The Monty Hall Paradox--eggheads explain?

Gillette

Sexy Member
Joined
Apr 2, 2006
Posts
6,214
Media
4
Likes
95
Points
268
Age
53
Location
Halifax (Nova Scotia, Canada)
Sexuality
100% Straight, 0% Gay
Gender
Female
Try this rationalization:

Taking away one option (e.g. revealing 1 goat) doesn't change the fact that your first choice had only a 33% chance of being correct, and is STILL only 33% likely to be correct.

Still likely to be 33% correct out of three options.

Once the third option is removed the chance is 50%. It's not possible to have a 1:2 probability ratio of two random options.

If I asked you to guess if I were a blonde, brunette or redhead, but told you that I am not a blonde, what are your chances of guessing my hair colour correctly?
 

Gillette

Sexy Member
Joined
Apr 2, 2006
Posts
6,214
Media
4
Likes
95
Points
268
Age
53
Location
Halifax (Nova Scotia, Canada)
Sexuality
100% Straight, 0% Gay
Gender
Female
The door you originally picked has a 66% of being the wrong door. Once there are two doors left, the odds that you picked the right door from the beginning goes down. Therefore switching will increase the odds of winning on average, and in the long run, but not every time.

Sorry, if the door you picked is not the door he opens that means your odds go up. If this weren't true the scientific process of elimination goes straight out the window. What the hell are "controlled" variables for if not that?
 

rob_just_rob

Sexy Member
Joined
Jun 2, 2005
Posts
5,857
Media
0
Likes
43
Points
183
Location
Nowhere near you
Still likely to be 33% correct out of three options.

Once the third option is removed the chance is 50%. It's not possible to have a 1:2 probability ratio of two random options.

If I asked you to guess if I were a blonde, brunette or redhead, but told you that I am not a blonde, what are your chances of guessing my hair colour correctly?

It's not the same question, Gillette, because there are three different results, not 2.
 

Guy-jin

Legendary Member
Joined
Oct 1, 2007
Posts
3,836
Media
3
Likes
1,368
Points
333
Location
San Jose (California, United States)
Sexuality
Asexual
Gender
Male
Still likely to be 33% correct out of three options.

Once the third option is removed the chance is 50%. It's not possible to have a 1:2 probability ratio of two random options.

If I asked you to guess if I were a blonde, brunette or redhead, but told you that I am not a blonde, what are your chances of guessing my hair colour correctly?

Problem is, you're assuming the second choice is separate from the first choice, and it isn't. That's why you can boost your odds by always switching the second time.
 

Gillette

Sexy Member
Joined
Apr 2, 2006
Posts
6,214
Media
4
Likes
95
Points
268
Age
53
Location
Halifax (Nova Scotia, Canada)
Sexuality
100% Straight, 0% Gay
Gender
Female
It's not the same question, Gillette, because there are three different results, not 2.

Help me, please, because I'm not seeing the difference.

Three options are given initially then one is removed leaving only two options. Out of two options the random chance of guessing correctly should be 50%.

I understand what the probability was but now with one option eliminated it is a different question and a different probability.
 

Gillette

Sexy Member
Joined
Apr 2, 2006
Posts
6,214
Media
4
Likes
95
Points
268
Age
53
Location
Halifax (Nova Scotia, Canada)
Sexuality
100% Straight, 0% Gay
Gender
Female
Problem is, you're assuming the second choice is separate from the first choice, and it isn't. That's why you can boost your odds by always switching the second time.

How is it not separate? It is a new choice with a different number of options, hence a different probability.
 

rob_just_rob

Sexy Member
Joined
Jun 2, 2005
Posts
5,857
Media
0
Likes
43
Points
183
Location
Nowhere near you
Help me, please, because I'm not seeing the difference.

Three options are given initially then one is removed leaving only two options. Out of two options the random chance of guessing correctly should be 50%.

Not if you already made a choice before one option was removed.

Again, imagine there being 10 doors and 8 of them being opened revealing the goat.
 

