Game on!

I've been reading a great book by two economics professors (Avinash Dixit from Princeton and Barry Nalebuff from Yale) called The Art of Strategy: A Game Theorist's Guide to Success in Business and Life.  It's an update to their earlier classic, Thinking Strategically.  I've also been listening to one of the Great Courses called "Games People Play: Game Theory in Life, Business, and Beyond", which is taught by Dr. Scott Stevens, and economics professor at James Madison University.  The two books and "Great Course" cover a number of topics in game theory which I would like to explore a little further.  

At its core, game theory is a branch of economics that deals with the science of strategic decision-making.  The mathematician John von Neumann and economist Oskar Morgenstern helped develop some of the important concepts of game theory in the early 1940's, with the mathematician John Nash (whose story was told in the 2001 movie, "A Beautiful Mind" starring Russell Crowe and Jennifer Connelly) building upon von Neumann's and Morgenstern's early work.  One of the most well-known problems in game theory is the so-called Prisoner's Dilemma (which I explained in my post, "Tit for Tat"), which was developed by scientists at the RAND Corporation in the 1950's.

While there are a number of different elements to every "game" (and these are not "games" in the everyday sense), the basic building blocks common to all are the players, their preferences, the available strategies (which depend upon their preferences), and how these strategies affect the outcome of the game itself.  Importantly, as I mentioned in the last post,("Pascal's Triangle"), understanding even just the basic elements of game theory will help leaders make better decisions.  So as I look to finish out calendar year 2022 (wow, that seems incredible!), I will spend some time on some of these basic elements of strategic decision-making using some more famous problems from game theory.

Pascal's Triangle

Last time I talked about the famous problem in probability known as the "problem of points" (also known as the "division of stakes").  Again, suppose that Bobby and Patty are playing a game of "Heads or Tails" by flipping a quarter (Patty wins with "Heads" while Bobby wins with "Tails").  They've decided that whoever is the first to win ten games will win the prize, in this case a dollar's worth of change that they found on the sidewalk.  Unfortunately, their parents call them in for dinner before they can finish the game.  Patty has 9 wins, while Bobby has only 7 wins.  What's the fairest way to divide up the dollar's worth of change now?  

I discussed how the mathematician Pierre de Fermat solved this problem, which while relatively straightforward could get very cumbersome.  Fermat's solution relied upon the fact that if one player needs r more rounds to win and the other needs s rounds to win, the game will have been won by someone after r + s - 1 rounds.  By the simple laws of mathematics, these rounds have 2^(r + s - 1) possible outcomes.  So, to use our example, Patty needs only 1 more win, while Bobby needs 3 more wins.  If the game had not been interrupted, there would have been a winner after three rounds (1+3-1=4-1=3), leading to 8 possible outcomes (2^3 = 2 x 2 x 2 = 8).  But what if the game had been interrupted when Patty needed 5 more wins and Bobby needed 7 wins?  We would have had to calculate the possible outcomes for (5+7-1) 11 more games, which would equate to 2^11 = 2,048 possible outcomes!  That would take forever to calculate the respective probabilities that Patty or Bobby eventually win the game.

Pascal's method of solving the "problem of points" definitely makes things a lot easier and introduces two concepts, specifically expected value and the arithmetic triangle that now bears his name (Pascal's Triangle).  Briefly, the expected value in a game of chance is the probability of that outcome multiplied by the value of the reward.  As an example, if Patty has a 25% chance of winning one thousand dollars, her expected value is 25% x $1,000 = $250.  Pascal suggested (and Fermat agreed with him) that the the division of stakes would be fair if the expected value to the player did not change.  

Pascal also further elaborated on the arithmetic triangle that now bears his name, though the concept was studied much earlier by mathematicians in India, Persia, and China.  Basically, the triangle is constructed by placing the number 1 in row zero (the top row).  Each entry of each subsequent row is constructed by adding the number above and to the left with the number above and to the right, treating blank entries as zero.  Here are the first eight rows of the triangle:

Notably, Pascal's triangle determines the coefficients which arise in any binomial expansion (i.e. when the a binomial such as x + y is raised to a positive integer power of n).  Importantly, the entry in the nth row and kth column is denoted by the following notation and determined by the rule below:

In general, according to the binomial theorem, when a binomial expression such as (x + y) is raised to a positive integer power of n, we have the following expression, where the coefficients a_k are precisely the numbers in row n of Pascal's Triangle!

