Deceptive math?

[Jnyusa]

To make this situation work out the way intuition says it should, the rats and cars would have to be randomly reassigned every time a door was opened.

The probabilities do not have to be reassigned. The only requirement for 1:2 is that the host be allowed to open any door at random. For probability to rise to 2:3 the host has to be constrained as to which doors he can open.

Jn

[Faramond]

The fact he opens a wrong door doesn't change the fact that you're suddenly getting to choose both doors you didn't choose initially.

That's the best phrasing for the solution I've ever seen, hal. Thanks!

No, that's not correct. The way the game is presented above, one of your choices always remains your initial choice. As soon as they open the door that you chose, the game is over.

I don't understand your objection to hal's explanation.

The reader has to assume that the host always opens a door with a goat, as Jn noted.

Yes, and it's my contention that no one would make this assumption.

Well, I would make that assumption, if given the question in a math setting. But I do agree that no one would make this assumption when confronted with this situation in real life.

[Jnyusa]

don't understand your objection to hal's explanation.

I understood hal to be saying that your own wrong choices are eliminated. That's not the case. It is a wrong choice that you didn't make that is eliminated. But maybe I misunderstood him.

Jn

[Faramond]

To make this situation work out the way intuition says it should, the rats and cars would have to be randomly reassigned every time a door was opened.

The probabilities do not have to be reassigned. The only requirement for 1:2 is that the host be allowed to open any door at random. For probability to rise to 2:3 the host has to be constrained as to which doors he can open.

Jn

If the host opens either door the contestant hasn't chosen at random, sometimes she'll ( women can be sleazy game show hosts too! ) open the car door, so in those cases the odds of getting the car are zero. But sure, if the host randomly opens a door, and the door did happen to have a rat, then the odds for both remaining doors are 1/2.

[Jnyusa]

If the host opens either door the contestant hasn't chosen at random

Actually, the host must be able to open the door the contestant chose as well.

Hm ... I worked this out once before and I'm a bit lazy to do it again, but I think that if the host can open at random either of the two door the contestant did not choose, then the contestant raises their odds to 2:3 by never switching.

When I saw this game I was immediately curious how the different rules might affect the probabilities so I drew out a table for all the options. But I don't recall now what all the results were. I could do it again, I suppose .....



Jn

[axordil]

But I do agree that no one would make this assumption when confronted with this situation in real life.

Actually, if we go back to the eponymous Monty Hall, in situations like this he NEVER opened a curtain/box/etc. with the car-type prize, since that took away suspense. That's the other variable: the host's interest. Generally, it didn't matter to Monty (or other game-show hosts) so much if someone won or lost per se, only that it was "good" drama. He wasn't trying to rip off the contestant, he was trying to entertain.

So in some real-life situations, the problem makes even more sense. If one wants to call a game show real life...

jny--

The way I read hal was: you had one of the three possible chances, and now you have two. The fact that one of those two is a wrong chance doesn't change the fact that it was one of the three and remains so.

[axordil]

From Wikipedia, the free encyclopedia

The Monty Hall problem is a puzzle involving probability loosely based on the American game show Let's Make a Deal. The name comes from the show's host, Monty Hall. A widely known statement of the problem is from Craig F. Whitaker of Columbia, Maryland in a letter to Marilyn vos Savant's September 9, 1990, column in Parade Magazine (as quoted by Bohl, Liberatore, and Nydick).

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

While this is a common presentation of the problem, it is problematic as it leaves important conditions of the problem unstated.

The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive, although the problem does not yield a logical contradiction.

Once the host has opened a door, the car must be behind one of the two remaining doors. Since there is no way for the player to know which of these doors is the winning door, many people assume that each door has an equal probability and conclude that switching does not matter. However, as long as the host knows what is behind each door, always opens a door revealing a goat, and always makes the offer to switch, opening a losing door does not affect the probability of 1/3 that the car is behind the player's initially chosen door. As there remains only one other door, the probability that this door conceals the car must be 2/3 (that is, 1 - 1/3). The "equal probability" assumption, whilst being intuitively seductive, is actually incorrect; switching increases the chances of winning the car from 1/3 to 2/3.