Guy-jin

Legendary Member
Joined
Oct 1, 2007
Posts
3,836
Media
3
Likes
1,368
Points
333
Location
San Jose (California, United States)
Sexuality
Asexual
Gender
Male
Help me, please, because I'm not seeing the difference.

Three options are given initially then one is removed leaving only two options. Out of two options the random chance of guessing correctly should be 50%.

The choice of which door to remove is not random. It is always going to be a booby prize.

To rework your example to be equivalent, you would have to have two choices, say blonde and brunette. I'd tell you there are blondes behind two doors and a brunette behind the third.

Then I ask you to find the brunette.

You pick one, I must choose to open a door you didn't choose that has a blonde behind it or else I give it away.

See the difference? It isn't like I randomly eliminated a door. I had to eliminate a door depending on what your initial choice was. That's why they're not independent.
 

jason_els

<img border="0" src="/images/badges/gold_member.gi
Joined
Dec 16, 2004
Posts
10,228
Media
0
Likes
163
Points
193
Location
Warwick, NY, USA
Sexuality
90% Gay, 10% Straight
Gender
Male
And that's how the game is won by switching. Because you have to wait until after Monty has increased your odds by getting rid of one of the doors that isn't the car and letting you choose again.

If you stick by your original choice, you're still playing as if you had a one in three chance, and you're not upping your chance of winning at all.

Ever play craps?

But because you didn't choose door three, that means you've already chosen one of two to be correct because three was eliminated for you. Left with one or two, your choice is just as valid:

Let's take two different parallel universes here. You and your doppleganger are playing this game at the same time and in each turn the car is behind the same door in both universes. Every time you and your doppleganger choose, the first door never reveals the car so you get the option to switch. If, when each of you is asked to stay or switch, each chooses to switch, and You1 always chooses a different door from You2, both should of you should tend to win more than the other over many chances though that should be impossible.
 

Mem

Sexy Member
Joined
Jul 4, 2006
Posts
7,912
Media
0
Likes
54
Points
183
Location
FL
Sexuality
99% Gay, 1% Straight
Gender
Male
Take an Ace card and two Jack cards.

Ask your friend to set them down and remember where one Jack card is. Either one, it doesn't really matter.

Pick one card face down. Remember it's position.

Tell your friend to show you where one of the Jacks are and turn it over.

(You are looking for the Ace.)

Switch your original choice. More times than not the card you switched to will be the Ace.

Try it 50 times switching and tell me how many times the Ace comes up.

Try it 50 times by keeping your original choice, and tell me how many times the Ace comes up.

The law of percentages says that if you switch your choice you will find the Ace more often.
 

rob_just_rob

Sexy Member
Joined
Jun 2, 2005
Posts
5,857
Media
0
Likes
43
Points
183
Location
Nowhere near you
Let's take two different parallel universes here. You and your doppleganger are playing this game at the same time. Every time you and your doppleganger choose, the first door never reveals the car so you get the option to switch. If, when each of you is asked to stay or switch, each chooses to switch, and You1 always chooses a different door from You2, both should of you should tend to win more than the other over many chances though that should be impossible.

Why should it be impossible? You know that your original choice has only a 33&#37; chance of being correct. Logically, the other remaining choice is at 67%.

If I gave you a 10-question true or false test and then handed you a sheet with the correct answers to the first 5 questions, of course you'd get more than half correct, on average.
 

Guy-jin

Legendary Member
Joined
Oct 1, 2007
Posts
3,836
Media
3
Likes
1,368
Points
333
Location
San Jose (California, United States)
Sexuality
Asexual
Gender
Male
But because you didn't choose door three, that means you've already chosen one of two to be correct because three was eliminated for you. Left with one or two, your choice is just as valid:

Let's take two different parallel universes here. You and your doppleganger are playing this game at the same time. Every time you and your doppleganger choose, the first door never reveals the car so you get the option to switch. If, when each of you is asked to stay or switch, each chooses to switch, and You1 always chooses a different door from You2, both should of you should tend to win more than the other over many chances though that should be impossible.

No, actually. You1 and You2 will, after doing it an infinite number of times, have both won 66% of the time. Because some of the time, they will end up picking the same door.