So, let's go back to our original problem of points described in the previous post, where Patty needs to win 1 more round in order to win the game and Bobby needs to win 3 more rounds.  Using Fermat's equation (r + s - 1), we know that there will be a winner after 3 additional rounds.  We go to the 3rd row of Pascal's Triangle, which contains the following numbers 1, 3, 3, and 1.  Note that these numbers add up to 8, which is the number of possible outcomes that we determined above for the next 3 rounds of play.  The first number represents the outcome where Bobby wins (again, he needs 3 to win and Patty only needs 1), i.e. 1 chance out of 8 (which is the same as what we calculated above using Fermat's method).  The sum of the next 3 numbers (3, 3, and 1) add up to 7, which is the number of chances that Patty will win out of 8 (again, the same that we calculated using Fermat's method!).

Isn't mathematics fun!?!?  There are a number of online videos explaining how to use Pascal's Triangle to solve the "problem of points", but hopefully my explanation is sufficient to at least give you a rough idea of Pascal's method.  There's also a great book discussing the history of probability, as told through Pascal's and Fermat's correspondence about the "problem of points" by Keith Devlin called "The Unfinished Game".  

For the last 3 posts, I've been talking a lot about mathematics, specifically mentioning a few of the more famous problems in probability and statistics, namely the birthday problem, the Monty Hall problem, and the problem of points.  The question that I am sure all of you are asking is, "What does all of this have to do with leadership?"  The answer might surprise you - these problems have a lot to do with leadership!  These three famous mathematics problems provide a great foundation in the science of probability, which is needed in order to start talking about the science of game theory.  Game theory is a branch of economics that deals with strategic decision-making.  In other words, game theory can help your organization make better decisions!  So at least for the next few posts, I want to discuss a little bit more of the fascinating discipline of game theory, specifically as it pertains to making better decisions! 

The Problem of Points

Suppose that Bobby and Patty are playing a game of "Heads or Tails" by flipping a quarter (Patty wins with "Heads" while Bobby wins with "Tails").  They've decided that whoever is the first to win ten games will win the prize, in this case a dollar's worth of change that they found on the sidewalk.  Unfortunately, their parents call them in for dinner before they can finish the game.  Patty has 9 wins, while Bobby has only 7 wins.  What's the fairest way to divide up the dollar's worth of change now?  

Patty argues that because she is winning, she should take home all the money.  Bobby cries out "No fair!" and states that they should just split the money 50/50.  After a few more minutes of arguing, Patty suggests that because she has won 9 out of the 10 games needed to win, she should receive 9/10 of the dollar.  Bobby again cries out "No fair!" and states that using that argument, he should take home 7/10 of the dollar (and not 1/10 as Patty proposes).  Who's correct here?

Actually, mathematicians have been arguing about how to divide up the winnings going back as far as the 15th century.  It's called the "problem of points" or the "division of stakes", and it was first discussed by the mathematician Luca Pacioli (who was a contemporary of Leonardo da Vinci) in his 1494 book Summa de arithmetica, geometrica, proportioni et proportionalità.  Incidentally, this same book also contains the first published description of the double-entry system of accounting, so Pacioli is often referred to as the "father of accounting" (the book actually was plagiarized from the writings of  Piero della Francesca, but that is a topic for another day).  Pacioli solved the problem by dividing the stakes in proportion to the number of rounds won by each player (Patty would have been happy!).

Another Italian mathematician, Niccolò Fontana Tartaglia suggested that Pacioli's method would not be fair if the game was interrupted after the first round (in our example, Patty would have won the dollar after winning just the first game).  He proposed a different way of dividing the stakes based on the ratio between the size of the lead and the length of the game.  Unfortunately, mathematicians soon discovered that this wasn't a perfect solution either.

Here's where history gets even more interesting, at least for me.  The 17th century mathematicians Blaise Pascal and Pierre de Fermat (most famous for his eponymous last theorem) exchanged a series of letters in 1654 in which they discussed how to solve the "problem of points".  Pascal and Fermat started with the insight that the "division of stakes" shouldn't rely on what had already taken place during the interrupted game, but on the possible ways that the game could have finished if it had not been stopped prematurely.  For example, if Patty has a 9-7 lead over Bobby in a game to 10 versus a 19-17 lead in a game to 20, she has similar odds of winning in both games.  In other words, what matters is the number of rounds that each player still needs to win in order to win the game as opposed to the number of rounds he or she has won up to the point that the game is stopped.

