Of all of these shows, though, the one that most piques my mathematical interest is TLC’s Four Weddings.  Based on a British show with the same name, the premise is as follows: four brides-to-be, unknown to one another, meet and attend each others’ weddings.  When one bride gets married, the other three score various aspects of the wedding, and the bride with the highest score among the four wins a honeymoon (contingencies are in place in the event of a tie, though these are not always explained and seem to vary from season to season).  In order to make for good TV, the show frequently manages to bring out the worst aspects of these women, as they nitpick and pass judgment on everyone else’s wedding.  Here’s a short clip to give you a taste for what this show is all about:

How are the weddings scored?  This process is explained in detail during the course of each episode.  The wedding is broken down into four categories: dress, venue, food, and overall experience.  For overall experience, the other three brides in attendance give a score from 1 to 10 (though I’ve never seen a bride give another wedding a 10).  For the rest of the categories, though, the brides can only rank the weddings as being 1st, 2nd, or 3rd in the given category.  1st place gets 10 points, 2nd place gets 6, and 3rd gets 3.  The total possible number of points a wedding can score is therefore 120.  At the end of each episode, the scores for each bride are broken out in detail.  The wedding budget and headcount are provided as well.

After watching a few episodes, it seemed like the best way to ensure a win was to simply outspend your competitors.  Certainly a large budget can help improve the guest experience or score a hip dress, but I was curious as to what overall trend (if any) could be made between, say, money spent on a wedding and the wedding’s overall score.

With this noble goal in mind, I proceeded to DVR 28 episodes of this show.  After recording the scores for 112 weddings, some of my questions were as follows: is the amount a person spends on a wedding correlated to the score they receive from the other competitors?  What about the amount a person spends per guest?  Finally, did the most expensive wedding win more frequently than would be expected by pure chance?

Let’s look at some data.  Here is a scatterplot of each wedding’s budget, vs. the total points earned.

Click to embiggen!

As the dots suggest, there is a slight positive correlation between the amount one spends on a wedding, and the score one receives from one’s fellow competitors.  The coefficient of correlation here is approximately 0.286, though if we discard the two $150,000 weddings, this improves to around 0.348.

Of course, a $10,000 wedding for 10 people might be a much nicer affair than a $10,000 wedding for 1,000 people.  If you spend more money per guest, does this translate into a higher score as well?  The dots don’t lie; here’s another scatterplot:

Click to embiggen!

There is quite clearly an outlier in this set of data – this corresponds to a bride who spent $150,000 for a 120 person wedding, for a whopping $1,250 spent per guest (this particular bride did end up taking first place).  Eliminating this outlier, though, the correlation here is weaker than the correlation for actual cost, at a meager 0.098.  In other words, there may not be a linear relationship between cost per guest and total score.  This may be because certain fixed costs, such as the dress and the venue, won’t necessarily vary much with the guest total, unlike something like food.

This analysis, though, obscures a key point.  In order to win the honeymoon, you don’t necessarily need a high score; you only need a higher score than your three other competitors.  With this in mind, it may be simpler to just look at the scores episode-by-episode, and see how the amount spent on a wedding compares to the wedding’s relative ranking.

In the graph below, the blue bars count the number of times the most expensive wedding was given a certain rank.  The red bars count the number of times the wedding with the highest cost per guest was given a certain rank.  Note that if cost had no bearing on the rank, we would expect an equal number of weddings in each rank (in this case, about 7 per rank).

Click to embiggen!

As you can see, spending the most money seems to bestow an advantage: fully 50% (14 out of 28) of the most expensive weddings were ranked first.  This would be unlikely if cost had no impact on ranking.

The picture is a little murkier for cost per guest -the frequency for each ranking sticks pretty closely to 7, so it’s not clear that spending more per guest gives any advantage.

In conclusion, there is a small positive correlation between amount spent and score received, though this does not transfer to the amount spent per guest.  Compared only to one’s other three competitors, the amount spent appears to confer an even greater advantage, so if you are on this show and want to show your competitors that you are better than them, my advice would be to simply outspend them.  If all you’re interested in is a nice vacation, though, it may be cheaper to just stick to your budget, and plan a vacation on your own.

Our cat's third birthday is also this week. It is unclear which event he is celebrating, although the dilated pupils suggest he is celebrating a bit too hard.

I’d like to share with you some data on the geographic distribution of my US readers.  While there is a large California bias, people from all over the country seem to have stumbled upon this corner of the internet, and have hopefully enjoyed their time here.

This represents you, gentle reader. Darker green means more viewers.