If you throw in the facet that they always pick different doors in the first round, then you're adjusting the probabilities and making it so that the two people's choices aren't independent, and you'd have to recalculate the odds.
 

jason_els

<img border="0" src="/images/badges/gold_member.gi
Joined
Dec 16, 2004
Posts
10,228
Media
0
Likes
163
Points
193
Location
Warwick, NY, USA
Sexuality
90% Gay, 10% Straight
Gender
Male
Still, given the same deck, same lay of cards, but two different people making the choices, then both should outwin the other. For whatever reason, I always choose the far right card, Gillette always chooses the far left card. We always choose to switch. Then we both outwin each other and that can't happen.

Take an Ace card and two Jack cards.

Ask your friend to set them down and remember where one Jack card is. Either one, it doesn't really matter.

Pick one card face down. Remember it's position.

Tell your friend to show you where one of the Jacks are and turn it over.

(You are looking for the Ace.)

Switch your original card. More times than not the card you switched to will be the Ace.

Try it 50 times switching and tell me how many times the Ace comes up.

Try it 50 times by keeping your original choice, and tell me how many times the Ace comes up.
 

Gillette

Sexy Member
Joined
Apr 2, 2006
Posts
6,214
Media
4
Likes
95
Points
268
Age
53
Location
Halifax (Nova Scotia, Canada)
Sexuality
100% Straight, 0% Gay
Gender
Female
But once the door is opened your choices are diminished to two, just as in my example one option was eliminated before my hair colour was guessed.

Because he will always open a door showing the booby prize means realistically that no matter what you chose you always had a 50&#37; chance.

The game is played the same every time, correct?
One incorrect option is always removed, yes?

What that means is that even though you iniatially had three doors to choose from your end choice will have been, from the beginning, a 50/50 chance.
 

Guy-jin

Legendary Member
Joined
Oct 1, 2007
Posts
3,836
Media
3
Likes
1,368
Points
333
Location
San Jose (California, United States)
Sexuality
Asexual
Gender
Male
Still, given the same deck, same lay of cards, but two different people making the choices, then both should outwin the other. For whatever reason, I always choose the far right card, Gillette always chooses the far left card. We always choose to switch. Then we both outwin each other and that can't happen.

Then it's not random, and the events are not independent.

If they're independent, after an infinite number of times, you guys will both win the same number of times because you'll pick the same one some of the time.
 

rob_just_rob

Sexy Member
Joined
Jun 2, 2005
Posts
5,857
Media
0
Likes
43
Points
183
Location
Nowhere near you
Still, given the same deck, same lay of cards, but two different people making the choices, then both should outwin the other. For whatever reason, I always choose the far right card, Gillette always chooses the far left card. We always choose to switch. Then we both outwin each other and that can't happen.

That's not the same game, though. You're comparing apples to Audis.
 

Quite Irate

Sexy Member
Joined
Apr 4, 2007
Posts
701
Media
34
Likes
26
Points
248
Sexuality
50% Straight, 50% Gay
Jason, you suffer from a delusion known as "reality" - mathematics isn't about reality by its very nature. Having two options doesn't mean that there's a 50/50 chance that either is right. In the real world, it does. Not in math games, though.
This is the folly of the explanations typically given:
the probability involving the two real items behind the final choices isn't the same thing as the probability of picking the correct door. You're not influencing the probability of the doors themselves, you're influencing the probability of choosing a door. Shit, that didn't come out like I wanted it to. Let's keep going - One door is more equal than the other. Your problem is that you think the 1/2 probability of what's behind the doors is the percentage in use. It's not. It's secondary.


Your initial "uninformed"decision to pick a door gives you a one in three shot at getting it right. The unveiling of the goat doesn't change what's behind the doors, yes. There's still a goat, and still a car. The idea that you are betting on the objects is a logical fallacy. You are betting on the likelihood that the door you initially chose was correct. You have to go at this in a pure math (rather, theoretical) sense. The initial probability IS one third. You are impartial to every variable that might influence your initial choice. After the first goat door is shown, you are betting not that the likelihood of one door holding the car is greater than the other, but that your choice is now more informed than your initial one. The probability, if going at those two doors without any past experience in the matter, that you get the car is 1/2. Not the case if you started with three doors. Your choosing of door A actually *changes* the likelihood of that door holding the car, mathematically. How's that for chaos theory, eh?

Now, these numbers that people are throwing around for percentages *are not* case by case numbers. Going back to what I said earlier - if you have a single run at a percentages game, the mathematical values and possibilities will likely seem less apparent in the outcome than if you played that same percentages game 500 times. Making the switch in this game isn't particularly influential if you're only going one round. Ever flip a coin and keep landing on the same side? This stat stuff isn't that reliable for small numbers. You could play this game several times, and the benefit of switching versus not switching wouldn't make itself apparent to you. However, in general it does increase the likelihood of winning the prize.


As you can see, I'm not the greatest teacher. I do understand the math behind, though, but my difficulty lies in representing the outcome in non-math terms. I hope this makes sense to you. If you don't think the first part I mainly scratched doesn't meld with the second part, clear it from your mind. My wording wasn't accurate.
 

rob_just_rob

Sexy Member
Joined
Jun 2, 2005
Posts
5,857
Media
0
Likes
43
Points
183
Location
Nowhere near you
But once the door is opened your choices are diminished to two, just as in my example one option was eliminated before my hair colour was guessed.

Because he will always open a door showing the booby prize means realistically that no matter what you chose you always had a 50% chance.

The game is played the same every time, correct?
One incorrect option is always removed, yes?

What that means is that even though you iniatially had three doors to choose from your end choice will have been, from the beginning, a 50/50 chance.

You can't go by the end choice without considering the initial choice. The inference drawn from your initial choice makes your final choice a 2:1 chance.
 

jason_els

<img border="0" src="/images/badges/gold_member.gi
Joined
Dec 16, 2004
Posts
10,228
Media
0
Likes
163
Points
193
Location
Warwick, NY, USA
Sexuality
90% Gay, 10% Straight
Gender
Male
Not really. You can do this all on your own with 2 people. Give each the chance to switch and they always both switch, then both should always win more than the other if they're playing the exact same game in spacetime as the other. Pretend Monty had 2 separate contestants in isolation booths. Neither pick door 3 the first time, then both get the chance to switch. Gillette loves odd numbers, mem loves even numbers so they always initially choose their respective favorite doors, 1 and 2 respectively. Each might choose to always switch and, if they do, then it shouldn't matter what door they choose initially so long as it's the wrong door because the act of switching in the second round is what increases the chances, not the door (according to the theory).

No, actually. You1 and You2 will, after doing it an infinite number of times, have both won 66% of the time. Because some of the time, they will end up picking the same door.

If you throw in the facet that they always pick different doors in the first round, then you're adjusting the probabilities and making it so that the two people's choices aren't independent, and you'd have to recalculate the odds.
 

Guy-jin

Legendary Member
Joined
Oct 1, 2007
Posts
3,836
Media
3
Likes
1,368
Points
333
Location
San Jose (California, United States)
Sexuality
Asexual
Gender
Male
But once the door is opened your choices are diminished to two, just as in my example one option was eliminated before my hair colour was guessed.

Because he will always open a door showing the booby prize means realistically that no matter what you chose you always had a 50% chance.

The game is played the same every time, correct?
One incorrect option is always removed, yes?

What that means is that even though you iniatially had three doors to choose from your end choice will have been, from the beginning, a 50/50 chance.

That's the fallacy most people ended up buying, including myself at first. It seems like it's 50/50 because it seems like the first choice doesn't matter, but it does.

Because the first choice only gives you a 33% chance of picking the car. If you pick the car the first time, Monty's choice is unaffected, and then your decision to switch will always end up with you losing when you picked the car initially.

So you'll lose 33% of the time.

The first choice also gives you a 66% chance of choosing the booby prize. If you pick the booby prize the first time, Monty always shows you the other booby prize, then you switch and you always end up winning when you picked the booby prize initially.

So you'll win 66% of the time.

If you always switch.

Got it?