For a San Francisco 49ers fan like myself, most of the last decade has been spent in a fairly constant state of disappointment.  But after ten years without a playoff appearance, the team blossomed this season under the influence of new head coach Jim Harbaugh, and came within one game of their first Super Bowl appearance since 1995.

This poster hangs proudly in our apartment.

Despite a great season, in which the team won 13 of their 16 games for the first time since 1997, many people have still voiced doubts about whether or not this team is for real.  What if this season was just a fluke?  After all, quarterback Alex Smith, long jeered by 49ers fans since being drafted in 2005, was the starting quarterback for every game this season.  And while his stats this year were his best ever, they don’t compare favorably to quarterbacks like Tom Brady, Aaron Rodgers, or Drew Brees.  While the offense might not be the flashiest, this year it was relatively error-free; combined with stellar defense and special teams units, the 49ers were able to grind out wins fairly consistently from week to week.

The question remains, though, no matter how much the fans may try to ignore it: how much of a role did simple luck play in the 49ers season?  To answer this question, of course, one needs a way to measure the team’s luck, and compare it to other teams.  Alternatively, one needs to look at some measures of skill, and show that in fact, the 49ers simply were a more skilled team this year, both compared to other teams in the league and compared to the 49ers of previous seasons.  Here I’ll talk a little bit about luck, and in a follow-up post I’ll talk more about skill.

In what sense could the 49ers be considered a lucky team during the 2011 season?  Well, one way to look for a team’s luck is to measure the strength of the opposing teams during the course of the season.  If your team plays all of its games against terrible opposing teams, then it’s more likely your team will have a favorable record, even if it’s only mediocre, simply by virtue of the fact that the teams it plays against aren’t very good.  Conversely, if your team plays all of its games against very strong teams, it will be much harder to get a winning record.  Bearing this in mind, we ask: how lucky were the 49ers in terms of the strength of the opponents they faced?

The answer, as many football fans could probably guess, is fairly lucky.  Part of this stems from the fact that the 49ers played six of their sixteen games against opponents from their own division, which is not known for its strength.  Here’s a list of all teams in the 2011 season, ranked by the strength of their schedules.

Table 1: 2011 strength of schedule. Green indicates teams that made it to the playoffs.

As you can see, the 49ers schedule is ranked near the very bottom, tied with New England and just above New Orleans.  So the schedule was relatively easy.  Having said that, I’ve heard many people attribute the success of the 49ers solely to the ease of their schedule, but have not heard the same criticism made for teams like New Orleans, New England, and Green Bay.  To remark on the ease of the 49ers schedule without pointing out the collective ease of all of these teams’ schedules seems a bit unfair.

Let’s ask another question: how does the strength of the 2011 schedule compare to the strength of the 2010 schedule?  Here’s a new table, just like the one above, but with data from the 2010 season:

Table 2: 2010 strength of schedule. Green indicates teams that made it to the playoffs.

In the 2011 season, the 49ers were much closer to the middle of the pack as far as strength of schedule is concerned.  Comparing year over year, based on this data, it would not seem unreasonable to attribute some of the 49ers’ newfound success to their relatively easy schedule.

But wait! the careful reader may exclaim.  There’s a chicken-and-egg argument to be made here: did the 49ers improve in 2011 because they had an easier schedule, or did they have an easier schedule because they had improved?  Indeed, San Francisco’s 13 wins in 2011 correspond to an extra 13 losses for their opponents, as compared to only 6 wins for the team in 2010.  Any conclusions drawn from the above data must keep this in mind, since the opponents wins for any team are not independent of the team’s performance.

One way to try and correct this is to remove each team’s performance when considering the strength of the opposing teams.  So, for example, in 2011 the teams the 49ers faced won 115 games and lost 141, for a win percentage of just 0.449.  However, if we remove games played against the 49ers, who had a record of 13 wins and 3 losses, then among all remaining games, the teams the 49ers faced won 112 games, and lost 129, for a slightly higher win percentage of .467.

If we make this adjustment across all teams for the 2011 season, the following picture emerges.

Table 3: 2011 Adjusted 2011 strength of schedule. Green indicates teams that made it to the playoffs.

While San Francisco’s schedule still ranks near the bottom after this adjustment, a few other teams have slipped just below it. How does this compare to the previous season?  To wrap things up, let’s look at the adjusted strength of schedule for 2010.