Fermat's solution relied upon the fact that if one player needs r more rounds to win and the other needs s rounds to win, the game will have been won by someone after r + s - 1 rounds (work it out, he's absolutely correct!).  By the simple laws of mathematics, these rounds have 2^(r + s - 1) possible outcomes.   Fermat's solution then was to compute the odds for each player to win simply by writing down all of the possible outcomes and counting how many of them would lead to each player winning.  He would then divide the stakes accordingly.

So, to use our example, Patty needs only 1 more win, while Bobby needs 3 more wins.  If the game had not been interrupted, there would have been a winner after three rounds (1+3-1=4-1=3).  There would have been 8 possible outcomes (2^3 = 2 x 2 x 2 = 8), as follows:

In other words, Patty would win 7 out of the 8 possible outcomes (7/8), while Bobby would only win 1 out of the 8 possible outcomes (1/8).  Patty should therefore take home 7/8 of the prize, while Bobby should take home 1/8 of the prize.  But wait, that's kind of ridiculous too, isn't it?  For the first possible outcome in the Table above, why would they still continue to flip the coin since Patty already won?  Just to check that this still works, let's look at only the feasible outcomes:

Let's calculate the probability of Patty winning (remember, with a coin toss there is 1/2 chances of "Heads" and 1/2 chances "Tails"):

P(A) = 1/2

P(B) = 1/2 x 1/2 = 1/4

P(C) = 1/2 x 1/2 x 1/2 = 1/8

By the rules of probability, if you have mutually exclusive events, you add the probabilities.  Therefore, the probability of Patty winning, P(Patty) = 1/2 + 1/4 + 1/8 = 7/8, which is exactly what we found in the case above!

While Fermat's method of solving the "problem of points" works quite well, just imagine having to calculate the probabilities or listing out all of the possible outcomes when the number of rounds required to determine a winner becomes larger!  Here is where Pascal's solution makes things a little easier and involves something that is now called expected value and the arithmetic triangle that now bears his name (Pascal's Triangle).  We will cover Pascal's solution to the  "problem of points" next time.

Let's Make a Deal!

Our family started watching the 2008 movie "21" starring Kevin Spacey, Kate Bosworth, and Jim Sturgess the other night.  The movie is based upon the book Bringing Down the House and tells the story of how a group of MIT students won a lot of money playing Blackjack in Las Vegas.  We didn't make it to the end, but one of the opening scenes caught my attention.  The MIT mathematics professor played by Spacey challenges one of his students (played by Sturgess) with the classic "Monty Hall Problem", which is the subject of today's post.

The "Monty Hall Problem" is a brain teaser that is named in honor of Monty Hall, the host of the television game show "Let's Make a Deal".  The problem was originally submitted by Steve Selvin to the journal American Statistician in 1975, but it became famous from a letter sent by reader Craig Whitaker to Marilyn vos Savant (best known for having the highest known IQ reported by the Guinness Book of Records) for her weekly column "Ask Marilyn" in Parade magazine in 1990.  Here is the question as posed by Whitaker:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Let's break this down using simple rules of probability.  At the beginning of the game, the contestant has a 1/3 chance of winning a car, which means that the probability of winning a goat is 2/3.  When the game show host reveals a goat behind one of the two remaining doors (not selected by the contestant), the contestant can increase the probability of winning a car to 2/3 if he or she switches to the other door!  

When vos Savant first explained this veridical paradox, approximately 10,000 readers, including 1,000 who had a PhD in mathematics, wrote back to the magazine to tell her that she was wrong.  The famous mathematician Paul Erdős (read about the so-called Erdős Number here) remained unconvinced until someone showed him a simulation of the problem.  If a simulation worked for Erdős, it will probably work for all of you too!

Let's look at all the different permutations of this game, which gives us a total of 9 different scenarios:

If the contestant (in this case, you) adopts the strategy of not switching, he or she wins 3 times out of 9 total chances, or 1/3.  However, if the contestant does indeed switch doors, he or she will increase the chance of winning a car to 6/9, or 2/3!