Put quite simply, the player has a 2/3 chance of choosing a goat to begin with. The other goat gets revealed by the host, so by switching the player turns the 2/3 chance of having a goat into a 2/3 chance of having a car.

further down, they address hal's reading, and I have italicized the relevant connection:

The most common objection to the solution is the idea that, for various reasons, the past can be ignored when assessing the probability. Thus, the first door choice and the host's forced response are ignored. Because there are two doors to choose from, many people jump to the conclusion that there must be a fifty-fifty chance of choosing the right one.

Although ignoring the past works fine for some games, like coin flipping, it doesn't work for all games. In this case what should be ignored is the opening of the door. The player's choice is between the originally picked door and the other two — opening one is simply a distraction. There is only one car. The original choice divides the possible locations of the car between the one door the player picks (1/3 chance) and the other two (2/3 chance). It is already known that at least one of the two doors contains a goat. The revealing of the goat therefore gives the player no additional information about his own door. It doesn't change the 2/3 probability that the car is still in the block of two doors.

[superwizard]

What an interesting thread! I actually watch numb3rs and so I've already seen this before. So the way I understand it is if the game show host chooses a door without the prize and that there is only one good prize and the others are useless then it would make sense to switch doors. There is actually an online version of the game that you can play multiple times to satisfy your curiosity: http://www.shodor.org/interactivate/act ... &vendor=na
I don't know much about real life game shows though so I can't help you in deciding whether it really is applicable or not!

[Faramond]

If the host opens either door the contestant hasn't chosen at random

Actually, the host must be able to open the door the contestant chose as well.

Hm ... I worked this out once before and I'm a bit lazy to do it again, but I think that if the host can open at random either of the two door the contestant did not choose, then the contestant raises their odds to 2:3 by never switching.

When I saw this game I was immediately curious how the different rules might affect the probabilities so I drew out a table for all the options. But I don't recall now what all the results were. I could do it again, I suppose .....



Jn

No, I think this is wrong. You will also get to 1/2 odds for both remaining doors if the host can only open one of the two that the contestant did not open, so long as the host does happen to randomly open a rat door.

I'm going to represent each possible state of the system with a three digit number string. The doors are numbered 1-3. The first digit of the string is the door the contestant chooses, the second digit is the door the host randomly chooses to NOT match the contestant, and the third digit is where the car really is. Since both contestant and host are choosing randomly, each string has an equal chance of any other string of describing the state of the system. The only constraint is that the first and second digits don't match, because of my condition that the host not open the contestant's initial choice.

Below are the strings. I've bolded the strings where the host doesn't randomly reveal the car, that is, the strings where the second and third digits don't match. My contention is that when the host doesn't reveal the car, the odds for the two remaining doors are 1/2. For the bolded strings, I've colored blue if the contestant gets the car, and red if the contestant gets a rat.

121
122
123
131
132
133
231
232
233
211
212
213
311
312
313
321
322
323

If the host randomly chooses a non-contestant door that happens to have a rat behind it, then the odds the contestant chose correctly to begin with are 1/2.

[yovargas]

Isn't the point that, no matter how many doors, if the host opens every door except the last one, then the answer to "Is it to your advantage to switch your choice of doors?" is always 50/50?

Holy crap I'm confused. Can someone just tell me if this is right or wrong, please?

[Faramond]

What an interesting thread! I actually watch numb3rs and so I've already seen this before. So the way I understand it is if the game show host chooses a door without the prize and that there is only one good prize and the others are useless then it would make sense to switch doors. There is actually an online version of the game that you can play multiple times to satisfy your curiosity: http://www.shodor.org/interactivate/act ... &vendor=na
I don't know much about real life game shows though so I can't help you in deciding whether it really is applicable or not!

But you really need to know how the host chooses the door. If it's truly at random, and the car could have been exposed, then there is no abvantage to switching.

[Faramond]

Isn't the point that, no matter how many doors, if the host opens every door except the last one, then the answer to "Is it to your advantage to switch your choice of doors?" is always 50/50?

Holy crap I'm confused. Can someone just tell me if this is right or wrong, please?

Sorry, yova. Can't answer unless the actions of the host are clearly defined, such as always choosing randomly or making sure to always choose a rat door.

