4 ms·

But in the actual game, you are always given the chance to switch. Here are the possible outcomes of the game, with the third column being what the contestant s

But in the actual game, you are always given the chance to switch. Here are the possible outcomes of the game, with the third column being what the contestant should do to win: You Pick Car Switch? ------- ----- ------ 1 1 N 2 1 Y 3 1 Y 1 2 Y 2 2 N 3 2 Y 1 3 Y 2 3 Y 3 3 N Without knowing anything else, switching means that you win 2/3 of the time.

function_seven 11y ago
Your comment (EDIT: Sorry, not yours, but parent) was talking about a random flip of coin determining which door Monty shows. That's not the actual game. If Monty is randomly showing a door without regard to goat/car, then you have to also consider the scenarios where he reveals a car and you lose early.
- ghshephard 11y ago
  If, while randomly flipping a coin, he reveals the car then you have a 0% chance of winning. If he doesn't reveal a car then you have a 2/3 chance of winning if you switch doors.
  - function_seven 11y ago
    In this scenario, Monty's door choice and your door choice are completely independent actions. Because of this, you gain no new information from the door he reveals. In the classic version of this puzzle (where Monty always reveals a goat), you do gain information, because most of the time Monty's forced to choose the only other goat, and therefore avoid the car. http://c2.com/cgi/wiki?NotTheMontyHallProblem http://c2.com/cgi/wiki?NotTheMontyHallProblem The switching strategy is effective because it relies on Monty's avoidance of the car. When Monty isn't trying to avoid the car, his revelation doesn't help you out at all. It just tells you that you either (A) now have a slightly better chance (50%), or (2) you've already lost. (0%). EDIT: Also see the table on this wikipedia article. It lists probabilities for all the different variants of the MHP. Your scenario is referred to as the Monty Fall or Ignorant Monty variant. https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_... "Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car (Granberg and Brown, 1995:712) (Rosenthal, 2005a) (Rosenthal, 2005b). Switching wins the car half of the time.
    - ghshephard 11y ago
      When you originally pick a door - there are three. (We all agree on this). Because you are picking at random, and the prize is only behind one door, there is a 1/3 chance that you have picked the right door. (likewise, We can all agree on that.) Therefore, I think we can all agree that there is a 2/3 chance that the prize exists behind one of the other two doors. Let us take your approach, and have Monty flip a coin, revealing ANY door at random. If he flips your door, and you have the prize - You have a 100% chance of winning. If he flips your door, and you have the goat, 0% chance of winning. Those two scenarios are easy, and we can ignore them. Likewise, if he flips the coin, and shows one of the other two doors, and they have the prize - then, once again - easy - 0% chance of winning. We can ignore these scenarios. I think up to this point, we can agree on all of the above scenarios. (They are win/lose) Where it gets complex, he flips a coin, selects one of the doors you aren't on, and reveals a goat. Given that you had a 1/3 chance originally of winning, and a 2/3 chance of not winning, and given that Monty has now revealed that one of the other two doors (By chance) does not have the car, your 2/3 chance of not winning is still the same, but now it applies to only a single door - I.E. You have a 66% chance of winning by selecting that other door - and the fact that Monty selected that door by chance, is irrelevant. Unfortunately, despite my (flawed) apparently logical argument - I'm completely wrong, and you are correct. from random import * def putPrizeInDoor(): doors=[0,0,0] door_winner=randint(0,2) doors[door_winner]=1 return doors trials=100000 testCount=0 win=lose=0 for x in range(trials): doors=putPrizeInDoor() userSelects=randint(0,2) montyReveals=randint(0,2) if montyReveals==userSelects: # Monty reveals the door user is on pass else: if doors[montyReveals]==1: # Monty Revealed the prize. pass else: # this is where it's interesting - Monty reveals, at random, door not winning that user not on. revealedDoors=set([montyReveals,userSelects]) newUserChoice=list(set([0,1,2])-revealedDoors)[0] testCount+=1 if doors[newUserChoice]==1: win+=1 else: lose+=1 print ("Tested: {} Wins: {} Lost: {} Percentage: {:.2f}".format(testCount,win,lose,win/testCount)) Tested: 44484 Wins: 22220 Lost: 22264 Percentage: 0.50 It's interesting that the deliberate selection by Monty of the goat, increases my chance from 33% to 66% if I switch, but the random selection by Monty of the goat, only increases my chance from 33% to 50% if I switch. I'm not really sure where the argument I made breaks down, but, empirically, it's clearly broken. I'll have to meditate on it a bit to see where I've gone awry.
Retric 11y ago
I changed to rules so in "3, 1, Y" he might open door 3 and show you a car. At that point switching is a meaningless choice.
deleted 11y ago
[deleted]

function_seven 11y ago

Your comment (EDIT: Sorry, not yours, but parent) was talking about a random flip of coin determining which door Monty shows. That's not the actual game. If Monty is randomly showing a door without regard to goat/car, then you have to also consider the scenarios where he reveals a car and you lose early.

