Learning Binomial Probability with Lucky Charms
St. Patrick’s Day rapidly approaches, and, this year, to get an early start on the festivities, I did what every fun-loving adult does to celebrate the holiday — that’s right, I bought a box of Lucky Charms to do a study of probability.
This particular box was of the 14.9oz (net weight) variety, and, having bought it at full price at a supermarket, it set me back a shade over $5.00. However, the goal is not to do a comparison shopping experiment… rather, it's to understand the nature of luck, and how it relates to the chances of finding a genuine chewy marshmallowy (Leprechaun-hat-embedded) four-leaf clover within a particular bowl-size pour of the Lucky Charms.
INTO THE MARSHMALLOW-VERSE
My assistant, along with one of my sons, carefully opened the box and interior bag, and did a methodical count of the number of each component within the population (math-speak for “whole box of cereal”). Their tally revealed the following:
Pink Hearts: 52
White-ish Unicorns: 50
Red Balloons: 44
Purple Horseshoes: 60 (but many were broken pieces, so revised down to 42)
Green Clovers Embedded in Yellow Leprechaun Hats: 30
Blue Moons: 39
Yellow Stars Shooting Through an Orange Sky: 36
Boring Cereal Pieces: 3,127
(Sidenote: Apparently, “horseshoe” is a relatively unstable, though admittedly appealing, shape for a marshmallow. So, after a delicate piece-matching, the team estimated the number of “whole horseshoes” that were dropped in by the General Mills marshmallow-loader prior to be jostled apart “in-box”.)
Ignoring the boring cereal pieces for the moment, the count revealed 335 total marshmallows -- so with 8 different types of marshmallows, that is an average of slightly less than 42 per type. Besides the above-mentioned horseshoes, the rainbows pretty much hit right on the average (as I guess one would expect from rainbows — known outside the leprechaun marshmallow-verse for their willingness to bridge the entire visible light spectrum).
Let’s look at the marshmallow “histogram”:
Hmm… I was expecting to find a more even distribution of chewy marshmallowy types within the box, but most importantly, look how few of those coveted hat-embedded clovers I received. Is it possible that clovers are indeed a rare find in the leprechaun marshmallow-verse? Or can it be that I just happened to be very unlucky here - that the leprechaun marshmallow-verse actually does contain an equal number of every type of marshmallow, and the great marshmallow-loader just happened to drop a disproportionately low number of clovers into my box?
Well, luckily, we can turn to probability to study this!
ANALYZING THE BOX
Assuming a random draw of 335 marshmallows (from a large enough marshmallow-verse that we can assume “replacement” - again math-speak here), and that there is a 12.5% (i.e. 1 out of 8) chance that each marshmallow drawn is a clover, what is the probability that we would draw 30 (or fewer) clovers? (And yes, one might contend that I should be looking instead at a Chi-squared test for uniform distribution, but because I’m specifically focused on clovers, the approach here is valid.)
The above is a “binomial probability” problem — that is, calculating the probability that Clovers ≤ 30 in a random draw of 335 marshmallows, given the assumption that P(Clover) = 1/8 in every individual draw. To solve this, we use the following formula:
The result is that the probability my box would have 30 or fewer Clovers is only 0.0261752. So, stated another way, if the macro leprechaun marshmallow-verse does indeed consist of 1 Clover out of every 8 marshmallows, I am one of the unlucky 2.6% of people who received 30 or fewer clovers in my 14.9oz box.
ANALYZING THE BOWL
OK, accepting my potential bad luck at the box level, I moved on to the next step of the investigation. Taking the consistency of my particular box as a given (i.e. 30 clovers among 335 total marshmallows - again assuming unbroken horseshoes), let’s now determine the probability that I will have at least 1 clover in each bowl.
Based on the suggested serving size of 27g, the box should contain 15.6 servings, but to make the math easier, I’ll go with the “approximately 15” serving suggestion beneath that. Including cereal pieces, we have 3,462 total pieces, so divided by 15, that’s approximately 231 pieces per bowl, and assuming an even distribution of pieces (ignoring the fact that it appears marshmallows may in fact skew slightly toward the bottom of the box), we can treat this as another binomial probability problem, where it’s easier to calculate the probability that we have no clovers in a particular bowl pour:
Where in this case…
because, there are 30 clovers out of 3,462 total pieces, and 30 ÷ 3462 = .00867.
The result is that there is a 13.4% chance that a particular bowl poured from my box will have no clovers in it, or the corollary, there is a 86.6% chance that a particular bowl will have at least one clover in it!
Taking that one step further, with yet ANOTHER binomial probability, the probability that every one of the 15 bowls poured from my box will have at least one clover is calculated as:
Which is just 86.6% to the 15th power. The result is 11.6%. Or, the easier-to-digest corollary — there is an 88.4% chance that at least one of the 15 bowls I pour from my particular box of Lucky Charms will have no clovers.
So, if I eat all 15 bowls of the Charms this Sunday, and find at least one chewy marshmallowy four-leaf-clover in every bowl, then I, my friends, will consider myself to have had a pretty lucky St. Patrick’s Day!