Of course, a California bias shouldn’t be all that surprising.  After all, California is the most populous state in the country, accounting for roughly 12% of the country’s total population, according to this 2010 Census data.  A more interesting thing to think about, then, is not the map pictured above, but how the map contrasts with actual population data for each state.  For example, New York is a hair darker than Texas on this map, even though Texas has a larger share of our population: roughly 8% as compared to 6% for New York.

One way to compare population data with Math Goes Pop visitor data is through rankings.  This table compares the rankings, and shows the relative difference for each state (feel free to play around with the data to your heart’s content):

State	View Proportion Rank	Population Proportion Rank	Rank Difference
California	1	1	0
New York	2	3	+1
Michigan	3	8	+5
Texas	4	2	-2
Illinois	5	5	0
Pennsylvania	6	6	0
New Jersey	7	11	+4
Massachusetts	8	14	+6
Florida	9	4	-5
Virginia	10	12	+2
Washington	11	13	+2
North Carolina	12	10	-2
Ohio	13	7	-6
Maryland	14	19	+5
Georgia	15	9	-6
Missouri	16	18	+2
Connecticut	17	29	+12
Minnesota	18	21	+3
Oregon	19	27	+8
Arizona	20	16	-4
Colorado	21	22	+1
Wisconsin	22	20	-2
Tennessee	23	17	-6
Indiana	24	15	-9
Washington DC	25	50	+25
South Carolina	26	24	-2
Louisiana	27	25	-2
Kentucky	28	26	-2
Iowa	29	30	+1
Utah	30	34	+4
Oklahoma	31	28	-3
Alabama	32	23	-9
Kansas	33	33	0
Nebraska	34	38	+4
Arkansas	35	32	-3
Nevada	36	35	-1
Mississippi	37	31	-6
New Hampshire	38	42	+4
Rhode Island	39	43	+4
New Mexico	40	36	-4
Idaho	41	39	-2
Maine	42	41	-1
West Virginia	43	37	-6
Hawaii	44	40	-4
Vermont	45	49	+4
Alaska	46	47	+1
Montana	47	44	-3
Delaware	48	45	-3
North Dakota	49	48	-1
South Dakota	50	46	-4
Wyoming	51	51	0

The rankings give us some information: we see that Indiana and Alabama are not as well represented in readership as one might expect given their population rankings (both states have Math Goes Pop readership rankings 9 levels below their population rankings).  On the other hand, folks from DC, Connecticut, and Oregon are visiting this site more than would be expected based on population numbers alone; readership rankings for these states are 25, 12, and 8 levels above their population rankings, respectively.

But while the rankings tell us some things, they leave a great deal out.  For instance, while California is ranked first in both the proportion of the US population and the proportion of visitors to this site, the rankings tell us nothing about how these proportions compare to each other.  In fact, while California boasts the number 1 proportion in both categories, the proportion of California viewers to Math Goes Pop is more than twice the proportion of California’s population (26% of my viewers vs. 12% of the US population).

Comparison of the proportions in this way allows us to get a better understanding of how visitors to this site are distributed across the country, as compared to the overall population distribution.  While the population is distributed more heavily in California, the proportion of California Math Goes Pop visitors is greater than can be explained by simple population data.

If we compare the state-by-state readership proportions to overall population proportions, we get the following picture:

Click to Embiggen!

Big ups to our nation’s capital for having the largest share of viewership relative to its overall share of the country’s population.  In addition to DC, the proportion of readership from 10 states is greater than the state’s proportional population: California, Massachusetts, Michigan, New York, Connecticut, Maryland, Oregon, New Jersey, Washington, and Vermont.  For example, while Massachusetts accounts for roughly 2% of the country’s population, thus far it has accounted for nearly 3.5% of Math Goes Pop readership.

I won’t go into an analysis of why some states might be over- or underrepresented in the blog’s readership according to this metric.  I just thought it might be appropriate to share some of this data in commemoration of Math Goes Pop!  I hope you will continue to enjoy the material posted here.  And if you live in Wyoming, Mississippi, Alabama, or any of the other 40 states I didn’t mention in the paragraph above, let’s see what we can do to get some mathematical love flowing in your neighborhood.

As a shining beacon of what is hip, MTV has a responsibility during its movie awards to highlight the most popular films of the year. This is in stark contrast to the priorities of higher brow award shows such as the Oscars, for which artistic achievement is placed on the highest pedestal. This is not to say that these two goals need be mutually exclusive; indeed, since the first MTV Movie Awards was broadcast in 1992, the “Best Film” has agreed with the Academy Award winning best film three times (1997′s Titanic, 2000′s Gladiator, and 2003′s Lord of the Rings: The Return of the King). Even so, a quick glance at the nominated films from these two awards shows each year reveals a fairly small overlap, in general. But if MTV’s goal is to prop up films from an angle focused more on pop culture, it is natural to ask how good of a job they do.

