Super Bowl Squares

It's that time of the year, when seemingly uninterested people run around feigning interest in what is now a true American celebration - the Super Bowl.  This year's Super Bowl features the Pittsburgh Steelers and the Arizona Cardinals.  Not that anyone outside of Phoenix and Pittsburgh actually cares who is in it.  That however won't prevent Super Bowl parties from cropping up across the nation on Sunday, February 1st.  And, at most of those parties, someone will usually run around trying to get the last few stragglers to participate in the second greatest tradition of the Super Bowl - the "Super Bowl Squares" (the first has to be watching the Super Bowl commercials).  

Basically, for a small donation, you get to place your name in a 10x10 square grid.  For multiple donations, you may be able to place your name on multiple squares.  Before the actual game starts, #s from 0-9 are drawn randomly and placed across each of the 10 columns.  The same process is repeated to the left of each row.  Then the name of team 1 is drawn and placed at the top, and the second team is placed on the left.  Now you have a grid that has each of the possible last-digits of the scores of each team.  When the game ends, you look at the score, then look at the last digits of each team's score, and see whose name corresponds to that permutation and that individual wins a pre-determined amount of the accumulated donations.  This process doesn't have to be limited to the game-ending score.  Many variations exist.  For example, frequently some smaller amounts could be won based on the digit permutations at the end of each quarter.

This year, as I almost always do, I participated in one of these Super Bowl Squares.  I donated for 2 squares.  After all the squares were filled out, the #s were randomly drawn, the team's assigned, the coordinator of the game handed me my sheet.  I had drawn the following 2 permutations:

Arizona 2 - Pittsburgh 5

Arizona 8 - Pittsburgh 2

I promptly tossed the sheet in the recycling bin.

Later on, I decided to see for myself the likelihood of my winning.  

The analysis below shows the aggregated game-ending digit permutations and combinations for every NFL game played in the Super Bowl era, including playoff games.  That's 9,509 games!  That also means that there's a reasonable likelihood that the probabilities shown are close to the true probabilities.  As a matter of fact, every single permutation has been "hit" at least once.  A game ending in the 2 2 permutation has only happened once - on Sunday, December 5th, 2004 the Buffalo Bills beat the Miami Dolphins in Miami 42-32.

First, a little math.  I refer to both permutations and combinations.  There is a difference between the two.  A combination refers to a sequence or collection without regard to order.  A permutation is a combination with a specific order.  Here's an example.  Take what we commonly (and mistakenly) refer to as a "combination" lock.  We say to unlock the lock, "use combination 472".  Well, that's actually only mildly helpful.  Knowing those three #s alone we wouldn't be able to open the lock.  We need to know the specific order of that combination of #s.  In other words, is it 274, 247, 427, 472, 724, or 742.  So in this example, there is one combination.  There are six permutations.

Ok, now on to the tables and charts below.

In TABLE 1, I show all 100 permutations of game-ending scores.  So for example, one can see that the likelihood of the game ending with the winning team's score ending in a 4, and the losing team's score ending in a 3 is 2.94% (to see this, in TABLE 1, go down to the row with the digit 4, then across to the column with the digit 3).  So, this specific permutation has a 2.94% likelihood.  

If you didn't care about whether it was the winning or losing team that had the 3 or the 4 in the last digit, as long as there was a 3 and a 4, then look to TABLE 3.  TABLE 3 shows the probabilities of each combination.  As such, the 3 4 or 4 3 combination has about a 3.67% likelihood of occurring.  CHART 2 graphically illustrates what's in TABLE 3.

TABLE 2 is not meaningful in and of itself, but simply shows the probability of any given digit occurring (note that in this table, the percentages add up to greater than 100.00% since "any 7" will include for example, the "1 7" combination, that will also show up under "any 1").  CHART 1 simply illustrates what's in TABLE 2.

Let's take a look at what my chances are.  It's a little complicated so bear with me.  Remember, I have 2 specific permutations.  Ari 2/Pit 5 and Ari 8/Pit 2.  However, since I don't know ahead of time who will win the game, I need to average the 2 permutations that yield the 2 5 combination for the first scenario and the two that yield the 8 2 combination for the second.  I can do this by either going to TABLE 1, and adding the respective likelihoods of each of those permutations, or I can simply go to TABLE 3.  From TABLE 3 I can easily see that the 2 5 combination shows a likelihood of 0.36% (this is made up by the 5 2 permutation likelihood of 0.23% and adding it to the 2 5 permutation likelihood of 0.13%).  Therefore, the average expectation for the specific 2 5 permutation is 0.18%*.  For the second scenario, from TABLE 3, I can see that the likelihood of this combination is 0.28%, and hence the average expectation for the specific permutation is 0.14%.

* (Technically, I shouldn't be averaging the permutations expectations.  What I should be doing is weighing each permutation by the likelihood of Arizona (or Pittsburgh) winning or losing the game.  So for example, if the likelihood of Pittsburgh winning the game is estimated to be 70%, then a truer expectation for my specific 2 5 permutation might be 0.7*0.23% +0.3*0.13% = 0.20%.  However, if you assume that each team's likelihood of winning the game is close to 50%, then averaging is fine).

So there you have it, the combined likelihood that I would win ANYTHING is about 0.32% (0.18%+0.14%).  

Hence why I tossed the sheet.  Hope you have better permutations!  Good Luck!