I'm sure most everybody knows this one. A contestant on a game show has to choose between three doors. Behind one is a car, and the other two, goats. When the contestant has chosen, the host reveals the first remaining door that has a goat behind it, because the contestant may have chosen a goat OR the car. Then the host asks the contestant if he/she wishes to change his/her mind. The question is... is it more, less, or equally probable that the contestant's choice is right versus changing his/her mind?

The contestant's chance of picking the car the first time is 1 in 3. The contestant's chance of picking a goat the first time is 2 in 3. If the contestant picked a goat the first time (2/3 chance), then they'll win the car if they do switch doors. If the contestant picked the car the first time (1/3 chance), then they'll win the car if they don't switch doors. Thus the contestant is twice as likely to win the car by switching doors.

Imagine a more extreme case. Suppose there are a million doors; one has a car behind it, and the rest have goats. After you pick your door, the host reveals 999,998 doors with goats behind them, leaving two doors closed, including the one you picked, and offering you the chance to switch doors. Your chance of picking the car the first time is 1 in 1,000,000. Your chance of picking a goat the first time is 999,999 in 1,000,000. If you picked a goat the first time (999,999/1,000,000 chance), then you'll win the car if you do switch doors. If you picked the car the first time (1/1,000,000 chance), then you'll win the car if you don't switch doors. So you're 999,999 times more likely to win the car by switching doors.

The more doors there are, the more likely you are to win by switching doors. However, even starting with just three doors, you're more likely to win by switching. In a single game with three doors, though (like on the game show), I would probably just go with my instincts, since a probability like 2:1 really doesn't carry a lot of weight for one event; it could easily go the other way. Probabilities are much more useful for a chain of multiple events, to maximize your winnings in the long run.
++++++++++++++++++++++++
08-29-02, 12:32 AM
Pin~Jinx
Now that one door is revealed,
only two options are left.

So, the probability that the player picks the right or the wrong door is 50-50.

If you want me to give it a statistical look, then...

1/2

Aaahh.....so I say it is equally probable that the the contestant's choice is right versus changing his/her mind.

Pin~Jinx roll eyes

08-29-02, 01:41 AM
silenteuphony
I respectfully disagree, Pin~Jinx. By your logic, if the host reduces the possibilities to two doors including the winning door, you have a 50% chance of winning whether you switch or stick with the same door, no matter how many doors you start with. So let's go back to my million-door scenario.

Whether you guess right or wrong, the host will always pick two doors which include the winning door, since the host knows which door is the winner. The host picking these two doors has no effect on your odds of picking the right door in the first place. So your odds of winning if you stick with your original guess are exactly the same as picking the right door in the first place: 1 in 1,000,000.

If we use your logic, then your odds of winning if you stick with your original door would be 50%. This means if you played the game 1,000,000 times and always stuck with your original choice, you would win 500,000 times. But this won't work unless your original guess was correct in the first place, which means you picked the 1 correct door out of 1,000,000 doors 50% of the time! Clearly, this is a contradiction, unless you're psychic.

Your odds of winning are completely determined by your odds of picking the right door in the first place, since this determines how the host selects the second door. If your original choice is wrong, then the second door is not random; it must be the winning door. So the odds of the second door being the winner are exactly the same as your odds of guessing wrong the first time, which are better than your odds of guessing right, whether there are three doors or a million. So, by the odds, you should always pick the second door.

08-29-02, 09:26 AM
FlyingHellfish
Good explanation, SE.

The general rule is that you should always choose the other door once you're presented with more information.

08-30-02, 07:57 AM
Byter
I agree with Pin~Jinx on this one.

After a wrong guess the first time where your odds were 1/3
you are now presented with a situation where the odds are now
1/2 of winning the car.

In effect there are two games going on here and what has happened
earlier has absolutly no effect on the outcome of the second.

If you take both together your chances of winning the car are going to
be 2/3, no matter wheather you change you pick or not.

Regards Mike b (AKA Byter)

08-30-02, 09:29 AM
FlyingHellfish
Here's another way of looking at it. There is actually a small bit of strategy involved for the contestant, so we'll assume that he makes the smartest decision. This is not a game of pure luck.

Start with the scenario that the contestant's first guess is the door with the car behind it. There's a 1/3 chance of that. Obviously, the host will show a door with a goat behind it, and there's a goat behind the last door. If the contestant switches, he loses, but if he stays, then he wins. Therefore, in this scenario, it would be best if he stayed.

On the other hand, what if the contestant's first guess was a door with a goat behind it? There's a 2/3 chance of that. Now, of the 2 doors left, one has a goat behind it, and the other has the car. The host HAS to show the door with the goat behind it, so the last door has to have the car. In this setup, if the contestant sticks with his first guess, he loses, but if he switches, then he wins.

