Tangled in the Rigging 2: Probabalistic Boogaloo

I can't believe I have to do this again.

It was two years ago, and the New Orleans Hornets had just won the NBA Draft Lottery. With the third-worst record in the league, the Hornets had a 13.7 percent chance to win the 2012 Anthony Davis sweepstakes. This is equivalent to a coin coming up heads three times in a row -- i.e., unusual but not super rare -- but of course everyone cried foul since the league owned the Hornets at the time and was working to sell them to a private owner.

The draft lottery being rigged is one of the oldest jokes known to man. That joke is so old, creationists have to insist it was created by the Devil to test our faith. That joke went stale so long ago, the mold growing on that joke was used to discover penicillin*. Enough.

* - If you like this construction, here are some more examples.

And even reasonable people -- people who love them some numbers! -- fall into this trap. Here's Grantland's Bill Barnwell on Twitter tonight:

Odds of the Cavaliers winning 2011, 2013, and 2014 lotteries: 13,467-to-1.

— Bill Barnwell (@billbarnwell) May 21, 2014

This is technically true:

P(Cavs win in 2011)*P(Cavs win in 2013)*P(Cavs win in 2014) = (2.8%)(15.6%)(1.7%) = .0000743,

or 13,467-to-1. But really, that's disingenous for a couple reasons. First off, the Cavs actually had two chances to win in the 2011 lottery: their own chance, based on their own abysmal performance (19.9%) and the Clippers' chance, acquired by trade (2.8%). So their actual probability of winning the first pick in 2011 was 22.7%, improving their odds to 1,661-to-1.

@Doctor_Bryan @Alex_Kienholz Yep. I should have said "with the picks they won with".

— Bill Barnwell (@billbarnwell) May 21, 2014

Okay, fine. But even then, we're still not talking about the odds of the Cavs winning three of four lotteries. We're talking about the odds of the Cavs winning those three lotteries in that order. If we really wanted to find the odds of the Cavs winning three of the last four lotteries, we'd need to calculate something like this:

P(2011)*P(2012)*P(2013)*P(~2014) + P(2011)*P(2012)*P(~2013)*P(2014) + P(2011)*P(~2012)*P(2013)*P(2014) + P(~2011)*P(2012)*P(2013)*P(2014),

where P(x) is the probability the Cavs won in year X, and P(~x) is the probability the Cavs didn't win in year X.

Also, this sequence of lottery selections isn't even the least improbable four-year stretch in the last decade. From 2005-2008, four straight lottery winners had less than a 10% chance of winning. The odds of those four teams winning is 200,194-to-1. But because the picks went to four different teams, no one thought it was that unusual.

That's because every lottery outcome is unlikely until it happens. Here's another example: the Rhode Island lottery has a four-digit numbers game, where you have to pick four single-digit numbers in order to win the grand prize. Sunday's winning numbers were 9-2-9-9. Now, the odds of drawing three nines in a four-number drawing could be expressed as 1,000-to-1, and that's correct, but are the three nines really so unlikely? No. They're no more or less unlikely than any other three-digit combination*; we're just wired to pick out those patterns.

* - We're assuming it's a fair lottery here, but it is Rhode Island, so who knows.

Sorting Through a Million Bags of M&Ms

As a kid, I used to sort bags of M&Ms by color. This was my first sort of data science project, and my parents' first clue that this one was a little off. Every once in awhile, I'll revert to that habit (especially with those fun-size bags you get around Halloween), which led to a long, drawn-out discussion with a friend about the probability of getting a fun-size bag of Skittles with no purples.

So when a recent trip to the vending machine produced a free bag of M&Ms, I found myself asking a number of questions about the distribution of the different colors in a bag of M&Ms. A quick Google search produced no official statement from the company, except this one from 2008:

Our color blends were selected by conducting consumer preference tests, which indicate the assortment of colors that pleased the greatest number of people and created the most attractive overall effect.

On average, our mix of colors for M&M'S MILK CHOCOLATE CANDIES is 24% cyan blue, 20% orange, 16% green, 14% bright yellow, 13% red, 13% brown.

Each large production batch is blended to those ratios and mixed thoroughly. However, since the individual packages are filled by weight on high-speed equipment, and not by count, it is possible to have an unusual color distribution.

