Talk:Monty Hall problem/Archive 2
I've moved the existing talk page to Talk:Monty Hall problem/Archive1, so the edit history is now with the archive page. I've copied back the most recent thread. Hope this helps, Wile E. Heresiarch 04:40, 10 Aug 2004 (UTC)
two envelope problem
I was talking to a friend about the Monty Hall problem today and he told me about a similar problem, and I don't think there is a Wikipedia article on it and I'm not sure how to present the solution either, but anyway here is the problem:
You are on a gameshow and the host holds out two envelopes for you to choose from A and B. So you choose an envelope (A) and it's got $2000 in it. The presenter then says that one of the envelopes has twice as much money in it as the other one and offers you the chance to switch. So you think about it this way... "If I switch I will go home with either $4000 or $1000, by not switching I will go home with $2000. There is a 50/50 chance that I will double my money by switching. A normal 50/50 bet results in me either doubling my money or losing it all, whereas here I will only lose half. Therefore this is a better than evens bet so I will make the swap." You are just about to swap envelopes when you think about the problem some more - "Surely this can't be right... ". Mintguy (T) 16:13, 15 Jul 2004 (UTC)
- Interesting, but not really similar. In the Monty Hall problem, there's just one possible positive payoff, so the only concern is maximizing your chance of getting it. This problem is more complicated. It is true that if you switch, the expected value of the new envelope is (1000*0.5 + 4000*0.5) = 2,500, so if all you care about is the average amount of money you'll take home, you should switch. However, most people in the real world are risk averse, meaning that they may prefer the sure 2,000. Isomorphic 02:58, 16 Jul 2004 (UTC)
- Yeah but you're wrong, you see, there is no advantage to switching. How can there be? . If you didn't know how much money was in the envelope, then you might make the same analysis and switch, but then after switching the same analysis would lead you to switch again.Mintguy (T) 07:41, 16 Jul 2004 (UTC)
- After I open the first envelope and choose to switch, but before I open the second, here is my position: The other envelope has $2,000. The one I'm holding has 1,000 or 4,000 with (we are assuming) equal probability. My expectation if I switch back is $500 less than if I hold. I'm holding. Dandrake 18:56, Aug 26, 2004 (UTC)
[Tired of nesting the blocks deeper and deeper, like bad C code.] Let's pause to list assumptions. My choice of an envelope is not correlated with the loading of the envelopes, so that I'm equally likely to have the good or the bad envelope. The host also is statistically unbiased: his telling me the news is not correeated with my initial choice of good or bad envelope. Oh, and he's telling the truth.
This is not to say that the expectation argument is right. This is the paradoxical part: that the argument has no apparent flaw in itself, but it gives the nonsensical result that one should choose and then change, even though no new information has come along to cause a change. The host's information is new, or seems to be, but how would my course of action be different if I had known it all along? Bottom line: the expectation argument leads to a silly result, but I don't believe that its flaw has been shown. Maybe this paradox deserves its own article. Dandrake 19:35, Aug 26, 2004 (UTC)
I think I'm not understanding this problem correctly. The way I read it, you either pick an envelope with X or 2X dollars in it with equal probability. Given the option to switch, this expands into four cases
| Picked X | Picked 2X | |
| Keep | X | 2X |
| Switch | 2X | X |
Doesn't this mean that each of these events occurs with equal probablity, and that it doesn't matter? I understand the expectation argument, but I can't reconcile it with this simple grid. Cvaneg 23:13, 26 Aug 2004 (UTC)
This problem now has its own page at Envelope paradox Mintguy (T)
I'm copying this to talk:Envelope paradox. Further discussion of this subject should continue there, but I will also leave the text above so that people can follow on from here. Mintguy (T)
Removed Assumptions
I decided to be a little bold and remove the assumptions section. If you read the problem as stated, then you don't need to make any of the assumptions listed. E.g., it doesn't matter whether or not you assume "Monty always opens a door," or that there is "always a goat behind the door Monty opens," because the problem clearly states that Monty opens a door and reveals a goat. That is all you need to know in order to determine the correct answer. Other points in this section also showed an incorrect understanding of the problem. --Simoes 15:53, 24 Jul 2004 (UTC)
- No, you entirely misunderstand. He is known to open the door this time, as you say, but calculating the probability requires knowledge, or assumption, about what happens in general. What, for instance, was the probability of picking an Ace out of a deck of cards, if you know that I took one, and it turned out to be an Ace? Maybe you think it's 1 in 13, but in fact I was using a pinochle deck. The specific probability calculation requires general knowledge of the conditions. For more detail, showing how the assumption is necesary for this problem, read the assumptions section, which I am restoring to its rightful place. However, if it is unclear, let us by all means fix it. Also, you have not addressed others, such as the obvious matter of the value of a goat. Dandrake 22:31, Jul 24, 2004 (UTC)
Monty's knowledge is critical to the problem
It's critically important to the problem that Monty knows what's behind each door. In the actual correct problem situation, there are three scenarios for what's behind each door:
| Scenario | Door 1 | Door 2 | Door 3 |
|---|---|---|---|
| Scenario A | Car | Goat | Goat |
| Scenario B | Goat | Car | Goat |
| Scenario C | Goat | Goat | Car |
Let's presume that the player always first chooses Door 1 (a safe assumption, since we can always just choose how the doors are numbered after the door is chosen.) If we assume that Monty knows what's behind all doors, and opens the door he does because it has a goat behind it, then our three scenarios look like this (with the revealed goat crossed out because the player knows to avoid it):
| Scenario | Player's first choice | Door 2 | Door 3 |
