doc: Monty Hall simulation for std::rand
A larger example for std::rand
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@ -71,6 +71,109 @@ use std::rand;
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let tuple = rand::random::<(f64, char)>();
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println!("{}", tuple)
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```
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This is a simulation of the [Monty Hall Problem][]:
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> Suppose you're on a game show, and you're given the choice of three doors:
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> Behind one door is a car; behind the others, goats. You pick a door, say No. 1,
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> and the host, who knows what's behind the doors, opens another door, say No. 3,
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> which has a goat. He then says to you, "Do you want to pick door No. 2?"
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> Is it to your advantage to switch your choice?
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The rather unintuitive answer is that you will have a 2/3 chance of winning if
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you switch and a 1/3 chance of winning of you don't, so it's better to switch.
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This program will simulate the game show and with large enough simulation steps
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it will indeed confirm that it is better to switch.
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[Monty Hall Problem]: http://en.wikipedia.org/wiki/Monty_Hall_problem
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```
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use std::rand;
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use std::rand::Rng;
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use std::rand::distributions::{IndependentSample, Range};
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struct SimulationResult {
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win: bool,
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switch: bool
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}
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// Run a single simulation of the Monty Hall problem.
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fn simulate<R: Rng>(rng: &mut R) -> SimulationResult {
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let random_door = Range::new(0u, 3);
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let car = random_door.ind_sample(rng);
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// This is our initial choice
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let mut choice = random_door.ind_sample(rng);
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// The game host opens a door
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let open = game_host_open(car, choice, rng);
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// Shall we switch?
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let switch = rng.gen();
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if switch {
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choice = switch_door(choice, open);
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}
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SimulationResult { win: choice == car, switch: switch }
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}
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// Returns the door the game host opens given our choice and knowledge of
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// where the car is. The game host will never open the door with the car.
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fn game_host_open<R: Rng>(car: uint, choice: uint, rng: &mut R) -> uint {
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let choices = free_doors(vec![car, choice]);
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rand::sample(rng, choices.move_iter(), 1)[0]
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}
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// Returns the door we switch to, given our current choice and
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// the open door. There will only be one valid door.
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fn switch_door(choice: uint, open: uint) -> uint {
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free_doors(vec![choice, open])[0]
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}
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fn free_doors(blocked: Vec<uint>) -> Vec<uint> {
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range(0u, 3).filter(|x| !blocked.contains(x)).collect()
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}
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fn main() {
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// The estimation will be more accuraty with more simulations
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let num_simulations = 10000u;
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let mut rng = rand::task_rng();
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let (mut switch_wins, mut switch_losses) = (0u, 0u);
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let (mut keep_wins, mut keep_losses) = (0u, 0u);
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println!("Running {} simulations...", num_simulations);
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for _ in range(0, num_simulations) {
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let result = simulate(&mut rng);
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match (result.win, result.switch) {
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(true, true) => switch_wins += 1,
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(true, false) => keep_wins += 1,
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(false, true) => switch_losses += 1,
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(false, false) => keep_losses += 1,
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}
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}
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let total_switches = switch_wins + switch_losses;
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let total_keeps = keep_wins + keep_losses;
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println!("Switched door {} times with {} wins and {} losses",
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total_switches, switch_wins, switch_losses);
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println!("Kept our choice {} times with {} wins and {} losses",
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total_keeps, keep_wins, keep_losses);
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// With a large number of simulations, the values should converge to
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// 0.667 and 0.333 respectively.
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println!("Estimated chance to win if we switch: {}",
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switch_wins as f32 / total_switches as f32);
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println!("Estimated chance to win if we don't: {}",
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keep_wins as f32 / total_keeps as f32);
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}
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```
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*/
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#![experimental]
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