Table 2: Adjusted 2010 strength of schedule. Green indicates teams that made it to the playoffs.

Notice that in absolute terms, the difference in the 49ers’ schedule is quite small comparing the adjusted 2010 season to the adjusted 2011 season.  In 2011, the number of adjusted games opponents won was 112, versus just 115 in 2010.  So from this standpoint, much of the 49ers’ progress in 2011 appears to be due less to having a lucky schedule, and more do to with actual improvements of the team.  As promised above, next time I will dig a little deeper in an attempt to quantify the 49ers’ improvement during this season.  For now though, I must pick out an appropriate outfit to wear this Sunday, in mourning of the absence of this team from the Super Bowl.

(A final note for the more statistically minded: In 2011, the correlation between a team’s winning percentage and the opponents’ combined losing percentage is quite high – the correlation coefficient is roughly .766.  With the adjustment mentioned above, this drops to .529, which agrees with the notion that there should be a weaker correlation between one team’s record and the combined record of its opponents when the team’s record is removed from the combined record.  For 2010, the correlations are a bit weaker: .574 and .333, respectively).

Currently, the playoffs consist of three rounds.  The first round is the Division Series, in which eight teams compete in a best-of-five match-up (equivalently, a first-to-three match-up, i.e. the first team to win three games wins the series).  The second and third rounds, better known as the Championship Series and World Series, are composed of four and two teams, respectively, but are both best-of-seven (equivalently, first-to-four).  Because of these three rounds of several games each, the playoff season is already quite long; therefore, the new proposed playoff round, it has been suggested, would be composed of either a single game between competing teams, or a best-of-three (first-to-two) series between the two teams.

Many people take issue with such a short series on the grounds of fairness.  In a season where each team plays 162 games, they say, it’s not fair for a team’s World Series hopes to ride on a single game, or even a short series composed of at most three games.  There are even those who suggest that the Division Series is too short, and that all three of the current rounds should be a best-of-seven.  These are noble sentiments, but are they reasonable?  We can use mathematics to try and answer this question.

Suppose two teams are meeting for a playoff series, and the probability that one team (call it, I don’t know, the Giants) will win a single game is p (this model is fairly simple, and does not take into account advantages associated with the starting pitcher, for example, but let’s keep things basic for now).  Then the probability that this team will win a one game series is again p, since the series consists of a single game.

What if the series is three games long?  In this case, the Giants will win if they win the first two games, or split the first two games and win the third game.  So there are three outcomes: WW, WLW, or LWW.  The probability of the first event is , while the probability of the second and third events are both (probability p of success is the same as probability 1-p of failure).  Adding these three probabilities gives the total probability that the team will win a best-of-three series:

In a best-of-five series, the Giants will win if they win three in a row, two of the first three and the fourth, or two of the first four and the fifth.  Using combinations to count the possibilities, we see that in this case, the probability of the Giants winning the series is equal to

With the same type of argument you can calculate the probability that the Giants win a best-of-seven series.  I’ll spare you the details: the result is .

In each case, the probability of winning the series is a polynomial in p, the probability of winning a single game.  But how to these polynomials compare?  Let’s turn to technology to lead the way!

Probabilities for a one, three, five, and seven game series.

Above is a graph of these four functions – the x-axis represents the probability p, while the y-axis represents the probability of winning the series.  The dark blue graph is for a single-game series (the function is p), the light blue graph is for a three-game series (the function is , the light green graph is for a five-game series (the function is ), and the red graph is for a seven-game series (the function is ).  What can we deduce from the picture above?

First, note that a longer series benefits the stronger team more than the weaker team – this makes intuitive sense, if you think about it.  Also, for teams that are perfectly evenly matched (i.e. p = 0.5), the length of the series doesn’t affect the probability of winning the series, which is also 50% in each case.

But what about teams with a slight, moderate, or strong advantage over their competition?  How does the length of the series affect the probability of winning the series?  Let’s look at a small table of values, in the cases p = .55, p = .6, and p = .7.

As you can see from the table (or the graph), the more evenly matched the teams, the less of a difference the length of the series makes.  If your team has a 55% chance of winning a given game, the advantage in a seven game series is increased by a little less than 6 percentage points.  With a 60% chance of winning a given game, the advantage in a seven game series is increased by 11 percentage points, and with a 70% chance of winning a given game, the advantage in a seven game series is increased by over 17 percentage points.