Analyzing these 2 situations, 1/3 of the time (the first setup) it would be best for the contestant to stay with his original guess, while 2/3 of the time, it would be best for the contestant to switch. Since we don't know if the contestant is right or not after the first guess, he should go with the odds (2/3 to 1/3) and switch.

08-30-02, 10:10 AM
FlyingHellfish
EDIT to my post:

Oops...I didn't see that SE had already explained it this way. Well, it never hurts to see it twice smile

08-30-02, 12:06 PM
silenteuphony
I realize my answer is somewhat anti-intuitive, but probability often is, and my math is correct. Since there is still some dissension, perhaps an appeal to authority will help. Marilyn vos Savant answered this question the same way; she has a question-and-answer column in Parade magazine, and according to Parade, she's listed in the "Guinness Book of World Records Hall of Fame" for "Highest IQ."

Here is a link to a discussion of the problem, including her answer: The Monty Hall Problem. This discussion also makes the excellent point that Marilyn's answer (and mine) only work if the game show host is always required to follow the same procedure. In the real game show that inspired this problem, Monty Hall was allowed to change his procedure, which means to truly optimize your strategy you would need to analyze the host's behavior/motivation (or be psychic), not just rely on probability.

However, assuming he does always reveal a goat and offer the chance to switch, my answer is correct. If you'd like another expert opinion, here's a link to Dr. Math's answer. Dr Math is not actually one person, but a moderated question-and-answer forum in which mathematical experts from all around the country submit answers to questions, and an expert panel from Drexel university, the site host, reviews the answers and posts the best answers.

Whether you agree with me or not, you're in good company, since thousands of experts, including professors in mathematics, took both sides of the debate after seeing Marilyn's answer. The best way to settle this is probably to prove it for yourself. The second link includes computer simulations, so you can actually play the game and check the results against the predicted probability, or you can always recruit a friend and simulate the game yourself.

I just tried one of the computer simulations myself (The Let's Make a Deal Applet), always sticking with the same door, and out of 100 games, I won 26. This is much closer to the 1/3 chance I predicted than the 1/2 chance from the alternate solution.


08-30-02, 01:08 PM
Pin~Jinx
SE,
This is how I actually comprehended the simple facts. With not much time at my disposle, I prefer keeping things as simple and straight as possible.......(perhaps that is why I enjoy doing Math!)

I Frankly tell you that I didn't go through your part of problem, no matter how tempting it may be. As far as WOK's problem goes, I think my answer is valid enough.
(Aren't we forgetting something, this has been posted more of like a poll.....with options and all)

THNX Mike (a.k.a.Byter) for siding with me!

Take Carez ALL,

Pin~Jinx (a.k.a anarchist!) big grin
08-30-02, 05:07 PM
Byter
It looks as though you guys are correct about switching. razz razz razz razz razz razz

After looking at silenteuphony's links it's starting to make sense.

The only time you'll lose by switching is if you quessed the
correct door in the first place & your odds of doing this are
only 1/3.

I tried that applett and got 36/100 by staying & 65/100 by switching.

GREAT Question WOK !
I think I'll dust of my GWBASIC skills & try running this for a while.
Should be interesting after a million or so tries.

Will let you know the outcome if I can make it work.

Regards to all Mike (AKA Byter) red face

08-31-02, 12:00 PM
Byter
Well here are the results done in GWBASIC:
I only let it run to 1 million 4 times, watching the computer do this
was as fun as watching paint dry. roll eyes
These are the results of staying with your original choice.

328261 first run
328428 second run
328582 third run
328400 fourth run

These are a little lower then expected that I imagine is
caused by the way the computer generates random numbers
(not compleatly random)

Thanks again for an interesting question WOK



09-01-02, 09:37 AM
WiteoutKing
Perfect! That was exactly the response I was looking for. This is a very controversial topic to this day. It is understandable where Pin~Jinx is coming from, but let's look at it like this:

If the contestant picks the car (1/3), then the host reveals a goat and gives you the opportunity to switch (L) or stay (W).
If the contestant picks a goat (2/3), then the host reveals the OTHER goat and gives you the opportunity to switch (W) or stay (L).

Now, here's the thing. Both Pin~Jinx and s.e. are correct, but in different ways.

The odds of WINNING are in fact 1/2, as was said by Pin~Jinx... sort of.
However, the odds of winning BY SWITCHING are 2/3, as was said by s.e.
This is probably the most famous example of conditional probability.

09-01-02, 08:06 PM
silenteuphony
Forgive me for extending this thread, but being an old-school mathematician, I felt the need to pin down some terminology. In the previous post, WK said the odds of winning are 1/2, "sort of." At first I thought this didn't make sense, but it does make sense if the player's strategy is random.

That is, if the player switches doors half of the time and doesn't switch the other half of the time, then their overall odds of winning will be 1/2 in the long run (a series of games). This is true no matter how many doors you start with, as long as the host reduces the possibilities to two doors, including the winning one.

