Monty Hall, Monty Fall, Monty Crawl

***Ace Mercury*** · 06-23-2009, 08:08 AM

Coding Horror: Monty Hall, Monty Fall, Monty Crawl

A variation on a classic probability riddle. Wikipedia has a good writeup on it, but if you haven't seen it before, I suggest trying to solve it first. Hopefully, we'll have a million people yelling at each other why they're wrong.

Monty Hall Problem: Suppose the contestants on a game show are given the choice of three doors: behind one door is a car; behind the others, goats. After a contestant picks a door, the host, who knows what's behind all the doors, opens one of the unchosen doors. The door the host opens is intentionally a goat; he would never pick the door with the car. He then asks the contestant, "Do you want to switch doors?" What are the probabilities for winning the car for switching doors or staying with your first pick?

Monty Fall Problem: In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win, either by sticking with your original door, or switching doors?

Monty Crawl Problem: Once you have selected one of the three doors, the host then reveals one non-selected door which does not contain the car. However, the host is very tired, and crawls from his position (near Door #1) to the door he is to open. In particular, if he has a choice of doors to open, then he opens the smallest number available door. (For example, if you selected Door #1 and the car was indeed behind Door #1, then the host would always open Door #2, never Door #3.) Now what are the probabilities that you will win the car if you stick versus if you switch?

***CaptHayfever*** · 06-23-2009, 01:32 PM

The Monty Hall problem is well-known already.
Monty Fall appears to have the same answer, as its conditions are ultimately identical.
Spoiler Below

The official answer is 2/3 for switching. I still have issues with this, as no one has ever adequately explained to me why the switch decision is not a new choice.

Monty Crawl
Spoiler Below

depends entirely on which door Monty opens. If he opens #1, then you are under the same conditions as the previous two, so it's 2/3 odds for switching. If he opens #2, it's 100% odds that you should take #1, whether it was your original pick or not.

Of course, you're not asking for the mathematician's (casewise logic) answer; you're asking for the statistician's (total probability estimate) answer, which is, funnily enough, stil 2/3 for switching.

And remember, "I'm-a Luigi, number one!"

**Deku Trii** · 06-23-2009, 02:43 PM

Monty Crawl

Spoiler Below

If he opens the lower numbered door, as he will 2/3 of the time, you win by switching 50% of the time. If he opens the higher numbered door, as he will 1/3 of the time, you win by switching 100% of the time. So you win by switching 1/3*1+1/2*2/3=2/3.

***CaptHayfever*** · 06-23-2009, 10:43 PM

^That answer is right, but the reasoning is a bit off:
Spoiler Below

If he opens door #1, then the usual Monty Hall Problem conditions are in effect, where the probability is 2/3 for switching, not just 50%. (Though I still think it should be 50%, but that's a rant I don't wanna get into right now.)

If he opens door #2, though, that means the car is absolutely behind door #1, so if you've already picked door #1, the odds are 0 for switching, while if you've picked #3, the odds are 100%.

But it does all add up to 2/3 like you have.

And remember, "I'm-a Luigi, number one!"

**Deku Trii** · 06-24-2009, 07:28 AM

^
Spoiler Below

I don't think so, actually. The usual monty hall conditions aren't in effect because he wasn't equally likely to pick either of the other doors. In the monty crawl variant, the information you know about the methodology by which he picks the door makes it equally likely that the car can be won by switching or not switching if he picks the lower numbered door.

I'm not sure how your explaination adds up to 2/3. There's only a 1 in 3 chance he'll pick the higher numbered door, and a 2/3 chance he'll pick the lower numbered door. If your odds were 2/3 for switching when he picked the lower numbered door, you'd win by switching 1/3*1+2/3*2/3=7/9 times.