|---|---|---|---|
| Scenario A | Car | Goat | |
| Scenario B | Goat | Car | |
| Scenario C | Goat | Car |
If on the other hand we leave the problem too vague, by just saying "Monty opens a door that has a goat behind it", we may create the impression that it is just chance that, this time, the door had a goat behind it -- and this changes the problem completely. Because it simply makes it look as if some of our original scenarios could not have happened:
| Scenario | Player's first choice | Door Monty opens | Door 3 |
|---|---|---|---|
| Scenario A | Car | Goat | |
| Scenario C | Goat | Car |
The faulty logic thus goes "If Scenario B had been correct, there would have been a car behind the door Monty opened; but there wasn't, so therefore Scenario B is not correct. The only remaining scenarios are A, where the player's first choice contains the car, and C, where the player's first choice contains the goat; therefore the chances are 50/50 whether the player switches or stays." -- Antaeus Feldspar 20:44, 21 Nov 2004 (UTC)
Standard Solution is Illogical
The standard solution seems illogical for 2 reasons. First, probabilities describe chances, and since Monty has removed one of the uncertainty by opening up a door, he changes the conditions which the original set of probabilities described. This is to say, the 2/3 probability should not be transferrable, and the new probability designed to describe the new conditions should state that the car has a 50-50 chance of being behind either of the unopened door. Second, say the standard solution is logical, and say the car is behind door 1, and the player will always choose between door 1 or 2, and Monty will always open up door 3: if the player had originally chosen door 2, then he should change his selection to door 1; if the player had originally chosen door 1, then he should change his choice to door 2: the results of the 1st situation contradicts that of the 2nd, and vice versa; the contradiction shows that the standard solution is wrong, and it doesn't make a difference whether to stick to the original choice or not.
- There's no contradiction. Door 1 and Door 2 are just labels, without any further information they are possesed of exactly the same qualities. Whichever door he chooses he should switch. Swap the labels on the doors and the solution is the same. Once more, by far the simplest way to look at the problem is to recognize that if the player randomly chooses a goat, and switches, he will win the car every time, and he has a 2/3 chace of choosing a goat. Now I wish someone could explain the two envelope problem to me. Mintguy (T) 17:22, 25 Aug 2004 (UTC)
Remove "assumptions" and rewording for clarity
I found the section on "Assumptions" to be a little confusing and to obscure the true spirit of the puzzle/problem. I have therefore removed that section and reworded the problem in the introduction to make it clearer that the host knows what is behind the doors and that he always opens a door with a goat. Then the assumptions are redundant and the intentions of the host, benign or otherwise, are not a factor. I think it is obvious that the contestant will want a car, not a goat, and there is no need to state this. I have also reworded the answer to try to make it clearer and more concise.
P S K --213.122.39.201 17:00, 25 Oct 2004 (UTC)
- I'm in agreement with the changes you made; the article was wandering into original research on variations of the problem. I think your recent edits have substantially tightened and clarified the article. By the way it would be terrific if you would make a user name and log in. Regards & happy editing, Wile E. Heresiarch 21:40, 25 Oct 2004 (UTC)
Hi P S K, you made the much needed changes to one of our important articles. -- Sundar 04:40, Oct 26, 2004 (UTC)
A clearer explanation?
I have written up what I think is a clearer explanation of the solution, at Monty Hall problem/temp1. I'd like others' feedback, to find out if it's as clear as I think it is and can go in the article. -- Antaeus Feldspar 01:38, 1 Dec 2004 (UTC)
Everyone's making a mountain out of a molehill, or a Lexus out of a goat
The Monty Hall problem is not a problem in people's skill at using probability and mathematics. It is not a problem of gullible people falling prey to tricks. It is not even a paradox. It is a problem in communication and rhetoric.
As commonly stated, the Monty Hall problem is ambiguous as far as how the host chooses the second door.
Whether information about the third door can be inferred by the host's actions, depends on whether the host uses information about what's behind the third door to make his decision.
If the host chooses the second door randomly among the two doors not already selected by the player, and it has a goat behind it, then there is no advantage to the player switching.
If the host chooses the second door because it has a goat behind it, then the probability that the third door has a car behind it jumps to 2/3, and the player should switch.
If someone were to ask me the Monty Hall problem, and it were not clear whether the host used information from behind the third door (or, equivalently, information from behind both the first and second doors) to make his decision, then I would come back and say that the problem was ill-defined. Neither of the two common solutions would be acceptable.
Actually, a third scenario is possible: If the host chooses the second door because the third door contains a goat, then the player should never switch.
The point is that you cannot infer information about the third door unless you can infer information about the host's decision-making process. Unless this is spelled out clearly (which it most often isn't), the problem is ill-defined.
Neither side in this "debate" has done bad mathematics, and mathematicians who righteously scoff at those who express contrary views, calling it bad mathematics, or mathematical illiteracy, add nothing to our understanding.
I'd like a humanities professor's analysis of this problem, because it is not fundamentally mathematical. I'd like to see a detailed analysis of all of the different precise phrasings of this problem, and how they parse against accepted English grammar. What assumptions are socially "valid"?
An analysis of rhetoric is more important than an analysis of probability, in this problem.
(FWIW, I'm a computer scientist who writes mathematical software.)