So, yes, in this sense both answers are correct, but I think the discussion was mainly centered on figuring the odds of winning by switching vs. winning by not switching. The odds of winning with a random strategy are a whole different question. So the 1/2 answer is correct, but only if you reinterpret the question.

It's interesting to note that the player who realizes they have a 2/3 chance of winning by switching will probably always switch and win 2/3 of the time. However, the player who believes they have a 1/2 chance of winning whether they switch or not may decide to use the random strategy (switch half the time, don't switch half the time), and in the long run, will win half the time just like they expect, but not for the reason they expect. Thus, to some extent, their belief about their odds could become a self-fulfilling prophecy.

[This message was edited by silenteuphony on 09-01-02 at 08:35 PM.]

09-02-02, 02:58 AM
Pin~Jinx
what
exactly
did you mean by

quote:
"...
It is understandable where Pin~Jinx is coming from..."



Anyhow!



Pin~Jinx

09-02-02, 09:07 AM
chanceygardner
Wooooooooooaaah....what is the host doing after opening one door which does not contain the car that increases your probability of picking correctly between 2 doors from 0.5 to 0.666 if you switch?

09-02-02, 12:10 PM
silenteuphony
What the host is doing is reducing your possible outcomes, or sample space, in the game. Suppose the host didn't reveal a goat behind one door, but left all three doors closed, then asked if you wanted to switch to either of the other two doors. Here, your odds are clearly 1/3 of winning if you don't switch.

When the host reduces the doors to two, he doesn't change the odds of your initial choice being right: they are still 1/3. But probability always adds up to 1, so the other 2/3 is now represented by the single remaining door.

So the host didn't change the odds from 1/2 to 2/3. A better way to think of it might be to say he started with the odds from three doors (your original decision) and changed the sample space (the number of doors) without changing the odds. The odds never changed; it was the sample space that changed.

Here's a different example. Suppose you buy a lottery ticket with a 1% chance of winning. Then a mysterious stranger who you somehow know is telling the truth shows you another lottery ticket and says that one of the two tickets (yours or his) will win. Their offer hasn't made you psychic; the odds of your ticket being a winner are the same as they always were: 1%. Therefore, since the probability adds up to 100%, the stranger's ticket (assuming they're telling the truth) has a 99% chance of winning.

In this situation, you would probably assume the stranger has inside knowledge and switch tickets. This is just like the game show. The host has inside knowledge, and is using this knowledge to home in on the winning door. He hasn't just picked two doors at random; he's picked two doors that he know include the winner, which is a more informed choice than you made originally.

09-02-02, 04:24 PM
FlyingHellfish
What you have to remember is that the 2 choices that the contestant is presented with-picking one door out of three, and then picking one door out of two-are not independent. At the second stage, it is not a 50/50 chance. The probability of winning at this point is dependent upon the contestant's first choice, and since the contestant's choice is variable, so is the probability immediately thereafter that he will win by switching doors. Once again, this is a game of strategy-the overall probability of winning has nothing to do with with what is behind which door. It has to deal with how well the contestant understands the probabilities involved. Assuming no other influences, a contestant who understands the probabilities will win 2/3 of the time, because he knows that 2/3 of the time he will win by switching. A contestant who thinks that there's no difference between either door will win 1/2 of the time because they assume that switching doors makes no difference, so that they will stay with their first choice some and they sometimes will switch. It is not possible to say at any given point what the probability of winning is because it depends on what the person knows.

This is similar to what is called a Markov Process, where current events (in this case, deciding whether or not to switch doors) are influenced by recent occurences (in this case, the contestant's first choice).

09-03-02, 06:42 AM
WiteoutKing
First off, Pin~Jinx, what I meant was I could see where you got your answer. When I said "sort of" I meant that the second stage alone would be a 50/50 shot. However, here we have conditional probability. Let's simplify this problem. On the first pick, if you choose the car, the other door will have a goat. If you pick a goat, the other door will have the car. Since the odds are 1/3 for picking the car first and 2/3 for picking either goat first, the odds that you were right first is 1/3, so the odds you need to change is 2/3.

09-03-02, 09:12 AM
chanceygardner
Noooooooooooooo.....this solution is only correct IF:

1. The host always knows which door contains the car.
2. The host never opens a door which contains the car.
3. The host always asks the contestants if they would like to change their selection.


...and these clauses are not in the original question. These all affect the probability.

For example, if the host only asked the contestants if they wanted to change their selection when he knew they had selected the car, then they would indeed be foolish to change their selction since they have a probability of being correct first time = 1 in this case.

This message has been edited. Last edited by: DorianGreyed,

Chancey,

You're absolutely correct in stating the assumptions required for the 1/3-2/3 answer:

1. The host always knows which door contains the car.
2. The host never opens a door which contains the car.
3. The host always asks the contestants if they would like to change their selection.

If you read the fine print, though, these assumptions were stated at the beginning of the thread. WiteoutKing stated in the original post:

"When the contestant has chosen, the host reveals the first remaining door that has a goat behind it, because the contestant may have chosen a goat OR the car. Then the host asks the contestant if he/she wishes to change his/her mind."

WK never uses the word always, but since no alternative is presented, we can assume that this represents a standard procedure, thus satisfying your third condition. As for your first two conditions, the second one implies the first one, and WK also satisfies the second one (and thus the first). Since the host always reveals a goat, he never reveals the car.

In my third post (Appeal to Authority), I also address the assumption that the game show host always uses the same procedure:

"Marilyn's answer (and mine) only work if the game show host is always required to follow the same procedure. In the real game show that inspired this problem, Monty Hall was allowed to change his procedure, which means to truly optimize your strategy you would need to analyze the host's behavior/motivation (or be psychic), not just rely on probability. However, assuming he does always reveal a goat and offer the chance to switch, my answer is correct."

I realize this is a long thread, but if you look over all the posts carefully, I really think we've covered all the bases.
+++++++++++++++++++++++++++
09-03-02, 12:52 PM
methos
yes, your original probability is 1/3, because you had 1 choice of 3 doors to chose from, but if one door is opened, there are only 2 left. assuming that the door that was opened was a goat, this means that you now have 1 choice of the two remaining doors. It doesn't matter whether you switch or not, your chances are the same. Not switching is the same as choosing the same door again (1 out of your two options... so 1/2 probability), and switching is choosing the other door (one out of your two options, so also 1/2 probability).

09-03-02, 01:05 PM
MrSensitive
The question is... is it more, less, or equally probable that the contestant's choice is right versus changing his/her mind?

To answer the question, one must first eliminate the variable (in this case, "versus changing his/her mind"), leaving us with the question:

"Is it more, less or equally probable that the contestant's choice is right"

The answer is: It is equally probable. (1 in 3)


But by eliminating the 3rd door, and being given a whole new set of varibles, the contestant is given a 50/50 chance of getting the car. The odds are no longer 1:3, but rather 2:3, with a new guess, that the car is behind the chosen door.
But the odds of the car being behind the initially chosen door are still 1:3, regardless of which door is chosen.

In other words, the odds of the car's presence behind a certain door remain the same. The only odds that change are the odds of the contestant guessing correctly.



Aw hell. what do I know. I'm a lover- not a figurer


Mr(statistical mathematics hurt my head)Sensitive

09-04-02, 01:08 PM
WiteoutKing
Yes, the host does in fact follow a rigid procedure: Let him/her guess, reveal a goat, ask to change. Let's make this REALLY simple big grin

If you first chose a goat, you win if you switch... 2/3.

btw, chancey, re: the first rule. If the host didn't know, he may accidentally reveal the car, defeating the purpose of the doors at all!

09-08-02, 10:41 PM
Professor
Sorry to be joining the party late, but there's a lively discussion of this subject (a whole chapter, in fact) in The Man Who Loved Only Numbers by Paul Hoffman (Hyperion, 1998) -- an entertaining biography of Paul Erdos, the colorful and eccentric Hungarian mathematical genius who died in 1996.

Here are some excerpts that may shed some light:

quote: "I told Erdos that the answer was to switch," said Vazsonyi [a mathematician friend of his], "and fully expected to move to the next subject. But Erdos, to my surprise, said, 'No, that is impossible. It should make no difference.'...Vazsonyi wrote out a decision tree, not unlike the table of possible outcomes that vos Savant had written out, but this did not convince him. ..."An hour later he came back to me really irritated. 'You are not telling me why to switch,' he said. 'What is the matter with you? I said I was sorry, but that I didn't really know why and that only the decision tree analysis convinced me. He got even more upset." Vazsonyi...hardly expected [this reaction] from the most prolific mathematician of the 20th Century..."Physical scientists tend to believe in the idea that probability is attached to things," said Vazsonyi...So why did I change my mind? Because my mind has been upgraded with information. This is the Bayesian view of probability. It took me much effort to understand that probability is a state of mind. My hypothesis is that Erdos had this idea of probability as being attached to physical things and that's why he couldn't understand why it made sense to switch doors."

This may help those who, in good company, are being misled by their intuitions.
smile

[This message was edited by Professor on 09-08-02 at 10:57 PM.]

09-29-02, 11:47 PM
K.K.
Unless the shows crew is moving things behind the other two doors,(which would actually make little difference) the odds are one in three, only.

?door=Goat
??door=Car
???door=trip to CanCun

Since the selection is made from the beginning, the player has a one in three chance of being the winner of the car. If the host asks if he/she wants to change their choice after the first door opens revealing the other prize, the contestant is STILL selecting from only one of the three original doors.

Something to consider is this. On a game show, the host doesn't neccessarily care if the player wins or not, and in fact would rather see a player win, as it is good for ratings, which draws advertisers, who in fact are paying for the prizes in the first place. However, the show will always retain twice as much as they give away, because the odds are one in three for the player and two in three for the game show.

And any fool on "the strip" in Las Vegas can tell you that the billions of dollars worth of casino's and hotels in that city were built by people who hosted games in which the house had the odds on their side, and in fact, usually not nearly as favorably as the people from "Let's Make a Deal" Nobody can BEAT the odds. For everyone who wins, there are far more who lose, and losing for a player, is winning for the host of the game, whether it is a TV game show, or a Black Jack table in Vegas. The odds are that you will win by hitting 16 or lower, more times than if you hit 17 or higher. So the house stands at 17 and hits 16. You have to beat their score, and this simple odds analysis creates a situation in which the player is forced to hit hands that are not as favorable to try to win, so the house cannot lose, in the long run.

10-08-02, 11:39 AM
PhysicsFan
I have gone over this with several people and they all get stuck on the hosts intent and that he has to know what is behind each door etc....

These things and a thousand more possibilities were not addressed and therefore should not be considered.

The question is simple: You approach a "magician's" stand and he has an ace and two other cards the he shuffles and lays out on the table. He asks if you want to bet money and you bet 1$ that you can pick the ace. You guess and then he picks up one card and it is not the ace then he asks you wether or not you want to guess agian for double-or-nothing. What do you do?

assuming: the cards aren't marked, he doesn't know which cards are which, they are all from the same deck, there isn't any magic, the table isn't a mirror, etc...

10-08-02, 07:27 PM
WiteoutKing
Well, it IS assumed that the host knows the door with the car, but moving on...

Here are all the possibilities:
x = not the Ace
a = the Ace

three possibilities at the start: axx, xax, xxa
Let's use axx.
If the first card chosen is an Ace (1/3), then the magician lifts one card, either non-Ace (1/3). If you switch (1/6), you lose $1. If you stay (1/6), you win $1.
If you choose a non-Ace (2/3), then the magician lifts one card, either the Ace (1/3) or the non-Ace (1/3).
If the magician lifts the Ace (1/3), you lose $1. If he lifts the non-Ace and you stay (1/6), you win $2 (you have to double your wager to double the bet). If you pick the non-Ace (1/6), you lose $2.

So... the odds of the following are...
losing $2 3/18 or 1/6
losing $1 9/18 or 1/2
winning $1 3/18 or 1/6
winning $2 3/18 or 1/6

Three-Card Monty, when randomized, actually gives you WORSE chances of getting it right. The odds of getting -$2, $1, and $2 are IDENTICAL, whereas losing $1, the given bet, is 1:1. Good way to make quick cash.

10-08-02, 09:26 PM
PhysicsFan
This is my first time posting here and I didn't realize that there were multiple pages. I was responding to the last post on the first page. I apologize, but I still feel my point is valid.

I don't see how everyone gets caught up on what the host knows or does. That is why I tried to change to something not so familiar (cards). My point is that the hosts knowledge or intent just clouds the issue.

I gave a scenario in which I say the host did pick a loser (randomly) now what do you do. (chances are you own a loser)

using your example
axx
if you selected the ace you lose if you switch, if you select 2 you win if you switch, and if you select 3 you win if you switch.

I will play the game with anyone and I will even play it with an ignorant host. Two people, another person and myself, they pick the door and if the host reveals the winner we both lose and if the host reveals a loser I switch they stay. My advantage will be diluted, but it will certainly show through after a couple of games.

10-09-02, 11:21 AM
PhysicsFan
In the situation given I am saying that the host doesn't know where the winner is, but he does know what card you pick. I am also not saying that you will win more often than lose I am saying that you will win more often if you switch than if you stay.

10-16-02, 03:24 AM
tsaeb
I skimmed all these answers, and I think that I still failed to see the one correct answer, which follows.

If the contestant picked Door A, and if the car was behind Door A, the odds are 1/3. If the contestant picked Door B, and if the car was behind Door B, the odds are 1/3. If the contestant picked Door C, and if the car was behind Door C, the odds are 1/3. I know that you agree. Here is where you begin to be wrong.

The host opens Door C, everyone sees no car, and the host asks whether the contestant, who is still capable of winning, wants to switch from Door A to Door B or from Door B to Door A. I know that you agree, although I never told you which door the contestant chose.

In come the bookies. Out go the bookies. The odds are even. What are the odds? 1/2 Why? Since we knew beforehand that the host would eliminate one losing door, all along the odds were 1/2, and these odds never changed.

Anyone participating in this website who so little as implied that tsaeb is nuts, go eat crow, you idiot! I'm talking to you people in "Western Religions" who doubted that tsaeb has spiritual gifts, including the gift of prophecy, and persecuted her by calling her "delusional" with "visions of grandeur" or "paranoid." Ha, ha!

[This message was edited by tsaeb on 10-16-02 at 03:41 AM.]

10-16-02, 06:41 AM
WiteoutKing
Let's go over this again, shall we?. Let's say the car is in #2.