Delicious metal popcorn. (Photo: Jason McDonald/MTV)

This question begets another one: how can we best measure a film’s popularity? My first thought was to consider the rankings on IMDB. There, users can give any film a score from 1 to 10; as a prime example of a range voting system, this seemed like a good place to measure the public’s reception of a film.

The results were mixed. With this metric, comparing the 20 years that both awards shows have been around, the MTV best film scored higher than the Oscar winning best film only 5 times. Oscar trumped 12 times, and the two awards tied three times. The trend is also worth mentioning – after 5 consecutive years of beating or tying the Oscars in IMDB score from 1999-2003, the Oscar winning film has bested the MTV winning film ever since. The disparity has become even larger in recent years (I call this the Twilight effect, as the Twilight films have won best film at the MTV awards for three years running). Here’s a graph of the scores over time (the list of MTV Best film winners is here; Oscar winners can be found here):

While you may argue there isn’t enough data here to draw much of a strong conclusion, the recent trend is fairly convincing. By this metric, it seems like Oscar winning films, at least over the past few years, seem to have been more popular.

Rather than looking at only the winner, though, you might expect to get a better sense of the popularity of films on display by looking at all nominated films, rather than just the winners. If we take the average IMDB score for all nominated best films at each awards show, rather than just the winning film, we get the following picture:

It’s come close, but the average IMDB score of MTV nominated films has never been greater than the average IMDB score for Oscar nominated films. We see the Twilight effect among the averages as well, though it was dampened somewhat this year due to the inclusion of critical darlings Inception, The Social Network, and Black Swan on the MTV nominee list.

“Hold up,” you may be thinking to yourself, “this is all a bunch of hooey.” You may think that IMDB scores are not a very good measure of a film’s popularity. It may be quite likely that IMDB scores are biased towards those same features of a film that make it more likely for Oscar consideration. Perhaps the type of person who goes onto the website to rate films is more likely to be somewhat of a connoisseur, and therefore the tastes reflected by the IMDB community are more likely to reflect the tastes of the Oscars. At the very least, it seems likely that teenage girls are not voting on the website in droves; how else can one explain the Twilight series’ limp average of only 4.9?

What else can we use to measure a film’s popularity? Well, to return to teenage girls, they don’t show their support for the Twilight series by rating it highly on the internet. They show their support by going out and seeing the movie multiple times. So perhaps we should look at box office receipts rather than IMDB score (and, of course, by picking sides in the bitter feud between Edward and Jacob). What sort of picture do we see in this case?

If we only consider the winning film from each awards show, the data looks like this (I’m only considering US box office numbers here):

(I’ve cut the graph so that you can’t see the leap in 1997, when Titanic took top prize at both shows.) Things look a little more erratic, but if you look closely, you’ll see that the MTV award winner has taken in more money than its Oscar winning counterpart 14 times out of 20. The Oscar has favored the larger cash cow only three times.

As with the IMDB rankings, we can try to smooth things out by looking at the average box office returns among nominees, rather than just returns for the winner. This yields the following graph:

These numbers aren’t adjusted for inflation, which may explain in part the growth trend (2009 numbers are bumped up because of Avatar, as well). I’m less interested in the actual numbers than I am the difference between the two graphs. And here we see, in contrast to the IMDB case, that the roles of the two awards shows have flipped. While the average IMDB score of Oscar nominees has always been higher than the average IMDB score of MTV nominees, the average box office return of MTV nominees has always been higher than the average box office return of Oscar nominees.

To return to the original question: do the MTV movie awards highlight more popular films than the Oscars? Well, it depends on how you define “popular.” If popular means highly rated, the claim is somewhat dubious. But if popularity is measured by the almighty dollar, then this seems like a fair conclusion to draw. Whatever your line of thinking, though, I’m fairly confident that current trends will continue, at least until the last of the Twilight films has exited theaters.

It was the first time in 18 years that such a quirky thing happened with a full schedule. On Aug. 10, 1993, all seven NL games featured one team scoring precisely two runs, STATS LLC said.

The last time it occurred with five or more runs was July 20, 1955, when all four AL games had at least one team score exactly six, STATS LLC said.

When I read this article, some questions immediately came to mind: exactly how rare is it for one team in a collection of 7 baseball games to have a common score of 5?  Also, if 7 teams in 7 games have the same score, which score are they most likely to share?  Are the 7 games with a common score 0f 2 more or less likely to occur than the 7 games with a common score of 5?

We can answer these questions with some (relatively) simple probability models, given some caveats.  I’d like to estimate these probabilities using only one parameter: the average number of runs a team scores during a game.  Of course, that average will vary from team to team, and also from year to year (in particular, runs per game have declined from the heyday of steroid-mania that gripped baseball at the turn of the millennium).  Due to different rules, there may also be variation between the American and National Leagues.  Let me ignore this, though, and consider only an average number of runs per game overall – what we lose in precision we will more than make up for in clarity.

Ahh, the late 90's, when it was easier to sock a few dingers.

The question remains: how many runs are scored on average in a baseball game?  I found some data online which is somewhat outdated, but I’ll stick to it for convenience (and, more importantly, out of laziness) – any alteration in this number is easy to propagate throughout the following discussion.  In this article from 2005, the author tabulated the average number of runs per game in MLB over a 5 year span from 2000-2004 (that’s over 12,000 games!).  He has a nice looking graph of the distribution of scores as well:

A savvy probability student might see the long tail of this probability distribution and liken it to the Poisson distribution, a distribution encountered in many probability courses, and which is frequently motivated by a desire to model “rare events.”  I put the term in quotations since what constitutes “rare” is frequently left undefined, and in any event, is not really pertinent to this discussion.

Let us suppose, then, that the number of runs scored per game by each team follows a Poisson distribution.  French aside, this means that the probability a team will score n runs is equal to

,

where A is the average number of runs scored per game – in this case, 4.82, and e is the unsung hero sometimes known as Euler’s number.  Don’t worry too much about this formula; if you prefer, the graph of the function looks like this (courtesy of Wolfram Alpha):

Note that the fit isn’t perfect – this graph starts much lower at 0 than the graph of the actual data pictured above, for example – but there is precedence for using the Poisson distrubtion to model runs in a baseball game (this article provides one such example, but a subscription is required to view it in its entirety).  More careful analysis is possible, and can be found in resources like this one, but again, I want to keep things relatively simple.

So, let us suppose that the probability that a team scores n runs is .  What then, is the probability than in a baseball game, one of the teams will score n runs?  Either team A can score n runs or team B can score n runs, but they can’t both score n runs since baseball games can’t end in a tie.  This means that the probability of A or B scoring n runs is simply the probability that A scores n runs plus the probability that B scores n runs, or

For the odds that this happens 7 times, we then multiply this number by itself 7 times (lurking under this is the assumption that runs scored in different games are independent, which seems like an entirely reasonable assumption to make).  To summarize, we estimate the probability that one team in each of 7 games scores n runs is

If n = 5 (as it did earlier this month), the probability is roughly .064%.  In other words, if 7 AL games were played every day, you would expect this outcome once every 1,560 days or so.  Having said that, with more careful analysis it’s possible to show that in fact, if 7 games will have teams scoring the same number of runs, 5 is the most likely number.  For comparison, when n = 2 the probability is only a paltry 0.00812%, making what happened on May 8th over 75 times more likely than what happened on August 10, 1993.  Of course, it’s not fair to compare these records to the 6 run record in 1955, since in that case only 4 games were played, rather than 7.  Nevertheless, it’s not difficult to adjust this model from 7 games to 4 games (or an arbitrary number of games).

So, rather than some murky intuition telling us this event should be unlikely, with a little more effort we can attempt to quantify exactly how unlikely this event should be.  More sophisticated models for runs could be used, but perhaps that is a topic I will save for another day.

An unlucky coin? Unlikely.

Um…what?  Who cares?  While 20/43 is slightly less than the expected 50%, this difference is not even close to being statistically significant.  Actually, the fact that this ratio is only 1 1/2 games shy of the mean is pretty good.  Matt Springer has posted an article that discusses why we shouldn’t really care about this difference.

This is the third in a series of posts about pools used for betting on the outcome of football games (part one can be found here, and part two here).  Let me briefly recall the setting, which is probably familiar to anyone who has been to a Super Bowl party.  Typically, one bets on the outcome of a football game using a 10 x 10 grid.  People can buy any number of the 100 squares on the grid, and when all the squares have been purchased, each row and each column is assigned a random digit from 0 to 9.

Suppose, for example, that you buy four squares, and after the rows and columns have been labeled, you find that you own square 3-7, square 2-5, square 9-0, and square 6-6.  You will win money if, at the end of any one of the four quarters, the last digit in each team’s score matches your pair.  For example, if the score after the 3rd quarter is 13-27, you will win some money, since the last two digits are 3 and 7, and you own square 3-7.  There are variants of this: some pools only pay out every half, not every quarter, and usually the payouts vary by quarter, so that having the right square at the end of the game wins you more money than having the right square at the end of the first quarter.

Here's an example of a football pool which has been tagged in the four squares mentioned above.