Not also that the change from a best-of-five series to a best-of-seven series isn’t really very large.  Even if your team is heavily favored (70% probability of winning each game), the change from a best-of-five series to a best-of-seven series is less than four points.  With more evenly matched teams, the difference is even smaller, suggesting that expansion of the Division Series from a maximum of five to a maximum of seven games isn’t necessarily a great idea.

On the other hand, the largest change in probabilities is between the jump from a best-of-one series to a best-of-three series.  While the change isn’t so significant for evenly matched teams (and more evenly matched teams would be most likely to play each other in this round under the suggested rule changes), for match-ups in which one team is heavily favored, the difference can be more significant.

Whether or not one wants longer series or shorter series depends, I suppose, on one’s baseball philosophy.  It certainly seems like having everything ride on a single game after a season of more than 150 games is a little unbalanced, but from a mathematical standpoint, the stronger team will most likely gain only a small advantage by moving to a three game series.  Of course, this simplified model can only tell us so much, and it’s possible that the advantages of a longer series are being underrepresented here.  To err on the side of caution, I’d be more inclined to support a best-of-three series, though whether or not this is possible without stretching the season too long is something that the folks who are paid better than me to think about these matters will have to decide.

With a payroll roughly a third the size of the Yankees’, Beane understood that the playing field was not a level one from an economic standpoint.  What’s more, at the end of the 2001 season, three of the A’s star players left Oakland for bigger paychecks.  To fill the void, the film (and book) show how Beane took a more analytic approach, and used statistical analysis to uncover players who were undervalued and could be purchased for much less than they were worth.  Beane, together with Paul DePodesta (Peter Brand in the film, and played by Jonah Hill), used a sabermetric approach to lead the A’s to a league-leading 103 wins for the season. While their first-place ranking for number of wins that year was shared with the Yankees, they spent much less per win than their New York counterparts (the A’s spent the least per win, while the Yankees spent the third most).  Here’s a table comparing the teams; the payroll numbers are taken from here, and differ slightly from the numbers that appear in the book.

Team	Wins	Losses	Payroll	Cost Per Win (millions)
Oakland Athletics	103	59	$40,004,167	$0.388
Minnesota Twins	94	67	$40,225,000	$0.428
Montreal Expos	83	79	$38,670,500	$0.466
Florida Marlins	79	83	$41,979,917	$0.531
Cincinnati Reds	78	84	$45,050,390	$0.578
Pittsburgh Pirates	72	89	$42,323,599	$0.588
Los Angeles Angels	99	63	$61,721,667	$0.624
Tampa Bay Rays	55	106	$34,380,000	$0.625
San Diego Padres	66	96	$41,425,000	$0.628
Chicago White Sox	81	81	$57,052,833	$0.704
Philadelphia Phillies	80	81	$57,954,999	$0.724
Houston Astros	84	78	$63,448,417	$0.755
Kansas City Royals	62	100	$47,257,000	$0.762
St. Louis Cardinals	97	65	$74,660,875	$0.770
Colorado Rockies	73	89	$56,851,043	$0.779
San Francisco Giants	95	66	$78,299,835	$0.824
Seattle Mariners	93	69	$80,282,668	$0.863
Milwaukee Brewers	56	106	$50,287,833	$0.898
Baltimore Orioles	67	95	$60,493,487	$0.903
Atlanta Braves	101	59	$93,470,367	$0.925
Toronto Blue Jays	78	84	$76,864,333	$0.985
Detroit Tigers	55	106	$55,048,000	$1.001
Los Angeles Dodgers	92	70	$94,850,953	$1.031
Arizona Diamondbacks	98	64	$102,819,999	$1.049
Cleveland Indians	74	88	$78,909,449	$1.066
Chicago Cubs	67	95	$75,690,833	$1.130
Boston Red Sox	93	69	$108,366,060	$1.165
New York Yankees	103	58	$125,928,583	$1.223
New York Mets	75	86	$94,633,593	$1.262
Texas Rangers	72	90	$105,726,122	$1.468

Their new approach threw out many pieces of conventional baseball wisdom: stealing bases and bunting were strict no-no’s, for example.  Naturally, these changes brought about some tension, and it’s this tension that makes for the dramatic thrust of the film.  In particular, mathematics takes a backseat, though there are some little cameos for those who are paying attention.