In short:

1. If the host chooses another door randomly to reveal, then switching doesn't help ( or hurt ).

2. If the host knows what is behind the door and always chooses to reveal a rat door that the contestant didn't choose, then it is good to switch.

3. If it's real life and the host is milking the situation for drama and making decisions that can't be consistently defined as random or constrained, then you can't say much of anything about the odds.

[yovargas]

I was thinking situation #2. So in the end, you have two closed doors. One has a rat, one has a car. How is that not 50/50?

[Jnyusa]

Good-o. Thanks for working that out, Faramond.

I don't remember now what rule resulted in 2:3 for never switching. I'd have to rethink the puzzle. The point remains, of course, that if different rules result in different outcomes, and you don't know what rule the host is using, then there's no reason to proclaim one outcome 'right' and the other one 'wrong.' It seems to me that I also calculated what the odds would be if the host switched rules at random! So if you don't know the rule, then what should you choose? But again, I'd have to recreate the tables to remember what answer I got.

Might do that now that we're into the discussion!

Someone mentioned that they never open the car door right off the bat. Yes, that's correct. It would destroy the suspense. But not every game had the fabulous prize. Most of the games had nice prizes but not spectacular prizes - sports equipment or cooking equipment or something like this. You can always generate random outcomes in a table like the one Faramond showed above, and then assign the car prize to the outcome that is most suspenseful.

Jn

[Faramond]

I was thinking situation #2. So in the end, you have two closed doors. One has a rat, one has a car. How is that not 50/50?

Because the way you got down to those two doors destroys the inherent randomness of the system. The host knows exactly where the car is, and acts on this knowledge every time.

[halplm]

Isn't the point that, no matter how many doors, if the host opens every door except the last one, then the answer to "Is it to your advantage to switch your choice of doors?" is always 50/50?

Holy crap I'm confused. Can someone just tell me if this is right or wrong, please?

No, I Don't believe it's correct.

A correct statement would be... if you walk in on a friend playing this game, and see all the doors open but two... and you don't know which door your friend picked...then YOU would have a 50/50 chance of choosing correctly.

Because the host eliminates a lot of bad choices... you gain a lot of information. Wikipedia has some good explanations, but they also go off a lot on the math.

Another bit they have, though that's good is this: Opening a bunch of doors with rats in them... gives you no NEW Information about you're own door. IT will still always only have a 1/3 (or 1/1000) chance of having the car... because you made that choice randomly. IF all the other wrong choices are removed... your choice is still low in probability of being right... while the chances of the remaining choices keeps going up.


An interesting not I had to figure out... the show "Deal or No Deal" that's on these days, eliminates this problem entirely because no one involved knows what's in each case...

[Faramond]

An interesting not I had to figure out... the show "Deal or No Deal" that's on these days, eliminates this problem entirely because no one involved knows what's in each case...

Right, I watched it once. The host doesn't do any choosing, I believe. There are about 20 cases, all with dollar amounts ranging from one cent to a million dollars. The contestant does all the choosing, and one by one they reveal cases and eliminate them. Periodically a "banker" will offer the player a sum of money to walk away from the game. As far as I could tell this sum was always somewhere below the expected value of the player's winnings based on the cases remaining. However, it might still be more than the player would likely make if they went down to the last case.

For example, if the cases left had 1000000, 100, and 1 dollars in them, the expected value of the contestant's winnings is (1000000+100+1)/3 = 333367 dollars. The "banker" might offer 200000 for the contestant to walk away from the game. This is less than when the contestant would make by opening the cases on average, and yet 2/3 of the time the contestant would make far less than that by opening cases. So it would seem smart to take the banker's offer.

[halplm]

Good-o. Thanks for working that out, Faramond.

I don't remember now what rule resulted in 2:3 for never switching. I'd have to rethink the puzzle. The point remains, of course, that if different rules result in different outcomes, and you don't know what rule the host is using, then there's no reason to proclaim one outcome 'right' and the other one 'wrong.' It seems to me that I also calculated what the odds would be if the host switched rules at random! So if you don't know the rule, then what should you choose? But again, I'd have to recreate the tables to remember what answer I got.

Might do that now that we're into the discussion!

Someone mentioned that they never open the car door right off the bat. Yes, that's correct. It would destroy the suspense. But not every game had the fabulous prize. Most of the games had nice prizes but not spectacular prizes - sports equipment or cooking equipment or something like this. You can always generate random outcomes in a table like the one Faramond showed above, and then assign the car prize to the outcome that is most suspenseful.

Jn

If there is no set of rules for the host, then there's no improvement by switching. If the value of each choice is similar, or you don't know if what you see is "good" or "bad" there's no value in switching.

If the prizes vary in value, you gain no information by something being revealed.

Perhaps added information is a bad way of saying it.

The clearly stated rules (as faramond stated them) are really the only case where you always switch. It defines a clear set of rules such that you always know the outcome. switching will result in winning 2/3 times, and not switching will result in winning 1/3 times.

When you aren't clear about the rules, you don't have enough information for the probabilities to work like that.

Lets remove all the doors and rats and cars...

Lets say you have envelopes with money in each. Two of the envelopes have $1, and the third had $100. Here's the sequence of events:

A. you pick an envelope, the host keeps the other two.
B. The host tells you that one of the envelopes he is holding only has $1. (you already knew this).
C. You're offered the choice of taking the envelope you chose, or both envelopes he's holding.

You should always take the two he's holding. There's a 2/3 chance you'll get the $100.

That is the game as it's defined with the rats and cars too... although I'm assuming you don't want to keep the rats you might win...

[halplm]

An interesting not I had to figure out... the show "Deal or No Deal" that's on these days, eliminates this problem entirely because no one involved knows what's in each case...

Right, I watched it once. The host doesn't do any choosing, I believe. There are about 20 cases, all with dollar amounts ranging from one cent to a million dollars. The contestant does all the choosing, and one by one they reveal cases and eliminate them. Periodically a "banker" will offer the player a sum of money to walk away from the game. As far as I could tell this sum was always somewhere below the expected value of the player's winnings based on the cases remaining. However, it might still be more than the player would likely make if they went down to the last case.

For example, if the cases left had 1000000, 100, and 1 dollars in them, the expected value of the contestant's winnings is (1000000+100+1)/3 = 333367 dollars. The "banker" might offer 200000 for the contestant to walk away from the game. This is less than when the contestant would make by opening the cases on average, and yet 2/3 of the time the contestant would make far less than that by opening cases. So it would seem smart to take the banker's offer.

Yes, and no, it's a little more compicated because you have to open a set number of cases each time... so it's still adventagious to play if you can improve his offer. It IS always less than the expected value.

Also, every time I've seen it get down to the final two cases, the host has offered to let the contestant switch cases... which is totally pointless, but adds drama

[Faramond]

An interesting not I had to figure out... the show "Deal or No Deal" that's on these days, eliminates this problem entirely because no one involved knows what's in each case...

Right, I watched it once. The host doesn't do any choosing, I believe. There are about 20 cases, all with dollar amounts ranging from one cent to a million dollars. The contestant does all the choosing, and one by one they reveal cases and eliminate them. Periodically a "banker" will offer the player a sum of money to walk away from the game. As far as I could tell this sum was always somewhere below the expected value of the player's winnings based on the cases remaining. However, it might still be more than the player would likely make if they went down to the last case.

For example, if the cases left had 1000000, 100, and 1 dollars in them, the expected value of the contestant's winnings is (1000000+100+1)/3 = 333367 dollars. The "banker" might offer 200000 for the contestant to walk away from the game. This is less than when the contestant would make by opening the cases on average, and yet 2/3 of the time the contestant would make far less than that by opening cases. So it would seem smart to take the banker's offer.

Yes, and no, it's a little more compicated because you have to open a set number of cases each time... so it's still adventagious to play if you can improve his offer. It IS always less than the expected value.

Also, every time I've seen it get down to the final two cases, the host has offered to let the contestant switch cases... which is totally pointless, but adds drama

I agree, there are cases where it's advantageous to keep playing.

In fact, to really analyze how one should play this game, you have to know how much each value each dollar amount has to the player. To give an extreme example, if the player has to get a million dollars for an operation to save his life, then it's never wise to take the banker's offer.

Do people usually switch at the end, or stay put?