The "Monty Hall Problem" has made its way into pop culture.  For example, the problem was featured in a 2011 episode of the television show "MythBusters", in which the hosts confirmed that when posed with a version of the "Monty Hall Problem", most people will not switch their choices (when the better strategy, again confirmed by the hosts, would be to switch their choice).  The television show "Survivor" also used a version of the "Monty Hall Problem" in seasons 41 and 42.  Surprisingly, in both cases, the contestant chose not to switch and still won the game!  It just goes to show that playing the odds is not always a winning strategy!  Next time, I want to talk about one more famous problem in probability, and then I promise I will explain what all of this has to do with leadership!

Happy Birthday!

Have you ever shared a birthday with someone you know, either at school or work?  The chance that this occurs is a lot more common than we realize.  As a matter of fact, the probability of sharing a birthday in a group of only 30 people (which is around the size of an elementary school classroom) is over 70%!  

A few years ago, I posted about a concept by Douglas W. Hubbard known as the "rule of five", which he discussed in his book, How to Measure Anything.  At that time, I mentioned a famous problem in probability and statistics known simply as "the birthday problem", which seems counterintuitive until you actually take the time to work out the mathematics.  Let's take a closer look.

Let's first review a couple of rules of simple probability.  Remember that if you are rolling a standard die, the chance of coming up with the number "3" is 1 chance out of six, or 1/6.  Similarly, the chance of rolling an even number (2, 4, or 6) is 3 chances out of six, or 3/6 (which of course simplifies into 1/2 or 50%).  What happens if you roll two dice?  What is the chance that you will roll "3" with the first die and "4" with the second?  The simple rules of probability state that the chance of two independent events occurring is a product of their individual probabilities.  In other words, we multiply 1/6 (the chance of rolling a "3") by 1/6 (the chance of rolling a "4"): 1/6 x 1/6 = 0.03, or 3%.  What's the probability of not rolling a "3" on one die?  It's just 1 minus the probability of rolling a "3", or 1 - 1/6 = 5/6.  You now have all the tools to answer the birthday problem!

Suppose you are a student in a classroom with 29 other students.  What is the probability that you share a birthday with at least one other student in the class?  The best way to tackle this problem is to first calculate the probability (P) that these 30 individuals do NOT share a same birthday, P(A).  Therefore, according to the simple rules of probability discussed above, the probability that at least one student shares his or her birthday with someone else in the class is P(B) = 1 - P(A). 

Let's also make things a little more straightforward and assume that there aren't any twins in the class (twins would share the same birthday) and that there are no leap years involved, which means that there are 365 possible birthdays in a year.  Imagine that we ask all 30 students to form a line, and each student will go and mark his or her birthday on a calendar, one right after the other.  The probability that the first student will have a birthday that is not shared with anyone else is 365/365.  The probability for the second student not sharing a birthday with the first student is 364/365.  The probability for the third student not sharing a birthday with the first or second students is 363/365.  These are all independent events, so in order to calculate the probability of all 30 students not sharing a birthday, we have to multiply the probability of the individual students:

P(A) = 365/365 x 364/365 x 363/365 x ... 337/365 x 336/365 = 0.293684

The probability that at least two individuals share the same birthday, P(B), is therefore:

P(B) = 1 - P(A) = 1- 0.293684 = 0.706 or 70.6%!

If you look the "birthday problem" up online, you can find a graph of P(B) as a function of the number of people in a group that looks something like this:

The "birthday problem" is a great example of what is called a veridical paradox, which is something that appears to be false but is nonetheless true.  Next time, we will discuss another famous veridical paradox (and I promise I will explain what all of this has to do with leadership next time too).

Where is everybody?

Everyone so often, I come across an interesting article or hear about something that I want to write about, even if the article or bit of information has nothing to do with leadership.  As my father has always told me, it's good to be well-rounded and knowledgeable about a lot of different things.  With my father's sage advice firmly in mind, I wanted to explore a recent article I came across that involves the Drake Equation, the Fermi Paradox, and something called the Great Filter.  It's all in the spirit of being well-rounded!