The most significant piece of mathematics making an appearance in the film is the Pythagorean Expectation, a formula discovered by Bill James that estimates a team’s win percentage in terms of its runs scored and runs allowed.  More specifically, the formula asserts that a team’s win percentage is approximately equal to

For example, the 2002 A’s scored a total of 800 runs, and allowed a total of 654 runs, for a Pythagorean Expectation of

(relevant stats can be found here). This compares to the team’s actual win percentage of 103/162, which is around 0.636.

In the film, Peter Brand applies this formula in order to estimate the number of runs the team needs to score, along with the maximum number of runs it can allow, in order to secure a playoff spot.  In one scene, he tells Billy Beane that he thinks the A’s will need to win at least 99 games to guarantee a playoff spot.  In a 162 game season, this equates to a win percentage of around 0.611.  In order to ensure that the Pythagorean Expectation is at least this large, we set

With a little algebra, this is the same as

Brand then informs Beane that in order for this to happen, the team needs to score at least 814 runs, and can allow no more than 645 runs.  This gives a runs allowed to runs scored ratio of 645/814, or around 0.793 < 0.798 (though, if I were being anal, I would point out that with 814 runs scored, the team could allow as many as 649 runs and still have a runs scored to runs allowed to runs scored ratio that is less than 0.798).

While the math formulas on display in the film are accurate, I would be remiss if I did not briefly discuss Hill’s portrayal of Peter Brand.  Overall, Hill does a good job; though Brand is clearly a nerd, Hill’s portrayal usually avoids caricature.

Like every other film featuring characters who are good at math, though, Moneyball can’t help itself from including a scene where we see how good Brand is at math because he can do mental calculations quickly.  This particular scene takes place when Brand is sitting in his first meeting with Beane and the rest of the baseball scouts, and though it serves to highlight the tension that exists between Brand’s new school of thought and more traditional baseball thinking, I think the scene could have been just as effective without the clichéd math exercise.

Also, in the interest of full disclosure, I should point out that there are some who feel the story told in Moneyball (both the film and the book) is an exaggeration.  More specifically, as this Slate article discusses, many people believe that the reason for the A’s success during the early aughts had less to do with sabermetrics, and more to do with the fact that they had awesome pitchers in Tim Hudson, Mark Mulder, and Barry Zito, none of whom feature prominently in the book or film.  While I don’t feel knowledgeable enough to weigh in decisively on this issue, the role of the defense certainly appears to be underrepresented here.

To try and convince you of this, recall that the A’s made it to the playoffs in four consecutive years, from 2000-2003.  Here is some data on how many runs they scored and how many runs they allowed during each of those years, and in 2004, when they did not make the playoffs:

Observe that especially from 2001-2003, while the A’s offense declined, their defense remained consistent in allowing relatively few runs.  Of course, this should not be viewed in a vacuum, but rather in relation to how baseball as a whole performed.  Therefore, it is better to consider not runs scored and runs allowed, but runs scored and runs allowed as a proportion of runs scored and runs allowed in the American League.  With this slight adjustment, we get the following picture:

Note in the above that a proportion of 1 means that the A’s were performing at an average rate, while a proportion greater than 1 indicates above-average performance, and a proportion less than 1 indicates below-average performance.  As we can see from the data, in 2001-2003, the A’s defense was allowing runs at a rate well below the average; in other words, the defense was relatively strong.  On the other hand, during the same period, the offense consistently weakened year-over-year, so that the number of runs the A’s scored was below the league average in 2003-2004.  In particular, during the 2002 season profiled in Moneyball, the number of runs scored took a sharp downturn relative to the league average, while the number of runs allowed still remained well below average.  This indicates, to me at least, that the role of the defense was certainly an important factor in the A’s playoff runs during the 2002 and 2003 seasons.  Note also that in the 2004 season the number of runs allowed rose sharply relative to the league average; without a corresponding uptick in runs scored, the A’s didn’t make it to the playoffs.

Nevertheless, I don’t think the issue is binary; excellent pitching and a sabermetric approach probably combined to help the A’s.  Even though Moneyball only explores one of these issues, it’s still a film well worth seeing.  If you’re no fan of mathematics, don’t worry, there isn’t much on display.  And if you’re no fan of baseball, surprisingly, I think you might enjoy the movie anyway.

