Kansas State is scheduled to meet UCLA in the 2015 Alamo Bowl on January 2^{nd}. We all want, or hope, or perhaps even wager that KSU will prevail, but can we do better? What insight can be derived from the 2014 data and extrapolated to predict the Alamo Bowl outcome? With what probability will KSU vanquish its scheduled foe? Read on, intrepid fan, *lux et veritas* await!

#### The Method Explained

I performed a probabilistic analysis of the 2014 season for KSU and UCLA, respectively. The analysis proceeds as follows: I consider points scored as one statistic, that is, a collection of numerical data indicating offensive performance. I do likewise for points allowed which indicates defensive performance. These statistics tend to follow a normal distribution, or a bell curve. Imagine both curves plotted on one set of axes, the horizontal axis comprised of equivalent "bins" and the vertical axis indicating how many outcomes corresponding to each bin size were achieved during the 2014 season.

The curves peak at the average values of points scored and points allowed and a narrower curve indicates greater consistency than a wider curve. In statistical terms, the average value is called a mean and the measure of wideness/narrowness is called the standard deviation. There are straightforward procedures for computing these values for any data set. Also note that if the average points scored is greater than the average points allowed, it follows that this team has probably won more games than it has lost. The converse is true as well.

There are several interesting things that we can do with this data mathematically. For example, it is relatively easy to compute the probability that points scored exceed points allowed, which in the parlance of athletic fanatics is called a victory, or a "win". Perhaps more interestingly, given two teams scheduled to meet in an athletic contest – say KSU and UCLA in the Alamo Bowl, a similar method can be applied to compute the probability that KSU will defeat UCLA. What does the data tell us?

#### The 2014 Campaign

First, I compute that KSU’s theoretical probability of winning (i.e. – the probability that points scored exceed points allowed) is 0.77, meaning that KSU should have won 9 games and lost 3 which, in fact, they did. Performing the same calculation for UCLA results in a probability of winning equal to 0.62. This indicates that UCLA should have won only 7 games and lost 5, although numerically the probability is very close to 8 wins and 4 losses. However, UCLA actually completed the season with 9 wins and 3 losses. You may be thinking to yourself, ‘What gives, Dr. Nerdenstein?’

Well, there are a couple of obvious points to make in defense of my mathematical machinations. First, my theoretical model may be inaccurate or worse, wrong. It's been said that all models are wrong, although some are useful. This model is accurate for KSU and relatively accurate for UCLA, tipping the balance toward useful. Second, perhaps this means that UCLA was lucky or over performed or is for some other reason (or possibly, no reason) an outlier while KSU toed the probabilistic line. How can we reconcile the discrepancy? Let's take a closer look at the data and the upcoming contest.

#### The January 2^{nd} Matchup

Statistically, KSU enjoys the following advantages over UCLA: higher average points scored (about 3 points) and lower average points allowed (about 6 points). However, UCLA is more consistent than KSU in terms of both points scored and points allowed. Consistency is seldom quantified and often overlooked in athletic contests which can lead to a vicious cycle of usurious loans and/or broken limbs to palliate statistical shortcomings in the betting fan’s so-called game. The other notable is that on average, UCLA played in closer scoring games than KSU. Coupled with consistent offensive and defensive performance, it's not surprising that UCLA "outplayed" their theoretical probability of winning - over half of their games were decided by a touchdown or less. With a record of 6-1 in these contests, this was exciting football for the fans and dangerous territory for opponents. What does it mean for the Wildcats?

After donning my lab coat and safety goggles, I replicated the statistical information on a set of punch cards and then fed them into my probability-generating computing machine (patent-pending). When I awoke the next morning, a computing error had occurred somewhere in the process and I was forced to repeat the analysis. Nonetheless, I eventually triumphed against my mechanical nemesis and am happy to report that KSU is also likely to prevail against its scheduled foe.

My calculations show that KSU has a 67% probability of winning the 2015 Alamo Bowl. Beware, brown bear! However, if the Bruins keep the score close, an exciting finish could leave the Cats licking their wounds.

*Quantum satis*: Cats beat Bruins in 2 out of 3 contests, ceterus paribus.

*Caveat venditor*: I know not which contest is scheduled for January 2, the 2/3 or the 1/3.

*Caveat emptor*: I do not speak Latin (fluently)!

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