I first heard about the so-called Drake Equation on Carl Sagan's 1980 television mini-series "Cosmos" (I remember getting the book that Sagan wrote as a companion to the mini-series that year for Christmas).  Frank Drake developed the equation bearing his name in 1961 to "estimate" the number of extraterrestrial civilizations that could theoretically exist in our Milky Way Galaxy.  Contrary to popular belief, the equation is more conceptual than numerical (in other words, the equation doesn't actually estimate the number of extraterrestrial lifeforms) and has been used to try to answer the questions "Is there life out there?" or "Are we alone?"  Here is the equation as originally formulated by Dr. Drake:

N = R* x fp x ne x f1 x fi x fc x L

where, 

N is the number of civilizations with which humans could communicate

R* is the rate of new star formation in the galaxy

fp is the fraction of those stars that have planets (like our solar system)

ne is the number of planets that could support life per star with planets

f1 is the fraction of life-supporting planets that actually develop life

fi is the fraction of planets with life where life develops intelligence

fc is the fraction of intelligent civilizations that develop technology to communicate with us

L is the length of time that civilizations can communicate

Again, the Drake Equation was never intended to actually come up with a quantifiable number of civilizations that which we could possibly communicate with, though several experts have provided estimates for each of these variables.  Drake originally estimated that N closely approximated L and concluded that there were probably between 1,000 and 100,000,000 planets with civilizations in the Milky Way Galaxy.  If that is indeed the case, then why haven't we been able to communicate with at least one of these civilizations?  

The great 20th century Italian physicist Enrico Fermi asked this very same question in the 1950's.  As the story goes, Fermi had a casual conversation in the summer of 1950 with fellow physicists Edward Teller, Herbert York, and Emil Konopinski in response to several reports of UFO sightings in the media.  While the exact quote is not known, Fermi is said to have blurted out, "But where is everybody?"  In other words, if there is intelligent life out there, why haven't we been able to communicate with them?  If, as revealed by the Drake Equation, that there are anywhere between 1,000 to 100,000,000 planets with civilizations in our own galaxy, why haven't we been able to contact them?  This, in essence, summarizes the Fermi Paradox.  

Here is where the article that I found comes in.  Scientists at NASA recently submitted a manuscript for publication that provides a possible explanation to the Fermi Paradox.  The manuscript has not been peer-reviewed (meaning that it hasn't been accepted for publication yet).  These scientists talk about something known as the Great Filter.  Robin Hanson, an economist at George Mason University argued in an essay he wrote in 1996 ("The Great Filter - Are We Almost Past It?") that our failure to locate extra-terrestrial life is due to some impenetrable barrier (or filter) that prevents intelligent life from developing to the point where it can communicate with us.  Hanson states that there are nine steps along the evolutionary path to intelligent life:

1. The right star system (potentially habitable planets that can support life - see the Drake Equation above)

2. Reproductive molecules (RNA, DNA, etc)

3. Simple (prokaryotic) single-cell life

4. Complex (eukaryotic) single-cell life

5. Sexual reproduction

6. Multi-cell life 

7. Tool-using animals with intelligence

8. A civilization advancing toward the potential for a colonization explosion

9. Colonization explosion (specifically, space colonization)

Hanson pointed out that life on planet Earth has made it successfully from step 1 through step 8, though we have yet to progress to the 9th and final step (so perhaps this is the ultimate barrier to finding out if life exists elsewhere).  He further argues that because we have yet to find evidence of extraterrestrial life, progression through these steps is very improbable, suggesting that there is a "Great Filter" along the path from the primordial soup to colonization.  He wrote then that "the fact that our universe seems basically dead suggests that it is very very hard for advanced, explosive, lasting life to arise."

Hanson also suggests the possibility of some sort of catastrophic social collapse, such as what happened to the dinosaurs here on planet Earth that effectively wipes out life's progression from steps 1-9.  Alternatively, intelligent life tends to destroy themselves before being able to make contact off-planet.  To that end, a group of investigators from NASA's Jet Propulsion Laboratory proposed several different ways that life on planet Earth could be wiped out, including nuclear war, pandemics, asteroid or comet impacts, climate change, and artificial intelligence.  I would argue that it doesn't take a creative genius to come up with a list of possible scenarios ending life as we know it, just watch any of a number of popular science fiction movies involving these apocalyptic scenarios.

There is one last possibility for why we haven't found evidence of intelligent extraterrestrial life (see my post "Weird Science" for a more detailed explanation).  Basically, this one comes straight from the science fiction movie "The Matrix" .  The Oxford philosopher Nick Bostrom proposed a few years ago  that at least one of the following statements is true - (1) the human species will become extinct before we progress to a post-human state (think artificial intelligence, virtual reality, robotics, and cyborgs!); (2) any post-human civilization is unlikely to actually run a computer simulation of real-world life and humans (basically because they either don't possess the ability or they are too ethical to do so); or (3) we are already living in a simulated world (which is to say that we all co-exist in the Matrix).  If we are living in a simulated world, then of course we wouldn't find evidence of life in outer space.  But then again, why wouldn't we?