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.

In a game of baseball, however, it’s more important to minimize time, not distance.  Given this, running a path that consists of four straight lines connecting each base is not optimal, because the runner must slow down to make the sharp turns at each base.  Of course, baseball players already know this, which is why they often swing out in their path before crossing first when they are confident that they can reach second or more.  But still, the question remains: are these trajectories optimal?

According to a trio of mathematicians from Williams College, maybe not.  According to an article originally posted in May, but which only now seems to be gaining some traction, recent graduate Davide Carozza wondered what would happen if runners began to curve immediately when leaving the batter’s box.  His conclusion may be somewhat surprising: using a circular path around the bases rather than a straight line path resulted in a 25% faster loop around the pitcher’s mound.

Brian Wilson likes those odds.

But of course, a circular loop doesn’t seem like the most efficient route either – after all, once you return to home you don’t need to run again, so it always makes sense to run as straight as possible from 3rd to home.  Upon further analysis, Professor Stewart Johnson found that the fastest loop indeed one that begins like a circle, but comes in closer to the bases on the last leg of the journey.

Here is the proposed optimal path.

As with any mathematical model, some caveats are in order.  The most significant one concerns acceleration – the red lines in the image above denote the acceleration, and you can see that all the lengths are the same, meaning that acceleration is assumed to be constant.  But of course, this isn’t really true when a batter is running the bases – as the latter article points out, “a runner may not be able to speed up as quickly while running along a curved path as he can along a straight one. In that case, some form of a banana path might make sense, allowing a runner to go straight for the first few critical seconds.”  Also, when the runner knows he hasn’t hit more than a single, it obviously makes little sense to run in a curved path.  Instead, it should be obvious that the wisest move is for the runner to get to first as quickly as possible.

Even bearing these warnings in mind, though, there still might be a good idea here worth exploiting.  If the ball is hit between two outfielders and the runner is fairly certain of a double (or better), how much of a curved trajectory should he take?  These are questions that could be answered during practices, and in a game where a fraction of a second can be the difference between a safe and an out, the answers might certainly have some applicability.

I would especially like to consider the San Francisco Giants to take these ideas to heart.  They need every run they can muster, and even though this team looks like a strong contender to bring San Francisco its first World Series title, a little mathematical advantage couldn’t hurt. After all, striking beard fear into the heart of your enemies can only take you so far.

My betrothed, sporting a most fearsome beard.

Presumably, those cancer odds are taken from The American Cancer society, which has the relevant stats posted here.  When it comes to some of the other claims in the ad, though, I couldn’t help but be skeptical.

Take the bowling claim, for instance.  This ad would have you believe that your odds of bowling a perfect game are 1 in 11,500.  This seems quite high, even when I consider the fact that I am not a bowling master.

Let’s try to reverse-engineer this statistic.  To score a perfect game in bowling, one must bowl 12 strikes in a row.  Let us suppose that your probability of bowling a strike on any given frame is some number p.  Furthermore, let’s suppose that your performance in any frame is independent of your performance in any other frame, so that you have a probability p of bowling a strike each time it’s your turn.  Of course, whether or not these probabilities are independent is up for debate.  On the one hand, bowling many strikes in a row may make you more anxious about keeping your streak going, which may in turn decrease your probability of another strike; but on the other hand, if you are an adrenaline junkie who thrives in the limelight that only a bowling alley can provide, perhaps such a chain would make it more likely for your streak to continue.  In any event, these are questions better suited to a psychologist rather than a mathematician, so for simplicity let us ignore them here.

If you have a probability p of bowling a strike, and a perfect game requires 12 strikes, then the probability you will score a perfect game is the product of 12 copies of p (one for each strike), or p¹².  If the above ad is to be believed, this probability must equal 1/11500.  In other words,

Taking the twelfth root of each side, we can then conclude that

In other words, your odds of bowling a perfect game are 1 in 11,500 if and only if the probability that you’ll bowl a strike is around 45.88%.  This seems like an extremely generous probability to give to the population at large.  After all, who among you or your circle of friends bowls a strike, on average, every other frame?  Perhaps I bowl exclusively with people who are not very good (myself included), but I would think a fairer probability for the entire population would be closer to 20% or 30% (maybe even this is too generous).

