

The Problem

 100 prisoners are individually numbered 1 to 100
 A room having a cupboard of 100 opaque drawers numbered 1 to 100, that cannot be seen from outside.
 Cards numbered 1 to 100 are placed randomly, one to a drawer, and the drawers all closed; at the start.
 Prisoners start outside the room
 They can decide some strategy before any enter the room.
 Prisoners enter the room one by one, can open a drawer, inspect the card number in the drawer, then close the drawer.
 A prisoner can open no more than 50 drawers.
 A prisoner tries to find his own number.
 A prisoner finding his own number is then held apart from the others.
 If all 100 prisoners find their own numbers then they will all be pardoned. If any don't then all sentences stand. 


The task

 Simulate several thousand instances of the game where the prisoners randomly open drawers
 Simulate several thousand instances of the game where the prisoners use the optimal strategy mentioned in the Wikipedia article, of:
 First opening the drawer whose outside number is his prisoner number.
 If the card within has his number then he succeeds otherwise he opens the drawer with the same number as that of the revealed card. (until he opens his maximum).


Show and compare the computed probabilities of success for the two strategies, here, on this page.




References

 The unbelievable solution to the 100 prisoner puzzle standupmaths (Video).
 wp:100 prisoners problem
 100 Prisoners Escape Puzzle DataGenetics.
 Random permutation statistics#One hundred prisoners on Wikipedia.


 
const _ = require('lodash');
 
const numPlays = 100000;
 
const setupSecrets = () => {
	// setup the drawers with random cards
	let secrets = [];
 
	for (let i = 0; i < 100; i++) {
		secrets.push(i);
	}
 
	return _.shuffle(secrets);
}
 
const playOptimal = () => {
 
	let secrets = setupSecrets();
 
 
	// Iterate once per prisoner
	loop1:
	for (let p = 0; p < 100; p++) {
 
		// whether the prisoner succeedss
		let success = false;
 
		// the drawer number the prisoner chose
		let choice = p;
 
 
		// The prisoner can choose up to 50 cards
		loop2:
		for (let i = 0; i < 50; i++) {
 
			// if the card in the drawer that the prisoner chose is his card
			if (secrets[choice] === p){
				success = true;
				break loop2;
			}
 
			// the next drawer the prisoner chooses will be the number of the card he has.
			choice = secrets[choice];
 
		}	// each prisoner gets 50 chances
 
 
		if (!success) return false;
 
	} // iterate for each prisoner 
 
	return true;
}
 
const playRandom = () => {
 
	let secrets = setupSecrets();
 
	// iterate for each prisoner 
	for (let p = 0; p < 100; p++) {
 
		let choices = setupSecrets();
 
		let success = false;
 
		for (let i = 0; i < 50; i++) {
 
			if (choices[i] === p) {
				success = true;
				break;
			}
		}
 
		if (!success) return false;
	}
 
	return true;
}
 
const execOptimal = () => {
 
	let success = 0;
 
	for (let i = 0; i < numPlays; i++) {
 
		if (playOptimal()) success++;
 
	}
 
	return 100.0 * success / 100000;
}
 
const execRandom = () => {
 
	let success = 0;
 
	for (let i = 0; i < numPlays; i++) {
 
		if (playRandom()) success++;
 
	}
 
	return 100.0 * success / 100000;
}
 
console.log("# of executions: " + numPlays);
console.log("Optimal Play Success Rate: " + execOptimal());
console.log("Random Play Success Rate: " + execRandom());
 