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 ak 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!
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