Variability

A friend texted me (Dr. Hutson) the other day while I was waiting in the queue for Space Mountain asking what are the odds that the Oklahoma City Thunder would upset the Golden State Warriors in a best-of-seven series, assuming that the Thunder had a 40% chance of winning any individual game of the series.  It got us to thinking how many times an underdog student team will beat the professors playing this Traveling Tourist Problem (TTP) Challenge that we had on Monday.  So far, we’ve finished third in 2014 and second this year (i.e., we are 0-2). 

Well, it may be hard for us to recover losing another TTP Challenge, but it is not hard to analyze my friend’s question, and it ties into the material that we are talking about in the course.  The problem of a best-of-seven series can be modeled with a walker on a grid who either moves right (the Thunder win) or up (the Warriors win).  The walker flips a weighted coin seven times and moves right after each flip with 40% probability and up with 60% probability.  We will assume that the result of each coin flip is independent, which means present and future coin flips are not affected by all of the previous coin flips (which may not be true in an NBA basketball series, but it makes the analysis simpler).  If we make that assumption then the results of a best-of-seven series could end at any one of the appendages in the following grid.

The probability calculations are made in the following manner.  Suppose we want to know the probability of the Warriors winning the series in five games.  This would require the walker to move four moves up at a probability of 0.6*0.6*0.6*0.6 and one move over at a probability of 0.4.  All of these are multiplied together due to the independence condition stated earlier.  Also, there are four possible ways for the series to end in a Warriors win in five games, namely TWWWW, WTWWW, WWTWW, and WWWTW.  So the probability that the Warriors win 4-1 is 5*0.4*(0.6)^4 = 0.2076.  The other probabilities are found in a similar manner.  So now if we want to know the probability that the Thunder will pull off an upset, we simply add the probabilities of each of the outcomes where the Thunder are victorious together to get 0.0256 + 0.0614 + 0.0922 + 0.1106 = 0.2898.  Thus, under the assumptions that we made, the Thunder have about a 29% chance of going to the finals.  Note that in any one game the Thunder have a 40% chance to win but that probability drops by over 11% for a seven-game series.  So in a one-game series (such as the NCAA Basketball Tournament or NFL Playoffs) the underdog has a better chance to pull off an upset as opposed to a seven game series which favors the favorite.

This idea brings us back to the one-game series we participated in this week, the TTP Challenge.  This challenge pits teams against one another to see which team can find the most efficient way to visit 19 different attractions in the Magic Kingdom.  It is loaded with gems from probability and network optimization as the groups found out.  One of the things that a lot of students pointed out after the competition was how lucky or unlucky they were throughout the day.  The first thing that affected all groups was that the crowd level at the Magic Kingdom was projected to be at a 5 for that day but turned out to be an 8.  So all groups were adversely affected by longer lines, especially in the afternoon.  Further, the data we had for projected wait times throughout the day (supplied by touringplans.com) was right on for a lot of the attractions, but for others, extenuating circumstances caused the data to be off.  For instance, Dumbo normally has two carousels running, but one was closed on Monday.  This caused one team to wait in line over 75 minutes to ride an attraction that was expected to have a wait of 20 minutes during that time of day.  Further, the variability in preferences that Disney queue operators give to FastPass lines as opposed to regular lines on different attractions can drastically effect how long it takes to get through a regular line.  It might be interesting for us to study the ratio of FastPass participants to regular line participants that are put through at the merge point of some lines.  This ratio seems to not be uniform across rides.  To see the consequences of this, a team waiting in a regular line, with a posted 35-minute wait, might actually wait 40 minutes or more if preferences are heavily-weighted to the FastPass line (we’re looking at you Buzz Lightyear Space Ranger Spin queue operator).  However, a team waiting in a regular line, with a posted 55-minute wait, might actually wait only 25 minutes if preferences are given to the regular line (as the winning group claims happened to them on Peter Pan).  Of course, there are other things that factor into this, such as the number of FastPasses Disney allocates to each time slot for each ride, but these are things we don’t have data for.  Another source of variability arises in ride break downs, which affect how long teams must wait to get on the ride, as several teams experienced when Splash Mountain went down for a while.  So, maybe we were all just random walkers on a grid flipping coins to see whether something would delay us or not throughout the day.  Those teams that flipped a ‘win’ more often found their journey easier than others.

Not to take away from the winning team’s strategy (which included a clever move to switch all of their FastPasses earlier in the day so as to be able to use more sooner and throughout the day), but variability played a large role in the outcome of each team, as it does with any competition.  When you think about it, there are a lot of sources that could cause delays in a theme park, such as congestion, ride closures, being attacked by birds, Morgan Freeman reading off all 44 Presidents’ names in order in the Hall of Presidents, etc.  All of these sources of variability forced teams to change their plans and strategies and adjust.  Those teams who adapted to this variability best finished sooner, or maybe they just got lucky.