November 6, 2010 Santa Fe Alliance for Science Professional Enrichment Activity The Problem of Points

Problem of Points: An Unfinished Game Game is between two players. Each player contributes an equal amount to the “pot”. First player to win n points (rounds) wins the game (e.g. “best of 7” means first player to win 4 points wins the game). Each point awarded by a random process (coin flip, etc.). If game is interrupted when one player is ahead, and can't be resumed, how should be pot be divided?

Differing Viewpoints (pre-1654) Player in the lead wins the pot. Game is called off; all bets are off. Pot should be divided proportionately, … according to points won (Pacioli); according to points remaining to play (Cardano); according to size of lead (Tartaglia).

Pascal and Fermat Tackle the Problem (1654) Antoine Gambaud, Chevalier de Méré, asked Blaise Pascal to answer puzzling questions on games of chance; problem of points is included. Pascal asked Pierre de Fermat to collaborate in finding solution to problem of points. Working together through letters, both men solved the problem, with slightly different – though mathematically equivalent – approaches.

Fermat's “Possible Futures” Player 1 has r 1 points remaining to win. Player 2 has r 2 points remaining to win. Enumerate all possible outcomes of (r 1 + r 2 – 1) rounds of play, even if one of the players would win before that many rounds had been played. Divide the pot proportionately, according to the ratio of outcomes in which each player would have won the game.

Fermat's “Possible Futures” - Example Each player put $10 in pot ($20 total). Player 1 has 2 points remaining to win. Player 2 has 3 points remaining to win. Enumerate all outcomes of 4 rounds of play. Player 1 wins in 11 of 16 outcomes, should be awarded 11/16 of the pot, or $13.75.

Pascal's “Expectation” Compute weighted sum of the returns (payoffs) associated with each of the outcomes; use outcome probabilities as the weights. Weighted sum of returns is the expected value of the game for the given player. Extra rounds of play need not be included. Divide the pot by awarding each player his or her expected value for the game.

Pascal's “Expectation” - Example Same problem as previous example, but this enumeration doesn't need to include extra rounds of play. Distinct outcomes with equal returns may be combined by adding probabilities (not shown here). Answer is equal to that given by Fermat's method.

Pascal's Triangle Pascal's triangle formed by starting with 1 at the apex (row 0), and increasing number of elements by 1 in each successive row. Outermost elements in each row have a value of 1. Each inner element has a value equal to the sum of the two adjacent elements immediately above.

Pascal's Triangle in the Problem of Points Fermat's “possible futures”, consisting of sequences of player 1/player 2 point wins, are enumerated completely by descent paths from apex to row (r 1 + r 2 – 1) of Pascal's triangle. The value of any element is equal to the total number of distinct descent paths from the apex to that element. Split row (r 1 + r 2 – 1) into two portions, one corresponding to player 1's winning outcomes, and the other to player 2's; the sum of each portion is the the number of winning outcomes for the corresponding player.

Pascal's Triangle - Example In our example, (r 1 + r 2 – 1) = 4; row 4 is bottom row here. All paths to left 3 elements in row 4 require at least 2 steps to left in descent from apex (2 points needed by player 1 to win). Sum of these elements is 11, as in Fermat's method. All paths to right 2 elements require at least 3 steps to the right in descent from apex (3 points needed by player 2 to win). These elements sum to 5.

References Keith Devlin, The Unfinished Game: Pascal, Fermat, and the Seventeenth-Century Letter that Made the World Modern, Basic Books, Leonard Mlodinow, The Drunkard's Walk: How Randomness Rules Our Lives, Vintage, Edward Packel, The Mathematics of Games and Gambling, Mathematical Association of America, “Pascal's triangle”, Wikipedia, Online: “Problem of points”, Wikipedia, Online: