[zippy2006]

Originally Posted by csuguy

I disagree - it wouldn't be 4/52, I would have an 80% chance of winning

Would you like to gamble?

What you don't understand is that the dealer is never going to show you a jack. There is no chance whatsoever that he will ever turn up a jack. You are being manipulated. It's like watching the rigging of the deck in front of your eyes. In the correct analogy the dealer isn't randomly turning up cards, he is only turning up non-jack cards.

Just so, the game show host does not randomly pick a curtain to reveal, he only reveals the non-car curtain!

[zippy2006]

zip's the dealer:

CSU picks the Ace of Clubs. Zip turns over every card except the 4 jacks. CSU loses

CSU picks the Ace of Spades. Zip turns over every card except the 4 jacks. CSU loses

CSU picks the Ace of Hearts. Zip turns over every card except the 4 jacks. CSU loses

CSU picks the Ace of Diamonds. Zip turns over every card except the 4 jacks. CSU loses

CSU picks the two of Clubs. Zip turns over every card except the 4 jacks. CSU loses

.
.
.

CSU picks the jack of Clubs. Zip turns over every card except for 3 jacks and an ace. CSU wins

.
.
.



This is an exhaustive case-by-case proof that you will lose 48/52 times and I will be rich

[csuguy]

Originally Posted by zippy2006

Would you like to gamble?

What you don't understand is that the dealer is never going to show you a jack. There is no chance whatsoever that he will ever turn up a jack. You are being manipulated. It's like watching the rigging of the deck in front of your eyes. In the correct analogy the dealer isn't randomly turning up cards, he is only turning up non-jack cards.

Just so, the game show host does not randomly pick a curtain to reveal, he only reveals the non-car curtain!

It doesn't matter that the dealer/host doesn't reveal any of the winning choices, so long as he doesn't remove the winning choices (removing jacks without telling you they did, or never putting the proper number of jacks in there). His removal of non-winning choices increases the odds of your getting a winning choice.

[csuguy]

Originally Posted by zippy2006

zip's the dealer:

CSU picks the Ace of Clubs. Zip turns over every card except the 4 jacks. CSU loses

CSU picks the Ace of Spades. Zip turns over every card except the 4 jacks. CSU loses

CSU picks the Ace of Hearts. Zip turns over every card except the 4 jacks. CSU loses

CSU picks the Ace of Diamonds. Zip turns over every card except the 4 jacks. CSU loses

CSU picks the two of Clubs. Zip turns over every card except the 4 jacks. CSU loses

.
.
.

CSU picks the jack of Clubs. Zip turns over every card except for 3 jacks and an ace. CSU wins

.
.
.



This is an exhaustive case-by-case proof that you will lose 48/52 times and I will be rich

You forgot a lot of cases in your example. For example:


CSU picks a jack. Zip turns over every card except the 3 jacks and the Ace of Spades. CSU wins

CSU picks a jack. Zip turns over every card except the 3 jacks and the Ace of Diamonds. CSU wins

CSU picks a jack. Zip turns over every card except the 3 jacks and a the 2 of Hearts. CSU wins

.
.
.

CSU picks a jack. Zip turns over every card except the 3 jacks and the 5 of Clubs. CSU wins


So you see, you are incorrectly modeling the situation.

[zippy2006]

Originally Posted by csuguy

You forgot a lot of cases in your example. For example:


CSU picks a jack. Zip turns over every card except the 3 jacks and the Ace of Spades. CSU wins

CSU picks a jack. Zip turns over every card except the 3 jacks and the Ace of Diamonds. CSU wins

CSU picks a jack. Zip turns over every card except the 3 jacks and a the 2 of Hearts. CSU wins

.
.
.

CSU picks a jack. Zip turns over every card except the 3 jacks and the 5 of Clubs. CSU wins


So you see, you are incorrectly modeling the situation.

No that's just bad probability. I am only saying that every time you pick a card, I will show you 48 non-jack cards because I know what each of the cards is.

I did conflate a bit and zoo brought this up on the first page.

If the dealer knows what each card is and purposely turns up non-jacks then you are clearly wrong, you will lose badly.

If the dealer doesn't know what any of the cards are and could just as possibly turn up jacks as non-jacks then it is a different story.

Either way the Monty Hall problem is equivalent to one where the dealer "knows" because the only possibility is that the blank curtain is shown.

[csuguy]

Originally Posted by zippy2006

No that's just bad probability. I am only saying that every time you pick a card, I will show you 48 non-jack cards because I know what each of the cards is.

I did conflate a bit and zoo brought this up on the first page.

If the dealer knows what each card is and purposely turns up non-jacks then you are clearly wrong, you will lose badly.

If the dealer doesn't know what any of the cards are and could just as possibly turn up jacks as non-jacks then it is a different story.

Either way the Monty Hall problem is equivalent to one where the dealer "knows" because the only possibility is that the blank curtain is shown.

It doesn't matter if the dealer knows that he is only revealing incorrect choices - since his knowledge of whether I picked the right card or not is irrelevant to my chances of getting the right card.

[zippy2006]

Originally Posted by csuguy

It doesn't matter if the dealer knows that he is only revealing incorrect choices - since his knowledge of whether I picked the right card or not is irrelevant to my chances of getting the right card.

...and your chances of getting the right card are 4/52, which is exactly where they remain.

The only reason you think your chances are improving is because you think he is randomly turning up cards. Really he is not. Really he is just manipulating you. Honestly come over to my house and we will gamble. Whenever you pick a non-jack you will lose, and whenever you pick a jack you will win. Nothing else matters. Every single time you will see 48 non-jacks face up.

Refer to my proof. I am a deterministic dealer who leaves 4 jacks down whenever you pick a non-jack and leaves 3 jacks and an ace of clubs down whenever you pick a jack. You see the exact same thing every time (48 cards up) but only win 4/52 times. This is because I am purposely turning up non-jacks. Even if I put a random card with the 3 jacks it wouldn't increase your chance of winning- you would still only pick a jack 4/52 times.

[csuguy]

Originally Posted by zippy2006

...and your chances of getting the right card are 4/52, which is exactly where they remain.

The only reason you think your chances are improving is because you think he is randomly turning up cards. Really he is not. Really he is just manipulating you. Honestly come over to my house and we will gamble. Whenever you pick a non-jack you will lose, and whenever you pick a jack you will win. Nothing else matters. Every single time you will see 48 non-jacks face up.

Refer to my proof. I am a deterministic dealer who leaves 4 jacks down whenever you pick a non-jack and leaves 3 jacks and an ace of clubs down whenever you pick a jack. You see the exact same thing every time (48 cards up) but only win 4/52 times. This is because I am purposely turning up non-jacks. Even if I put a random card with the 3 jacks it wouldn't increase your chance of winning- you would still only pick a jack 4/52 times.

Thinking it over, I will submit you are correct on this scenario. But I argue it is fundamentally different from the other scenario since I lack the ability to switch my choice.

The scenario above in the case of the doors would be like so: i pick a door, he removes one and I don't get to change. Thus the odds of being correct have not changed since I have no further choice, rather he will simply continue to reveal the other two doors as well.

[Ask Mr. Religion]

Originally Posted by csuguy

Thinking it over, I will submit you are correct on this scenario. But I argue it is fundamentally different from the other scenario since I lack the ability to switch my choice.

The scenario above in the case of the doors would be like so: i pick a door, he removes one and I don't get to change. Thus the odds of being correct have not changed since I have no further choice, rather he will simply continue to reveal the other two doors as well.

This is not a new problem. Well described in the link below (sometimes Wikipedia gets things right):

Monty Hall Problem

Us older geezers have remembered this from the Vos Savant brouhaha in Parade Magazine related to the scenario and already alluded to in some of the posts above.

AMR

[zoo22]

Originally Posted by zoo22

There's an important part to the question that (I think) you've missed. But whether it was intentional or not, everyone who's answered "change" has been wrong. It doesn't matter.

For the opportunity to switch based on what the host reveals to you to matter, the host must know what's behind the curtains. If that's not clarified, it makes no difference towards your odds what you do.

If the host knows what's behind the curtain he opens for you, then you should switch; your odds go from 1/3 to 2/3. If he doesn't know, you're being offered a coin flip.

Hi zippy,

Yes, it does matter if that the host knows what's behind the curtains.

Originally Posted by zippy2006

It's hard to think about on a step-by-step level. But 2/3 of the time you will pick the wrong curtain, then a wrong curtain is revealed and you can switch to the remaining one if you want. Since 2/3 of the time you were wrong, switching will get you to being right 2/3's of the time.

I agree that this is (obviously) a difficult puzzle to consider. Once you know the answer to the question ("switching increases your odds from 1/3 to 2/3," answering the question that clarifies that the host does know what's behind the curtains), it makes sense, but it's very counter-intuitive. By the same token, once you do understand it, making the jump to reconsider the answer if the host doesn't know what's behind the doors is just as counter-intuitive.

They're two different games, and they have different odds. To determine your answer(s), you have to consider the game you're playing, what can happen, what the various outcomes are.

If it's specified that the host does know what's behind the curtains, it means he's intentionally showing you an empty curtain.* There's only one thing that's going to happen when he makes his (non-random) choice: he'll reveal an empty curtain to you. There's no chance that you will lose based on his choice. He will only help you.

But if the host doesn't know what's behind the curtains, there are additional possible (random) outcomes, so the odds are different. If you choose the curtain with the car, he has zero chance of revealing the car when he randomly opens one of the two remaining curtains. If you chose an empty curtain, his chances are 50/50. His random choice could reveal the car and you lose right then and there. In this scenario, there's a chance you will lose based on his choice. He might help you, and he might hurt you.

A way to try reinterpret it so you can see how they are two different games is to consider that if the host doesn't know what's behind the curtain, he's actually "playing" in the game too. You're trying to pic a car; he's trying to pick an empty curtain.

Here's a simplified way to look at it using cards (this doesn't follow exactly with the three-curtains, but I think it might help):

(NOTE: I wrote this part out before seeing the long new string of card posts. But I'm posting it anyway)

- A full deck of shuffled cards. You want the Ace of Spades (of course).

- You choose a random card (you don't look). The chance you hold the AoS is 1/52.

- 51 cards left in the deck.

- The dealer now removes 47 cards that are not the AoS from those 51 cards.

- 4 cards are left.

- The chance of drawing the AoS from the remaining four cards is now 1/5 (we still don't know what you drew).

- You're holding a 1/52 chance; you're looking at a 1/5 chance.

BUT: If the dealer removes 47 random cards from the deck, the chance of drawing the AoS from those remaining 4 cards is 1/52. The odds haven't changed. Because the dealer didn't know what cards they were removing, they could have removed the AoS, and you've already lost.

You're holding a 1/52 chance, and you're looking at a 1/52 chance.

The host has to know what's behind the curtains for the switch to increase your odds.

Originally Posted by rstrats

zoo22,

re: “For the opportunity to switch based on what the host reveals to you to matter, the host must know what's behind the curtains. If that's not clarified, it makes no difference towards your odds what you do.”

You’re going to have to explain why the problem needs to include the statement that the host knows where the car is. The host only needs to know where the car is so that he doesn’t prematurely end the game.

Hi rstrats, I'd meant to comment that I'm glad you'd posted this thread. I'd actually just posted the question a couple of weeks ago (I got no responses ) ...I almost didn't respond, but then when I read the question, I wanted to.

Anyway, I hope the above explains it a bit. In response to your specific comment: It's not simply a matter of the game prematurely ending. What's important is that if the host "prematurely ends the game" with his random choice, you lose. Without the added clarification, there's additional randomness, additional possible outcome to consider, and because of that, the two scenarios have different odds. (Also see my related comment below).










*I'll point out that yes, even if it's clarified in the question that the host knows what's behind the curtains, he could still be making a random choice (he could have flipped a coin, or done any number of different things to determine his curtain decision). We don't now. But I think that would be mincing words. I believe that by specifying that the host knows what's behind the curtains is enough to imply that he is purposely revealing an empty curtain.

The reason that I don't also believe that "the question needs to clarify that the host knows what's behind the curtains" is mincing words is because people often don't consider that the two scenarios are specifically different, with different outcomes and odds, and that it's not as "simple" as always switching to increase odds.

[Spitfire]

This is still being debated?

Originally Posted by csuguy

You are incorrectly modeling it. It should be as so:

Picked door is correct and you stay: win
Picked door is correct and you don't stay: lose

Picked door is incorrect and you stay: lose
Picked door is correct and you don't stay: win

50-50

Here's why Zippy's model is correct, actually:

The host will have to open a door (or raise a curtain or whatever) that has nothing behind it before offering you the chance to switch.

If you chose the right door, the host will be able to open either of the other two remaining doors and show there is nothing behind it. There is a 1 in 3 chance of this.

However, if you did not choose the right door, the host will be forced to open the one of the two other doors that has nothing behind it. The chances of this are 2 in 3! Meaning 2 times in 3 the door the host doesn't open but offers you the chance to switch to will be the right door.

[Nydhogg]

Spitfire is right.


Damn it. A Catholic out-gambling a Heathen!
That can't be!


*hides, ashamed*

[Spitfire]

I never had to figure it out for myself, however. This riddle was mentioned in a book I was reading back in 2005, along with an explanation of how it works.

[Selaphiel]

The error people make is separating the events. Those who say it is 50/50 is not solving the Monty Hall problem, they are basically solving an event where there are two doors and you just pick one of them.

You must view the thing as one event, then it makes sense. Think of it this way:

To keep the door is the same as having one choice and when you make your first choice there are 3 doors which means there is 1/3 chance that you chose correctly and 2/3 chance that you chose wrong. Then the host reveals one of the doors you did not choose, leaving one. There is still a 2/3 chance that your first choice was wrong which means there is a 2/3 chance to win if you switch versus a 1/3 chance to win if you stay. The revealing does not affect the odds, the only thing that has changed is that the door you did not choose now carries the entire 2/3 chance.

Visually:

Let us say I initially choose A. Then A has a 1/3 chance of being correct and B AND C has a 2/3 chance.

[A]---------[B C]

1/3-----------2/3

Host opens C. And asks if I want to change:

[A]---------[B C]

1/3------------2/3

With C removed, B is the only part left of the set that makes up the 2/3 chance group. Which means there is one door carrying the 2/3 chance and one with 1/3 chance. If C was correct, the host would open B (Remember he can not open A even if it is wrong, he must open the part of [BC] that is wrong).

[rstrats]

zoo22,

re: "Yes, it does matter if that the host knows what's behind the curtains."

It makes absolutely no difference if the host does or doesn’t know where the car is other than the game being prematurely over if he picks the curtain with the car. As long as the game continues, though, you should always switch. You can prove this by doing the experiment suggested by The Graphite, .