Well, we have two bags of M&Ms here. We can check to see whether these proportions are still accurate using a chi-square goodness of fit test. Since there are six colors, we will be looking at a distribution with five degrees of freedom.

These tables show the result of the chi-square calculation for each bag.

BAG 1	Red	Orange	Yellow	Green	Blue	Brown	Total
Observed	8	12	4	10	12	8	54
Expected	7.02	10.8	7.56	8.64	12.96	7.02	54
(O-E)^2/E	0.137	0.133	1.676	0.214	0.071	0.137	2.369

BAG 2	Red	Orange	Yellow	Green	Blue	Brown	Total
Observed	8	7	2	14	12	11	54
Expected	7.02	10.8	7.56	8.64	12.96	7.02	54
(O-E)^2/E	0.137	1.337	4.089	3.325	0.071	2.256	11.22

For a 95% confidence interval (alpha = 0.05), x^2 = 11.0705 for a distribution with 5 degrees of freedom. This suggests that we have one normal bag and one outlier. This is not especially conclusive evidence for or against the 2008 distribution, but the significant lack of yellows makes me suspect the distribution has changed.

We can also ask questions about how unusual is a bag with only two M&Ms of a given color. Unfortunately, this is non-trivial to solve theoretically. But we can estimate these probabilities by simulating a large number of bags of M&Ms. I built a MATLAB script (available on request) to simulate an arbitrary number of 1.69-oz M&M bags. For convenience, I assumed each bag had a consistent number of M&Ms (54). I then drew 54 random numbers uniformly distributed on the interval [0,1], splitting up the number line to match the 2008 proprtions (i.e., a random number less than 0.13 meant a red M&M, a number between 0.13 and 0.33 meant orange, and so on). I repeated this process one million times, because "a million bags of M&Ms" sounded cool.


"A million bags of M&Ms isn't cool. You know what's cool? A billion bags of M&Ms."
Shut up, Justin.

Anyway, you end up with this graph showing the distribution for each color. It's discretized, of course, because 0.4 of an M&M is nonsensical. But, thanks to the central limit theorem, all of the distributions are normal. Note that the red and brown curves are essentially right on top of each other*.

* - Apologies to those of you, like my advisor, who are red-green colorblind, and thought M&Ms came in blue, yellow, orange, and a couple shades of brown.


Now that we have this data set, we can answer a whole bunch of questions.

What are the odds that I get a bag with less than N yellows?
For this, we can make cumulative distribution functions based on that figure. So, if we assume the 2008 distribution is accurate, the probability of getting four or fewer yellows in a bag is approximately 11%. And if each bag represents an independent sample (which might not be true, depending on the manufacturing process), the probability of getting two consecutive bags with four or fewer yellows is 1.2%.

What are the odds that I get a bag with less than N of one color?
Here we have to use a different curve. For instance, a bag with only four yellows seems rare from the previous graph, but remember: that just deals with the probability that you have four or fewer yellows, or four or fewer reds. This question deals with the probability you have four or fewer of one color, regardless of which color it is. And now we see that the probability you have a bag with no more than four of one color is about 48%. For two consecutive bags (assuming independence), the probability is a still-reasonable 23%. So, while getting two bags with a small number of yellows is unusual, getting two bags with a small number of any color is pretty common.


What are the odds that I get a bag missing a color?
I'm sure the process of setting those percentages involves minimizing this possibility: if you were six, and your favorite color was red, you might get upset if you went through a whole bag of M&Ms with no reds. As a result, this is a pretty uncommon occurrence: in my data set, the odds were approximately 1-in-690.

What are the odds that I get a bag that's entirely one color?
This never happened in the million trials I ran. In fact, the greatest number of any single color in one bag was 30 blues (out of 54).

What are the odds that I get a bag with equal numbers of all colors?
You would think this wouldn't be too crazy, but in fact it's very rare. I estimate that the odds are about 1-in-42,000.

What are the odds I get more blues than any other color?
Before I present these results, it's important to note that MATLAB's min and max functions don't deal with ties very well. Ideally, you'd had a function such that a two-way tie would count as 0.5 for each color, and a three-way tie as 1/3, but what actually happens is the left-most column gets 1, and everyone else gets zero. This means that red, orange, and yellow will be skewed a little high, and green, blue, and brown will be skewed a little low. But this is already a 1,000-word entry on candy-coated chocolate, so the min/max functions are good enough for me.