If you pick #1, then stay, wrong.
If you pick #2, then stay, right.
If you pick #3, then stay, wrong.
1/3 odds if you stay.
If you pick #1, then switch, right.
If you pick #2, then switch, wrong.
If you pick #3, then switch, right.
2/3 odds if you switch

The odds of winning altogether are 1/2, but the original question regarded SWITCHING or STAYING! For the last time, YOU WIN BY SWITCHING 2 OUT OF EVERY 3 TIMES!

10-17-02, 12:49 AM
tsaeb
WiteoutKing: Let's go over this again. You cannot switch from both #1 and #3, because one of those doors--either #1 or #3--will be opened and so unavailable for you to switch to it! You are asked if you want to switch after one goat has been removed! Do you want to win the goat that is on its way to the goat bathroom? Again, imagine the problem as with only two doors all along so that the third door is added for only hype or entertainment purposes.

Here is another example to make my solution clear. Imagine flipping a penny and that there are three possibilities of how it lands: heads, tails, and rim. After coming to one's senses, one realizes that since the coin will never appear flipped onto its rim, this never was and never should have been counted as a possibility. Therefore, there are only two possible ways to flip a coin: heads or tails. The probability was and is 1/2. I don't care what you say to deceive the audience: the probability here in both problems was and is 1/2, because we know beforehand that the third choice is no choice at all. In the door problem, the only change was in the elimination of a door which was extra in the first place.

Now, do you turn your back and take your hotdogs with mustard or sauerkraut or poison, trusting that the hotdog vendor will not serve you poison? Do you understand what a third "choice," which is no choice at all, means?

Notice that the example of the penny and the example of the "dressed" hotdog also just came to me: the penny last night and the hotdog tonight.

[This message was edited by tsaeb on 10-17-02 at 01:00 AM.]

10-17-02, 06:41 AM
WiteoutKing
This thread should've been dead last week, seeing as nearl EVERYONE agreed on 2/3. The question says, and I quote: "is it more, less, or equally probable that the contestant's choice is right versus changing his/her mind?" Think about it this way (if by the end of this post you still disagree I can't srgue too much more for my side). If you pick a wrong door FIRST (2/3), the switching wins, 2/3 of the time, and lose 1/3 of the time. If you originally chose the right door, you win 1/3 of the time if you switch, and lose 2/3 of the time. IT IS TWICE AS LIKELY YOU WILL WIN IF YOU CHANGE YOUR MIND!!! Your penny only starts out with two options, and you cannot choose. The hot dog, you must actually choose between mustard, poison, and poison! You don't know which one's which. You choose one, he takes a poison away and asks if you want to switch. THEN AND ONLY THEN DO YOU MAKE YOUR FINAL CHOICE.

Now listen, tsaeb, we've given you examples. We've given you trials. We've given you the RESULTS of trials. YES, you win 50% of the time given that your choice of door and choice between stay of switch is random, but you win BY SWITCHING 66% OF THE TIME!

10-17-02, 11:19 PM
tsaeb
WiteoutKing: Suppose that there are no doors. So, too, there is no car, and there are no goats. However, suppose that you are asked to choose blindfolded a letter, written on a slip of paper in a bag, from out of A and B and C, and suppose also that you are told that 1) one of the letters is a winner, 2) two of the letters are losers, and 3) one of the losing letters will be removed after you choose one letter. Realize that the choice has already been made for you by the game show folks, not by you, as to which letter is the winner!

Do you understand that it makes a difference whether you were instead asked to pick from A or B--that it makes a difference whether or not there is a C along for the ride--only if the game show folks were to change their winning choice? It also makes a difference whether you were instead asked to pick from A or C--that it makes a difference whether or not there is a B along for the ride--only if the game show folks were to change their winning choice? It also makes a difference whether you were instead asked to pick from B or C--that it makes a difference whether or not there is an A along for the ride--only if the game show folks were to change their winning choice?

In this example, where the doors, car, and goat are missing--but where it comes right down to the labeling--surely you can see that all along your choice, whether right or wrong and whether stay or switch, must always be 1/2, because we must assume that the game show folks do not change their winning choice of letter A in your hand or letter B in your hand or letter C in your hand, while there is only one losing letter left in the bag. This is true, because we were told that one of the losing doors would be removed, not switched: we see the removed goat! In fact, 8 or 98 losing doors could have been removed! We trust no hanky-panky behind the remaining two doors! Ahem!

Again, any number of trick to-be-removed doors can be supplied, but the choice must always be 1/2. The problem is one of common sense practicality, not make-no-sense probability.

P.S. Do we have to lay odds as to whether it is harder to quietly move a car or to quietly move a goat? I think not. I stay with 1/2; after all, isn't that the choice made for me from the start by the game show folks? It looks like it is the contestant who is along for the ride--by either car or goat.