***CaptHayfever*** · 06-24-2009, 01:41 PM

Spoiler Below

There are 3 possible situations.
(Case 1) Monty picks door 1.
This has a 2/3 chance of happening, & once it happens, your odds for switching are 2/3 by the original Monty Hall problem, since opening door 1 reveals nothing conclusive about doors 2 & 3.
2/3*2/3 = 4/9

(Case 2) Monty picks door 2 (1/3 chance of happening), & you had already picked door 2 (1/3 chance of happening).
You should switch to door 1 right now, 100%.
1/3*1/3*1 = 1/9

(Case 3) Monty picks door 2 (1/3 chance), & you had already picked door 3 (1/3 chance).
Again, switch to door 1, 100%.
1/3*1/3*1 = 1/9

(Case 4) Monty picks door 2, & you had already picked door 1.
The car is behind door 1; stay put.
Odds for switching = 0.

Total: 4/9+1/9+1/9+0 = 6/9 = 2/3.

And remember, "I'm-a Luigi, number one!"

Microphone_Kirby · 06-24-2009, 02:32 PM

Quote:

Originally Posted by CaptHayfever

Spoiler Below

There are 3 possible situations.
(Case 1) Monty picks door 1.
This has a 2/3 chance of happening, & once it happens, your odds for switching are 2/3 by the original Monty Hall problem, since opening door 1 reveals nothing conclusive about doors 2 & 3.
2/3*2/3 = 4/9

(Case 2) Monty picks door 2 (1/3 chance of happening), & you had already picked door 2 (1/3 chance of happening).
You should switch to door 1 right now, 100%.
1/3*1/3*1 = 1/9

(Case 3) Monty picks door 2 (1/3 chance), & you had already picked door 3 (1/3 chance).
Again, switch to door 1, 100%.
1/3*1/3*1 = 1/9

(Case 4) Monty picks door 2, & you had already picked door 1.
The car is behind door 1; stay put.
Odds for switching = 0.

Total: 4/9+1/9+1/9+0 = 6/9 = 2/3.

And remember, "I'm-a Luigi, number one!"

Spoiler Below

What if the car (in Case 4) is behind Door 3? Your saying that
D1: Car (Player)
D2: Zonk (Monty)
D3: Zonk
But it's still possible for this to happen:
D1: Zonk (Player)
D2: Zonk (Monty)
D3: Car

Remember that Monty doesn't pick the same door the Player picks. Therefore, it's still possible that the Car is behind Door Three.

And Your Case Two is impossible for the above reasoning.

***CaptHayfever*** · 06-24-2009, 05:03 PM

^The entire premise of Monty Crawl is that Monty goes to the closest door that isn't the car. If it's behind #3, he'll open #1.

And remember, "I'm-a Luigi, number one!"

Microphone_Kirby · 06-24-2009, 06:27 PM

Quote:

Originally Posted by CaptHayfever

^The entire premise of Monty Crawl is that Monty goes to the closest door that isn't the car. If it's behind #3, he'll open #1.

... I know that. What's your point?

**Deku Trii** · 06-24-2009, 10:28 PM

Spoiler Below

Let's say you picked door 1. The probabilities that he will pick door number 2 if the car is behind door #1, #2, or #3 are 100%, 0%, and 100%. The probabilities that he will pick door #3 if the car is behind door #1, #2, or #3 are 0%, 100%, and 0%.

If the host opens door 2, there is a 50% chance you can win by switching. If he opens door 3, there is a 100% chance you will win by switching. There is a 1/3 chance he will open door 3, and a 2/3 chance he will open door 2.

1/3*1+1/2*2/3=2/3

CuccoLady · 06-29-2009, 07:09 PM

@ Cap'n: I didn't get the Monty Hall solution at all either, but then I looked it up and it was explained and made much more sense.

Spoiler Below

There are three possibilities:

-you pick the car. The host reveals a goat, you switch and pick the other goat and lose.

-you pick a goat. The host reveals the other goat, you switch and pick the car and win.

-you pick the other goat. The host reveals the other goat, you switch and pick the car and win.

In other words, if you decide to switch, picking the car is the only way to lose... if you pick either goat the first time, you will win if you switch.