What to these two alternatives yield for the odds of bowling a perfect game?  Well, if p = .3, then p¹² is approximately 1 in 1,881,676; for p = .2, the odds plummet to 1 in 244,140,625.  Both of these are significantly lower than the odds cited in the ad (roughly 164 and 21,230 times lower, respectively).

It may be that the odds of witnessing a perfect game are around 11,500.  For example, when you go to a bowling alley, there may be experienced players practicing.  Moreover, there are many games occurring simultaneously at a bowling alley, thus increasing the odds that at least one of them will be a perfect game.  But saying “you have a 1 in 11,500 chance of seeing someone else bowl a perfect game” doesn’t sound as sexy as “you have a 1 in 11,500 chance of bowling a perfect game,” I suppose.

Some of the other odds are questionable as well.  For example, the National Weather Service has some data here that suggests the odds of being struck by lightning in a given year are about 1 in 500,000, not too far off from the ad’s claim of 1 in 576,000.  This isn’t an apples to apples comparison, though, because the ad does not specify that these are the odds you will be struck by lightning in a given year.  If we take into account the average lifespan in the United States (approximately 78 years), then the probability of being struck by lightning (in one’s lifetime, not in one particular year) is closer to 1 – (499,999/500,000)⁷⁸, which is around 1 in 6,411.  Much higher, you’ll note, than the odds of bowling a perfect game.  (Once again, of course, we are assuming that the odds of being struck by lightning don’t vary from year to year, and that the odds of being struck in one year are independent of the odds in any other year.)

If the point of the ad is to give us an intuitive understanding of how likely it is for us to develop cancer, then it seems important to give benchmarks that are accurate and relatable.  Most people have bowled, but few people will have a good intuitive understanding of what it means to face odds that are 1 in 1.8 million (the odds of bowling a perfect game if you get a strike 30% of the time).  I think the point of the ad is understood regardless, but it’s a shame that the claims leading up to this point weren’t checked more thoroughly.  Indeed, many of the odds quoted in the ad can be found here, a humor website that offers no sources for any of its statistics.

Perhaps Jon Stewart was in charge of fact-checking.  Given his lack of understanding about the nature of this program, this is perhaps the most reasonable explanation.

]]> http://www.mathgoespop.com/2010/09/standup.html/feed 2 http://www.mathgoespop.com/2010/07/lets-make-a-deal-with-paul-the-octopus.html http://www.mathgoespop.com/2010/07/lets-make-a-deal-with-paul-the-octopus.html#comments Wed, 14 Jul 2010 15:00:40 +0000 Matt http://www.mathgoespop.com/?p=486 As summer reaches its midpoint, we come to the end of another rousing year of World Cup soccer.  As with any international sporting event, fans all over the world have undoubtedly had their share of ups and downs.  Of all the countries in this year’s tournament, however, I think Germany may be receiving the most attention, . . . → Read More: Let’s Make a Deal with Paul the Octopus]]> As summer reaches its midpoint, we come to the end of another rousing year of World Cup soccer.  As with any international sporting event, fans all over the world have undoubtedly had their share of ups and downs.  Of all the countries in this year’s tournament, however, I think Germany may be receiving the most attention, for even though they didn’t make it into the finals, the Germans have one thing no other country has: a precognitive octopus.

At least, that is what the media would have us believe.  For the past several weeks, Paul the Octopus has captured the hearts, minds, and stomachs of people around the world.  He’s a charming octopus, to be sure, but it isn’t his good looks that have gotten him this far.  Instead, it’s his seeming ability to correctly predict the outcome of soccer matches.  As time has gone on and Paul’s predictions have continued to prove themselves accurate, the amount of press he has received has only increased.  Articles about him are everywhere on the internet: here‘s one discussing public outrage after he correctly predicted Spain to defeat Germany in the semifinals, and here‘s an article discussing his preference for Spain over the Netherlands in the finals.  Search for “Paul the Octopus” in Google News and you will find thousands of results.

This is one popular mollusk.

Of course, I suppose it’s possible that Paul really can see into the future, at least as far as soccer is concerned.  After all, he did correctly predict the winner of every game asked of him; an impressive feat, seeing as how his advice was requested a total of 8 times.  However, Occam’s Razor suggests that we should look for a simpler explanation.

A natural first choice is to guess that Paul was simply guessing randomly, and is very lucky.  The odds of this happening are small – assuming he has a 50% chance of picking correctly, the odds of him being right each one of the 8 times he predicted a winner in this World cup would be 1/2⁸, which is only approximately .39%. Very low odds indeed.

This analysis ignores biases that may be present – in particular, Paul’s octopus vision may bias him towards certain flag designs (which, given the fact that he frequently chooses Germany, seems plausible).  The Wikipedia article I linked to discusses other sorts of possible bias.  However, these biases would only influence which box he selected – they wouldn’t necessarily affect the odds that his selection would correspond to the winning team (although it is possible that he is being persuaded to throw his chips in for the favored team, which would increase the likelihood of his success).  Either way, questions concerning how Paul makes his selection are more interesting to me, so let me focus for the moment on that.

First of all, one could easily argue that unlike flipping a coin, the trials here (i.e. Paul’s selections) are NOT independent. Indeed, what’s going on here may be very similar to an article in the New York Times a couple of years ago that discussed the lurking presence of the Monty Hall Problem in a classic experiment from psychology (which I discussed here).

The idea is quite simple.  Paul chooses between two opponents in the World Cup by selecting a piece of food from one of two boxes.  Each box is labeled with a country’s flag, and this is the most obvious distinction between the boxes.  Suppose, for the sake of argument, that Paul has in his mind a ranking of his preferences for the flags, starting with the one he likes the most, and ending with the one he likes the least.  Assuming between any pair of flags Paul prefers one to another, one could theoretically determine his preferences by giving him sufficiently many pairings of different flags.  Moreover, assuming he does have preferences, the game selections are no longer independent, because each game gives us some information about his preferences.

Let’s dig deeper and look at his selections throughout the World Cup.  Paul gave predictions for 8 games, and those 8 games involved 9 separate teams (only one game did not involve Germany).  In the first game, Germany versus Australia, Paul selected Germany.  Let us note that as: Now, already this gives us information about Paul’s preferences.  There are 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9! = 362,880 possible ways to order these 9 teams, since you can choose any one of the 9 teams to be your favorite, any one of the 8 remaining to be your second favorite, and so on.  But given the above information, we already know that any choice with Australia ranked higher than Germany can not match Paul’s preferences.  This eliminates a surprising number of possible outcomes – half, in fact, since for every list in which Germany is ranked higher than Australia, we can flip these two countries to obtain a ranking in which Australia is higher than Germany.

This then affects the probability of all subsequent pairings!  Let’s take a look at the next game.  In the second game, Paul correctly predicted Serbia to defeat Germany.  We can write this as: This gives us even more information!  Now, not only do we know that Paul prefers Germany to Australia, we also know he prefers Serbia to Germany.

Moreover, suppose (as seems reasonable) we assume the probability that he prefers Germany to Australia is 50%.  Now, after the first match, we know he prefers Germany to Australia – the right followup question to ask is, what is the probability he prefers Serbia to Germany GIVEN that he prefers Germany to Australia?  In particular, we have a conditional probability question here – these two facts are not independent!  For example, the fact that Paul prefers Germany to Australia means that Germany cannot be last in his rankings – this should decrease the probability that he prefers Serbia to Germany.  And indeed it does: supposing that each of the 362,880 rankings is equally likely, the probability that someone will prefer Serbia to Germany given that they prefer Germany to Australia is not 50%, but is only 33%!

To see why, consider an arbitrary ranking of the 9 teams.  If we only consider the relative placement of Serbia (S), Germany (Ge), and Australia (A), there are 6 possible rankings among these three:

1. S > Ge > A
2. S > A > Ge
3. Ge > S > A
4. Ge > A > S
5. A > Ge > S
6. A > S > Ge.

However, if we also know that Ge > A, then rankings 2, 5, and 6 are eliminated, leaving us only with

1. S > Ge > A
2. Ge > S > A
3. Ge > A > S.

In particular, of these three remaining, Serbia is ranked higher than Germany only once.  Therefore, the probability that S > G GIVEN that G > A is only 1/3, or 33%.  (If you prefer, you can use a counting argument to show this as well.)

Of course, just as the first match gave us information, so did the second.  Therefore, when it comes to the third match, we have even more information at our disposal.  The third match was between Germany and Ghana, and Paul correctly identified Germany.  In other words: Now the appropriate question to ask, of course, is: what is the probability that Paul prefers Germany to Ghana, given that he prefers Serbia to Germany and Germany to Australia?  Well, we know that Germany can’t be his first or his last choice, because it must be preceded by Serbia and followed by Australia.  Therefore, among these four countries, Germany must rank second or third.

