[grraaaaahhh]

Those runs get ignored since that isn't the scenario. Specifically we're looking at the case where Monty opens a goat by chance (ie. he could have opened the car, but did not). It also helps that that result doesn't matter to the overall question, is it better to switch or to stay, since both options are irrelevant when Monty opens the car door.

[Dylan16807]

By deleting the universe 1/3 of the time, you change the nature of the problem. It starts to resemble a quantum immortality argument. And if I can delete universes where certain events happen then screw probability, I can guarantee a win despite picking randomly.

"Picks a random door out of all closed doors, but that door is never the winner" is a contradiction. It doesn't give you a different chance of winning because it's not a coherent scenario in the first place.

[OscarCunningham]

Conditional probability is a coherent concept with no need to talk about deleting universes. You can ask "If it's raining in the morning how likely is it to still be raining in the afternoon?" even though it's sometimes not raining in the morning.

[Dylan16807]

Conditional probability is fine. But you have to make the condition part of the question. You can't subtly omit certain results because it screws up the narrative.

This is a game where you start off with a certain layout, and then proceed forward. If you want conditions, they have to be conditions that you can apply before the game starts. You can't retroactively remove a significant chunk of samples.

If you added an actual outcome to him picking the winner, you could make a valid filter where the answer actually is 50:50. Perhaps they restart the game. Perhaps they never air the episode. And if you calculate only for finished/aired games then it's very clear then that you're solving a different math problem. You can't apply that answer to the original problem.

[semiotagonal]

RESCINDED: Error in my simulation! Apologies.

[sparky_z]

You're accidentally including the cases where montysPick == myPick. You have to discard those cases.

Monte has to show you what's behind a different door than the one you initially picked. The game doesn't make sense otherwise.

[semiotagonal]

Yes you are correct.

Corrected simulation gives 50/50 when Monty picks at random among 3 doors.

switchWins=33044 switchLoses=33581

#include <stdio.h> #include <stdlib.h> #include <time.h> #include <string.h>

#define NUM_RUNS 100000 #define NUM_DOORS 3 int main(int argc, const char* argv[]) { srand(time(NULL)); char doors[NUM_DOORS]; int switchLoses = 0; int switchWins = 0; for(int i = 0; i < NUM_RUNS; i++) { memset(&doors, 0, sizeof(doors)); doors[rand() % NUM_DOORS] = 1; int myPick = rand() % NUM_DOORS; int montysPick = rand() % (NUM_DOORS - 1); if(montysPick >= myPick) montysPick++; if(doors[montysPick]) continue; if(doors[myPick]) switchLoses++; else switchWins++; } printf("switchWins=%d switchLoses=%d\n", switchWins, switchLoses); return 0; }

EDITED TO ADD:

Just for further info:

3 doors: switchWins=33111 switchLoses=33470

4 doors: switchWins=50182 switchLoses=24698

8 doors: switchWins=74876 switchLoses=12687

100000 doors: switchWins=99998 switchLoses=2

[reidjs]

Hah, I wrote one as well to prove it to myself in the chrome dev tools console

  const pickCar = (switchChoice) => {
    const doors = [0, 0, 0];
    doors[(Math.random() * doors.length - 1) | 0] = 1
 let pick = (Math.random() * doors.length - 1) | 0
 let goat
 for(let i = 0; i < doors.length; i++) { if (doors[i] == 0 && i != pick) goat = i  }
 let result = pick
 if (switchChoice) for(let i = 0; i < doors.length; i++) { if (i != pick && i != goat) result = i }
 return result 
  }
  let switchWins = 0;
  let switchLosses = 0;
  for(let i = 0; i < 100000; i++) {
   pickCar(true) == 1 ? switchWins++ : switchLosses++
  }
  console.log('wins:', switchWins, ', losses: ',switchLosses)