That is the only explanation that actually made sense to me.

06-23-2009, 08:08 AM	#1
*Ace Mercury* Marshmallow Knight ☆ Supermod Join Date: Jun 2000 Location: Southern Ontario Gender: Posts: 23,274 Thanks: 568 Thanked 3,297 Times in 1,582 Posts Blog Entries: 1	Monty Hall, Monty Fall, Monty Crawl Coding Horror: Monty Hall, Monty Fall, Monty Crawl A variation on a classic probability riddle. Wikipedia has a good writeup on it, but if you haven't seen it before, I suggest trying to solve it first. Hopefully, we'll have a million people yelling at each other why they're wrong. Monty Hall Problem: Suppose the contestants on a game show are given the choice of three doors: behind one door is a car; behind the others, goats. After a contestant picks a door, the host, who knows what's behind all the doors, opens one of the unchosen doors. The door the host opens is intentionally a goat; he would never pick the door with the car. He then asks the contestant, "Do you want to switch doors?" What are the probabilities for winning the car for switching doors or staying with your first pick? Monty Fall Problem: In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win, either by sticking with your original door, or switching doors? Monty Crawl Problem: Once you have selected one of the three doors, the host then reveals one non-selected door which does not contain the car. However, the host is very tired, and crawls from his position (near Door #1) to the door he is to open. In particular, if he has a choice of doors to open, then he opens the smallest number available door. (For example, if you selected Door #1 and the car was indeed behind Door #1, then the host would always open Door #2, never Door #3.) Now what are the probabilities that you will win the car if you stick versus if you switch?

06-23-2009, 01:32 PM	#2
*CaptHayfever* Smash SuperMod Join Date: Jul 2002 Location: (n) - the place where I am Gender: Posts: 27,659 Thanks: 1,991 Thanked 2,486 Times in 1,513 Posts	The Monty Hall problem is well-known already. Monty Fall appears to have the same answer, as its conditions are ultimately identical. Spoiler Below The official answer is 2/3 for switching. I still have issues with this, as no one has ever adequately explained to me why the switch decision is not a new choice. Monty Crawl Spoiler Below depends entirely on which door Monty opens. If he opens #1, then you are under the same conditions as the previous two, so it's 2/3 odds for switching. If he opens #2, it's 100% odds that you should take #1, whether it was your original pick or not. Of course, you're not asking for the mathematician's (casewise logic) answer; you're asking for the statistician's (total probability estimate) answer, which is, funnily enough, stil 2/3 for switching. And remember, "I'm-a Luigi, number one!"

06-23-2009, 02:43 PM	#3
Deku Trii PP&R Mod Current Events Mod Join Date: Jun 2000 Location: Texas Gender: Posts: 9,031 Thanks: 521 Thanked 1,099 Times in 583 Posts	Monty Crawl Spoiler Below If he opens the lower numbered door, as he will 2/3 of the time, you win by switching 50% of the time. If he opens the higher numbered door, as he will 1/3 of the time, you win by switching 100% of the time. So you win by switching 1/31+1/22/3=2/3.

06-23-2009, 10:43 PM	#4
*CaptHayfever* Smash SuperMod Join Date: Jul 2002 Location: (n) - the place where I am Gender: Posts: 27,659 Thanks: 1,991 Thanked 2,486 Times in 1,513 Posts	^That answer is right, but the reasoning is a bit off: Spoiler Below If he opens door #1, then the usual Monty Hall Problem conditions are in effect, where the probability is 2/3 for switching, not just 50%. (Though I still think it should be 50%, but that's a rant I don't wanna get into right now.) If he opens door #2, though, that means the car is absolutely behind door #1, so if you've already picked door #1, the odds are 0 for switching, while if you've picked #3, the odds are 100%. But it does all add up to 2/3 like you have. And remember, "I'm-a Luigi, number one!"