[This message was edited by tsaeb on 10-17-02 at 11:48 PM.]

10-18-02, 12:08 AM
tsaeb
See post immediately above this one. A diagram of the initial choices would take into account the following:

A and B<>C, where A is the winning door and B and C are the losing doors. B<>C will later be called B or C.

B and A<>C, where B is the winning door and A and C are the losing doors. A<>C will later be called A or C.

C and A<>B, where C is the winning door and A and B are the losing doors. A<>B will later be called A or B.

Yes, there are three possible schemes on the part of the game show folks, but whichever the scheme employed for the game, the contestant must always have 1/2. Next time, we should all draw a diagram of some sort.

We can draw the three schemes above as three isosceles triangles in which the vertex is the car letter and the vertices of the bases are the two goat letters. In each scheme, the two equal sides collapse into a straight line, as one of the goats is removed. In this way, we can see that we are not deceived by going with 1/2 from the start. Again, the game show folks pull the strings (sides of the triangles), and either they let you win or lose: original vertex or remaining base point, respectively. Without switching, they cannot make the odds go above or below 1/2 at any time, no matter how many moving doors they supply from the start. (If they were to start with four doors, then we would draw mop threads (upside down letter "V" with an additional line down the center of it), which after removal, becomes a line, as expected for 1/2 all along. Imagine a hand fan, which makes a lot of wind when opened, but which is a line when closed. Of course, either they give you the car, or they don't give you the car.)

[This message was edited by tsaeb on 10-18-02 at 12:33 AM.]

10-18-02, 08:20 AM
Pin~Jinx
this is just a poll sort of thingie, not war & winners!?! frown

My initiall instinct was to say that
the options boil down to 1/2, or equal
as posted earlier.

and <
so,
what's the problem?

Isn't MATH basically about facts and logic?

I agree that by switching there is a 66% chance;
however;
WoK do review your initiall post which seems to mainly ask whether
the chances are equal or not.

Do you have some problem with ppl responding to that?!

Pin~(who is hurt noticing her harsh tone, as is usually gentle~)Jinx

10-19-02, 03:35 PM
WiteoutKing
For the last time tsaeb, listen. If you choose a door, one of the wrong doors is revealed. Then you choose, right? Agreed. When you are given the option to switch or stay the probabitlity, RIGHT THEN is 1/2 right? Agreed. HOWEVER, the question CLEARLY states to find if the odds of winning VIA SWITCHING were better than, equal to, or worse than staying! The answer is that switching wins 2/3 of the time, and staying wins 1/3 of the time. I quote from a previous post: "If you first chose a goat, you win if you switch... 2/3" This is my last post on this thread if it doesn't get through to you tsaeb, I'm tired of arguing.

10-21-02, 04:30 AM
tsaeb
My point is that the game is so contrived that you have two chances at picking, and only one of those chances, whether or not you switch, is going to yield a win or a loss--again, in a field of two possibilities even the first time choosing, since the third will be removed and so is not to be considered as a valid possibility or choice. Hence the 1/2.

If the field were 10 possibilities with 8 going to be removed, would you claim that the first is 1/10? No, it is still 1/2.

The whole problem is just one of two flips of a coin. Either you win by flipping correctly twice (no switching) or you lose by flipping incorrectly twice (no switching) or you win by flipping correctly once (switching) or you lose by flipping incorrectly once (switching). All this goes to show, again, 1/2.

10-21-02, 11:44 AM
PhysicsFan
The idea of the question is lost when you take the hosts intent or knowledge into account. In tsaeb's example of the letters in a hat lets pretend that the person drawing the second letter doesn't have any idea what letter they are going to draw, but they do by chance draw a loser. It is in this case better to switch from the one you have to the remaining one in the hat.

Set up the game
A=winner
B&C=loser
if you draw A first -> switch=lose stay=win
" " " B first -> switch=win stay=lose
" " " C first -> switch=win stay=lose
Above it is assumed that the host has drawn a loser each time, because their intent, motives and knowledge can change the issue-lets assume they have drawn losers in each case by chance.

We can also walk through all possibilities where the host draws the winner and you are immediately declared a loser, but the question just says the host reveals a goat - doesn't say why or how just that a loser is revealed. What is the best thing for you to do in this case?

This message has been edited. Last edited by: DorianGreyed,

PhysicsFan: My point all along has been that we can just ignore the removed door/goat from the start and assume that there are only two choices (car or goat), although there are also two chances to pick (first time, second time). My point is that since we are told that we in effect from the start can win or lose out of two doors, regardless of the number of doors which will be removed relative to our pick, the answer is always 1/2.

