of gross chewy shapes, and find out who is right…and who is dead. (Remake of The Princess Bride. I think this makes me Vizzini.)
Christopher, I think we want some kind of statement about how you will deal with the 6-9s. Are you going to call them all nines? Or sixes? Six-nines? Seven-and-a-halves? This could swing the whole thing entirely.
There are only 9s in these packs. No 6s. The value of each will be counted as 9. You want the over or the under, Simon?
OK, OK… so that’s an average digit score of 4.8 now. 9 x 4.8 is 43.2…. I’m going back over!
Christopher – How many gummy pieces should we be assuming? Your pack has nine pieces, and I found this site: http://www.taquitos.net/candy/Froot-Loops-Numbers, that also mentioned nine pieces/pack.
Awesome find, Monica. Thanks for linking us up! You know everything I know about the contents of these packs. I notice that the review mentions about nine pieces per pack.
Megan’s observation of the skewed nature of the sums is exciting, if we consider only 9’s and no 6’s. There are few great examples of skewed distributions to use in class experiments.
OR, can I poropose that every 6/9 creature be counted as a 7.5? If we work under the hypothesis that numbers are equally distributed, this is an acceptable work-around. But, do we have evidence that the 6/9’s occur twice as often the rest of the numbers in the pack?
There goes my prep….thanks for that….
Ah, that’s a point… there may only be as many six-nines as there are any other digit…
This is great! Here are my contributions. From my extensive experience with the jumbo boxes of fruit snack packages from Costco (which I, er I mean, my kids, love) I can say that there is a fair amount of variance among the number of snack pieces per packet. I think most of these things are packaged by weight and not by # of pieces (much easier to ‘count’ that way for a machine, I think). My Kirkland Signature snacks contain as few as 7 and as many as 11, but I will admit that I don’t count every package I open.
I’m going to agree with Bowen and assume the distribution is not uniform, and here is why, Christopher: Clearly, to come up with such a fantastically mathy product the team at Kellogg’s must consist of some pretty prodigious mathematicians and statisticians, who are most certainly aware of Benford’s law. This seems like a natural place for that law to apply, so I’ll assume 1, 2, and 3’s are more likely than 7, 8, 9. Plus, that would make for a great Easter Egg for all of us who have the time and energy to overthink this!!
Long story short, I’m betting the next pack has 10, but I’m still taking the under.
The obvious assumption to me seems to be uniform distribution o 0,1,2,3,4,5,7,8,9 (though choosing to not include sixes – they could just use a different color! – seems odd to me). I decided to let that be my null hypothesis, and went to see if I had any ground to reject it.
Were the sample larger, we could use a chi-square goodness of fit test, which compares observed counts of a distribution to an expected set of counts. Since we have 9 categories and only 9 snacks, the expected value for each category is only 1, which is not large enough for a chi-square to be accurate.
I ran one anyway, and got a p-value of 0.65. In other words, if we somehow assume a chi-square test makes sense in this case, then there is a 65% chance that our distribution could be this far off, given that the actual distribution is uniform. That obviously is not enough to reject the assumption that the distribution IS uniform.
However, since I don’t think the chi-square is a very good choice, I decided to calculate the EXACT probability of being “this bad or worse” under an assumption of uniform distribution. I decided that there were only four types of bags “bettter” – meaning more uniform – than this one: bags where you get one of everything, bags where you get 2 of one number and 0 of another (but everything else you get one), bags where you get 2 of two numbers and 0 of two numbers, and bags where you get 3 of one number and 0 of two numbers. I calculated exactly how many ways you could pull numbers out of the bag (in order) for each of these options, put it over the total number of possible bags (9^9), and found that only 25.1% of draws would be better than this one IF the distribution was uniform. Meaning there is a 74.9% chance of getting a bag this bad if the distribution is uniform, which is a very high p-value and certainly not one we should use to reject that assumption.
Therefore, I run on that assumption. Uniform distribution of 0,1,2,3,4,5,7,8,9 gives an expected sum of nine snacks to be simply 0+1+2+3+4+5+7+8+9 (very convenient that number of snacks in a pack = number of options) = 39. I choose under.
Probability work is here: https://drive.google.com/file/d/0B-C-lUvv4rQ4RzVYMjFUczJUT1k/view?usp=sharing
I’m going to apply some principles of parsimony and say that the one-bag sample serves more as an illustration of the configuration than giving us much data to go on. I think a uniform distribution is a reasonable “null hypothesis”; the big question seems to be whether they make twice as many 6-or-9s than the other numbers. If we figure they don’t (which I base on my many hours watching How It’s Made episodes about processed food), then I’d say the expected value is 39 so I’d take the under.
Two things I have to share: A similar mystery about M&Ms (http://www.npr.org/sections/money/2014/06/06/319509152/episode-544-the-m-m-anomaly) and a T-shirt you have to have (though apparently they’re not selling them new any more): http://images.cryhavok.org/v/Questionable+Content+Math+Is+Delicious.png.html
Christopher – I got impatient and found a box of Numbers at the store this afternoon. I immediately opened a pack – there were four zeroes – what are the odds?
And how many were in the pack, Monica? And how many 6 / 9s were there?
There were nine pieces in the pack. One 6/9. Since then I’ve opened one more pack, again nine pieces, with very different results.
You may have to switch your instruction to grammar. Then you can use these. http://www.wired.com/2010/09/eat-your-words-how-to-play-scrabble-with-cheez-its/
I’m going under 41 and saying the sum will be 39. I don’t believe the sum will be a prime number.
Here’s our answer.
So now I’m curious – have you opened all the packages in the box yet? Is the average “under”?
Wow, 39 – Señor Stadel got it on the head! Proof that all that estimation works!
I say it will be under because I think that Kellogs will manufacture more lower numbers than higher numbers. My specific vote is for 30,
Both 2’s are blue. Both 9’s (or 6’s) are orange. Even though the two 3’s are different colours, I feel like within a single “10 pouch pack”, the number-colour correspondence will be maintained more often than not.
As there’s also two greens, I’ll further extrapolate that the next pack will have more reds, yellows and purples to balance the excess of orange, green and blue here. Those colours will more likely correspond to “0” “1” (as here) “5” (as here) “7” & “8”. A total of 21. (This as opposed to 2, 3 (an anomaly), 4, and 9 twice (no 6), for a total of 27.)
Hence, I’m thinking under for the next – not only under 41, but perhaps under this 38. I also wonder whether Christopher would have set 6 and 9 as different had they been different colours!
I’m taking the under. You confirmed zeros are possible AND I think you have to count some of those “nines” as “sixes” if you have more than one in the pouch. It would be fair to count the first one out of the pack as a 9 the second as a 6 the third as a 9 etc. 🙂
A different approach: assuming that the numbers distribution is indeed uniform (although the point made by Bowen Kerins about the molds having a different amount of ‘1s’ than ‘8s’ made me doubt about this assumption) I created a small simulation using a spreadsheet and produced a thousand different packs (I made sure that the probability of each number to appear was the same and that there were no 6s, only 9s). The average sum of those thousand packs was 38.9 so I’m choosing ‘under’.