06-24-2009, 07:28 AM	#5
Deku Trii PP&R Mod Current Events Mod Join Date: Jun 2000 Location: Texas Gender: Posts: 9,031 Thanks: 521 Thanked 1,099 Times in 583 Posts	^ Spoiler Below I don't think so, actually. The usual monty hall conditions aren't in effect because he wasn't equally likely to pick either of the other doors. In the monty crawl variant, the information you know about the methodology by which he picks the door makes it equally likely that the car can be won by switching or not switching if he picks the lower numbered door. I'm not sure how your explaination adds up to 2/3. There's only a 1 in 3 chance he'll pick the higher numbered door, and a 2/3 chance he'll pick the lower numbered door. If your odds were 2/3 for switching when he picked the lower numbered door, you'd win by switching 1/31+2/32/3=7/9 times.

06-24-2009, 01:41 PM	#6
*CaptHayfever* Smash SuperMod Join Date: Jul 2002 Location: (n) - the place where I am Gender: Posts: 27,659 Thanks: 1,991 Thanked 2,486 Times in 1,513 Posts	Spoiler Below There are 3 possible situations. (Case 1) Monty picks door 1. This has a 2/3 chance of happening, & once it happens, your odds for switching are 2/3 by the original Monty Hall problem, since opening door 1 reveals nothing conclusive about doors 2 & 3. 2/32/3 = 4/9 (Case 2) Monty picks door 2 (1/3 chance of happening), & you had already picked door 2 (1/3 chance of happening). You should switch to door 1 right now, 100%. 1/31/31 = 1/9 (Case 3) Monty picks door 2 (1/3 chance), & you had already picked door 3 (1/3 chance). Again, switch to door 1, 100%. 1/31/3*1 = 1/9 (Case 4) Monty picks door 2, & you had already picked door 1. The car is behind door 1; stay put. Odds for switching = 0. Total: 4/9+1/9+1/9+0 = 6/9 = 2/3. And remember, "I'm-a Luigi, number one!"

06-24-2009, 05:03 PM	#8
*CaptHayfever* Smash SuperMod Join Date: Jul 2002 Location: (n) - the place where I am Gender: Posts: 27,659 Thanks: 1,991 Thanked 2,486 Times in 1,513 Posts	^The entire premise of Monty Crawl is that Monty goes to the closest door that isn't the car. If it's behind #3, he'll open #1. And remember, "I'm-a Luigi, number one!"

06-24-2009, 10:28 PM	#10
Deku Trii PP&R Mod Current Events Mod Join Date: Jun 2000 Location: Texas Gender: Posts: 9,031 Thanks: 521 Thanked 1,099 Times in 583 Posts	Spoiler Below Let's say you picked door 1. The probabilities that he will pick door number 2 if the car is behind door #1, #2, or #3 are 100%, 0%, and 100%. The probabilities that he will pick door #3 if the car is behind door #1, #2, or #3 are 0%, 100%, and 0%. If the host opens door 2, there is a 50% chance you can win by switching. If he opens door 3, there is a 100% chance you will win by switching. There is a 1/3 chance he will open door 3, and a 2/3 chance he will open door 2. 1/31+1/22/3=2/3

06-29-2009, 07:09 PM	#11
CuccoLady Anime and Cartoons Mod Join Date: Jun 2006 Location: THIS LOCATION REMINDS ME OF A PUZZLE, LUKE Gender: Posts: 9,077 Thanks: 2,825 Thanked 1,222 Times in 822 Posts	@ Cap'n: I didn't get the Monty Hall solution at all either, but then I looked it up and it was explained and made much more sense. Spoiler Below There are three possibilities: -you pick the car. The host reveals a goat, you switch and pick the other goat and lose. -you pick a goat. The host reveals the other goat, you switch and pick the car and win. -you pick the other goat. The host reveals the other goat, you switch and pick the car and win. In other words, if you decide to switch, picking the car is the only way to lose... if you pick either goat the first time, you will win if you switch. That is the only explanation that actually made sense to me.