ghshephard 11y ago
If, while randomly flipping a coin, he reveals the car then you have a 0% chance of winning. If he doesn't reveal a car then you have a 2/3 chance of winning if you switch doors.
- function_seven 11y ago
  In this scenario, Monty's door choice and your door choice are completely independent actions. Because of this, you gain no new information from the door he reveals. In the classic version of this puzzle (where Monty always reveals a goat), you do gain information, because most of the time Monty's forced to choose the only other goat, and therefore avoid the car. http://c2.com/cgi/wiki?NotTheMontyHallProblem http://c2.com/cgi/wiki?NotTheMontyHallProblem The switching strategy is effective because it relies on Monty's avoidance of the car. When Monty isn't trying to avoid the car, his revelation doesn't help you out at all. It just tells you that you either (A) now have a slightly better chance (50%), or (2) you've already lost. (0%). EDIT: Also see the table on this wikipedia article. It lists probabilities for all the different variants of the MHP. Your scenario is referred to as the Monty Fall or Ignorant Monty variant. https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_behaviors https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_... "Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car (Granberg and Brown, 1995:712) (Rosenthal, 2005a) (Rosenthal, 2005b). Switching wins the car half of the time.
  - ghshephard 11y ago
    When you originally pick a door - there are three. (We all agree on this). Because you are picking at random, and the prize is only behind one door, there is a 1/3 chance that you have picked the right door. (likewise, We can all agree on that.) Therefore, I think we can all agree that there is a 2/3 chance that the prize exists behind one of the other two doors. Let us take your approach, and have Monty flip a coin, revealing ANY door at random. If he flips your door, and you have the prize - You have a 100% chance of winning. If he flips your door, and you have the goat, 0% chance of winning. Those two scenarios are easy, and we can ignore them. Likewise, if he flips the coin, and shows one of the other two doors, and they have the prize - then, once again - easy - 0% chance of winning. We can ignore these scenarios. I think up to this point, we can agree on all of the above scenarios. (They are win/lose) Where it gets complex, he flips a coin, selects one of the doors you aren't on, and reveals a goat. Given that you had a 1/3 chance originally of winning, and a 2/3 chance of not winning, and given that Monty has now revealed that one of the other two doors (By chance) does not have the car, your 2/3 chance of not winning is still the same, but now it applies to only a single door - I.E. You have a 66% chance of winning by selecting that other door - and the fact that Monty selected that door by chance, is irrelevant. Unfortunately, despite my (flawed) apparently logical argument - I'm completely wrong, and you are correct. from random import * def putPrizeInDoor(): doors=[0,0,0] door_winner=randint(0,2) doors[door_winner]=1 return doors trials=100000 testCount=0 win=lose=0 for x in range(trials): doors=putPrizeInDoor() userSelects=randint(0,2) montyReveals=randint(0,2) if montyReveals==userSelects: # Monty reveals the door user is on pass else: if doors[montyReveals]==1: # Monty Revealed the prize. pass else: # this is where it's interesting - Monty reveals, at random, door not winning that user not on. revealedDoors=set([montyReveals,userSelects]) newUserChoice=list(set([0,1,2])-revealedDoors)[0] testCount+=1 if doors[newUserChoice]==1: win+=1 else: lose+=1 print ("Tested: {} Wins: {} Lost: {} Percentage: {:.2f}".format(testCount,win,lose,win/testCount)) Tested: 44484 Wins: 22220 Lost: 22264 Percentage: 0.50 It's interesting that the deliberate selection by Monty of the goat, increases my chance from 33% to 66% if I switch, but the random selection by Monty of the goat, only increases my chance from 33% to 50% if I switch. I'm not really sure where the argument I made breaks down, but, empirically, it's clearly broken. I'll have to meditate on it a bit to see where I've gone awry.

Retric 11y ago

I changed to rules so in "3, 1, Y" he might open door 3 and show you a car. At that point switching is a meaningless choice.

deleted 11y ago

[deleted]