Consider what the car and goat problem is and is not as follows. A man gives you his little black book of the names of 20 gorgeous dating prospects. He stipulates that only one prospect is still available, because all the others are happily married. You pick one name. He removes 18 names of happily married women. You are left with your choice and one other name. Do you stay with your choice or switch? From such a large field, maybe you should switch, because he may be trying to tell you something: who all along was very likely to be the 1 single out of 20. However, from a field of 3, do you really believe that the odds increase with your switching away from the name which you already chose? I think so. However, we cannot say the same for the car and goat problem, because the car and goat problem is different from this truly random problem in one respect: the car and goat problem is not truly random. In the car and goat problem, people most often pick Door A, less often pick Door B, and least often pick Door C.

Assume that folks pick Door A 40% of the time, Door B 35% of the time, and Door C 25% of the time. Only the game show folks took these odds and found them to be skewed so that the percentages which I chose are only illustrative. Then, it is the game show folks who have to lure contestants away from staying with Door A and into switching to Door C by putting the car behind Door C most often so that the game show folks can eliminate Door A. What am I saying? Anyone keeping a record of which door wins most often must come out with 1/3 for each door in the long run of record keeping. Maybe someone agrees with me that people do tend to pick Door A more than Door B more than Door C, because grade A is better than grade B is better than grade C by life-long conditioning! Therefore, since the game show folks must not appear to be fixing the outcomes, they must, in fact, fix the outcomes! By the time that the outcomes are fixed over the long run, the odds for any one contestant is 1/2, because "astute" viewers are looking at win or lose for each door and nothing else. Yes, I claim that game show folks fix the outcomes. What does this mean? The chances of the car being [behind Door A] < [behind Door B] < [behind Door C] on the first pick--to encourage the switching to eventually equalize the odds of the car being behind each door 1/3 of the time! It is not the probability which matters, but it is the public perception of each door producing an equal number of winning cars which is important. To this end, each contestant has a 1/2 chance due to fixing of the game! Does anyone here have wisdom concerning the game of life in the real world? My true first choice: trust God for a revelation! Then, argue with God for some understanding! Come on, folks, a revelation is what is usually the unthinkable, isn't it now? Ha, ha!

[This message was edited by tsaeb on 10-21-02 at 11:30 PM.]
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10-22-02, 09:24 AM
FlyingHellfish
Looking at a generic situation with n number of doors, with a car behind one door and a goat behind all the other n-1 doors, with it being known that the host will reveal n-2 doors with goats behind them, leaving 1 car and 1 goat unrevealed. We can even assume that the probability of the car being behind a certain door isn't necessarily equal to the probability of the car being behind another door (as a result of fixing the contest). Let xn be the probability that the car is behind Door n. Once again, the various xn don't have to be equal in this example, but they do have to add up to 1 (otherwise, it would be implied that sometimes there is no car at all). Now it's time to let the contestant make his/her first choice, and chooses door number a. The probability that door number a has a car behind it is xa, which is equal to the probability that the contestant will win if he/she decides not to switch doors. On the flipside, there is a 1-xa chance that the contestant is wrong, or equivalently, there is a 1-xa chance that the contestant would win if he/she could switch and pick all of the other doors simultaneously. At this point, the host reveals n-2 doors, all with goats behind them, and offers the contestant a chance to switch to the other door. Much like in drawing straws, there is still a 1-xa chance that the car is behind one of the doors other than door number a. However, the probability of all but 2 xn's has been reduced to 0. Hence, the probability that the car is behind the other door is 1-xa, and unless xa is greater than 1/2, it is in the contestant's best interest to switch.

10-22-02, 07:49 PM
WiteoutKing
tsaeb, listen to me. The question doesn't ask the odds of winning. That IS 1/2. The question asks the odds by switching and staying.
By switching always, the odds are 2/3.
By staying always, the odds are 1/3.

10-23-02, 01:03 AM
tsaeb
FlyingHellfish: I like your work very much. Now, we are talking some Nobel Prize winning potential here. Ha, ha! In your last line, it is my "strong suspicion (strong enough to go with it)" that in Door A the odds are > 1/2. Therefore, since also the field was only 3 doors total to start, yes, it does look to me to even up to 1/2. Thanks for showing enough patience to be blessed to be the one who comes up with something interesting. It is not just the answer but how we play the game (pardon my jokes).

WiteoutKing: Yeah, if I did not have such a strong revelation/hunch/intuition that there were more than meets the eye, I would have read those results a little more closely, but in skimming, I just did not like the explanations. Too simple. I like FlyingHellfish's work posted above yours. He allows for the unthinkable: that the cars vs. goats have to be "adjusted" so that in the long run it is true that each door wins approximately 1/3 of the time. Nevertheless, had there been many doors to begin, there would be no suspicion that what you wrote is correct. In science, I have to come along and be suspicious. Now, don't go teasing FlyingHellfish, because with a nickname like his, I won't allow him in the driver's seat!

All: Be suspicious when the problem is presented too simply. Be even more suspicious when the answer is presented too simply, but this part you already know, judging how you jumped on my challenges. Well, at least, you all will respond better in the future to a challenge.

This message has been edited. Last edited by: DorianGreyed,