All of this is very interesting to ponder.  And perhaps the takeaway lesson here is that we need to be careful about what we are doing to our planet.  It may be the only planet that we will ever have, and we need to be careful about not being the cause of our own destruction.

Happy Veterans Day 2022

On November 11, 1919, President Woodrow Wilson issued a message to all Americans on the one-year anniversary of Armistice Day, which marked the end of major hostilities in World War I at the 11th hour of the 11th day of the 11th month of 1918.  He issued a message to the people of the United States on that very first Armistice Day, in which he expressed what he felt the day meant to Americans:

“To us in America the reflections of Armistice Day will be filled with solemn pride in the heroism of those who died in the country’s service, and with gratitude for the victory, both because of the thing from which it has freed us and because of the opportunity it has given America to show her sympathy with peace and justice in the council of nations.”

Following that first anniversary of the end of World War I in 1919, Armistice Day was unofficially celebrated every year on November 11th, until Congress passed a resolution to officially honor and observe November 11th every year as Armistice Day in 1938.

Unfortunately, World War I didn't end up being "the war to end all wars" as everyone intended or maybe just hoped.  Soon, a little more than 20 years later, the world was again at war.  It was a veteran of that war (World War II) named Raymond Weeks who petitioned (then) General Dwight Eisenhower to expand Armistice Day to celebrate all veterans, not just those who died or served in World War I.  General Eisenhower, of course, supported Week's recommendation until he signed the bill that officially named November 11th as Veterans Day in 1954.  Incidentally, Weeks was later honored with a Presidential Medal of Freedom by President Ronald Reagan and was recognized as the “Father of Veterans Day" in 1982, just three years before his death.

Today, we celebrate Veterans Day, which honors all military veterans, those individuals who have served in one of the branches of the United States Armed Forces (Army, Air Force, Navy, Marine Corps, Coast Guard, and Space Force).  Veterans Day coincides with other holidays that are celebrated in other countries (Remembrance Day in the United Kingdom and Commonwealth countries, for example), though it is distinct from Memorial Day, a U.S. holiday honoring those who died while in military service and Armed Forces Day honoring those individuals who currently serve in the military.

While most calendars print November 11th as “Veteran’s Day” (with an apostrophe), the U.S. Department of Veterans Affairs states that the apostrophe is not necessary, “because it is not a day that ‘belongs’ to veterans, it is a day for honoring all veterans.”  The U.S. government also recommends that we honor all veterans at 2:11 PM Eastern Time with two minutes of silence.

I've always heard that during combat, soldiers fight primarily for each other.  Take a look at the following movie clip from the 2001 movie, Black Hawk Down.  The character "Hoot" (played by actor Eric Bana) is telling another character, First Sergeant Matt Eversmann (played by the actor Josh Hartnett) what motivates him to continue to serve.  Hoot, a Delta Force operator, says:

When I go home people'll ask me, "Hey Hoot, why do you do it man? What, you some kinda war junkie?" You know what I'll say? I won't say a goddamn word. Why? They won't understand. They won't understand why we do it. They won't understand that it's about the men next to you, and that's it. That's all it is.

I just recently finished the book Fairness and Freedom: A History of Two Open Societies, a comparative history of New Zealand and the United States by the historian and writer, David Hackett Fischer.  Fischer wrote about a study conducted by the nineteenth century French army officer Ardant du Picq, who interviewed veterans of the Algerian and Crimean wars and published his findings in the book Battle Studies.  He concluded that fear was nearly universal during battle, but the reason that soldiers overcame their fear and stayed to fight was the fear of letting down their comrades.  These findings were later confirmed and expanded upon in studies by the American journalist and military historian S.L.A. Marshall.  Following World War II, several studies led and conducted by the American sociologist Samuel A. Stouffer in the 1940's and 1950's confirmed that solderis fought primarily for each other, the soldiers fighting next to them.

The men and women who have served together in our nation's Armed Forces share a special bond, a bond shared even among those who have never been in combat.  But regardless of their motivations, there is still something truly special about their willingness to step forward, raise their hand, and potentially give their lives in the service of their fellow citizens and their country. 

If you are a veteran, today, we celebrate and honor you.  Thank you for your dedication and commitment to our country and thank you for your service!