Below is the syntax highlighted version of Pepys.java
from §1.3 Conditionals and Loops.
/****************************************************************************** * Compilation: javac Pepys.java * Execution: java Pepys k trials * * Which is more likely: at least one 1 in 6 rolls of a fair die, * or at least two 1s in 12 rolls of a fair die? * * This program takes two integer command-line arguments k and trials, * and repeats the following experiment trials times: * - roll 6k dice * - count number of 1s * - success = at least k 1s * * % java Pepys 1 10000000 * 0.6651856 * * % java Pepys 2 10000000 * 0.6186818 * * % java Pepys 1000 100000 * 0.50647 * * The exact answer is 1 - (5/6)^6 = 31031/46656 for k = 1, * 1346704211/2176782336 for k = 2, and * 1/2 as k approaches infinity. * * Reference: https://en.wikipedia.org/wiki/Newton–Pepys_problem * * Use for live coding. * - forget to type command line arguments * - to start: do experiment assuming k = 1 * - declare and initialize ones outside loop * (so that it never gets reset to 0) * - Put in extra ; after if statement. * - Integer division (r < 1/6) * - Insert System.out.println() in loop for debugging * - forget to recompile * - add command-line argument k (but use Integer.parseInt(args[0]) again) * - named constant for number of sides of dice * - forget curly braces around outer for loop * ******************************************************************************/ public class Pepys { public static void main(String[] args) { int SIDES = 6; // number of sides on a die // roll a fair die 6k times, seeking k 6s int k = Integer.parseInt(args[0]); // number of times to repeat each experiment int trials = Integer.parseInt(args[1]); // repeat experiment and count successes int count = 0; for (int t = 1; t <= trials; t++) { // number of 1s in 6k rolls int ones = 0; for (int i = 0; i < k * SIDES; i++) { double r = Math.random(); if (r < 1.0 / SIDES) ones++; } if (ones >= k) count++; } // print fraction of successes System.out.println(1.0 * count / trials); } }