If Germany ranks second, Serbia must be first, but we are free to make Australia third or fourth.  Similarly, if Germany is third, then Australia must be last, and we are free to make Serbia first or second.  In other words, we have four outcomes:

1. S > Ge > Gh > A
2. S > Ge > A > Gh
3. S > Gh > Ge > A
4. Gh > S > Ge > A.

Among these four choices, only 2 have Germany preferred over Ghana.  Thus the conditional probability that one would prefer Germany to Ghana is again 50%.

The interested reader can easily continue on in this fashion.  If you’re impatient, however, you can calculate the probability that Paul would have made the selections he did more directly.  All we need to know is who Paul selected in each match.  I’ll tell you that in the subsequent matches, Paul picked Germany over England, Germany over Argentina, Spain over Germany, Germany over Uruguay, and Spain over the Netherlands.

Given this information, suppose you want to know the probability that Paul selects Germany over Australia, Argentina, Uruguay, Ghana, and England, while he selected Spain over Germany and the Netherlands and Serbia over Germany.  As stated before, there are 9! possible lists of preferences.  In this case, it’s not hard to determine how many would lead to the behavior seen in this year’s World Cup.  Since Paul picked Germany over 5 teams, but behind 2 teams, we know that Germany can only be ranked 3rd or 4th (any higher and there wouldn’t be room for Spain and Serbia above, any lower and there wouldn’t be room for the 5 teams below).

If Germany is ranked 3rd, then we can choose to put either Spain or Serbia in 1st place.  Whichever one we don’t put in first place will then need to go in 2nd place, so that both countries are ranked higher than Germany.  After that, we are free to order the remaining countries however we like.  In other words, we see that there are 2 x 6! = 1,440 possible lists of preferences if Germany is ranked 3rd:

1. 2 choices
2. 1 choice
3. Germany
4. 6 choices
5. 5 choices
6. 4 choices
7. 3 choices
8. 2 choices
9. 1 choice.

Meanwhile, if Germany is in 4th place, we need to figure out how many ways there are to choose the three teams above it.  Notice that since Australia, Argentina, Uruguay, Ghana and England must all be ranked lower than Germany, the three countries ranked above it must be Spain, Serbia, and the Netherlands.  However, since we also need the Netherlands to be ranked below Spain, this only gives us three possibilities for the ranking of the first three teams: Spain, Serbia, Netherlands; Spain, Netherlands, Serbia; and Serbia, Spain, Netherlands.  Once we have made that selection, however, we are free to choose the 5 teams below Germany however we please.  In other words, if Germany is 4th, there are 3 x 5! = 360 possible lists of preferences.

Combining these, we see that there are 1,800 possible lists of preferences that would lead to the behavior shown by Paul.  Since the total number of outcomes is 9!, this gives a probability of only 1,800/9!, or roughly .49%.  This is the same value you will get if you calculate the remaining conditional probabilities and multiply them together.

Of course, if one wants a more impressive number, one can always try to correct for bias in Paul’s selection.  For example, suppose we assume that Paul will never choose England or Australia if given the option of a different flag – this seems reasonable, given experiments on how his species sees (the other flags have more contrast and are more focused on horizontal shapes, which apparently his species is drawn to).  If we make this assumption, the number of potential preference lists drops to 7! x 2 = 10,080, in which case the probability that Paul would choose as he did jumps up to 17.9%!

There are valid concerns about this model, though.  For instance, given the choice between two flags, why should we assume that Paul will always choose the same one over the other?  Equivalently, why should we believe that Paul’s decisions follow a prescribed preference list?  Indeed, when Paul was used to make predictions in 2008, he selected Germany over Spain, unlike his selection of Spain over Germany in 2010.  In 2008, it was Germany who was the victor, and so Paul guessed incorrectly – perhaps he learns from his mistakes, after all.

Whatever the case, I doubt this octopus has any special ability.  And even if he does, I don’t know that that would necessarily be a good thing.  For we all know that when 8 limbs are combined with super powers, nothing good can come of it.

Is this the future that Paul's powers portend? I believe it is.

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.