View Full Version : Monty Hall Open's a door.....
Your assumptions were almost on the mark, but not quite....
Here is the original question again:
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?
The basic point of this question is two bads and one good, and one of the doors is opened and you are asked if you want to change your choice. No where in this question is it stated that the host had the choice to display a door, only that he knew what was behind them. Neither is it much of a point NOT to open a door, as there is no change to the equation, and no reason for the person to switch, or to even ask the question, is there? The fact that the door was opened is the whole point of the question - does revealing one of the doors change your odds of having picked the car? Of course not.
I think this time there was too much read into the simple question.
University of Washington
Jeez, almost screwed up on that one - what I meant to say was that there was no change to your odds - it will always be 1 in 2 if one of the doors is opened. The door has to open, or there is not much point to this question. Thus, assuming that the host has a choice on whether to open a door or not is incorrect. If the host DOESN'T open another door, your odds remains the same at 1 in 3, and asking if you want to change doesn't affect that at all.
Your whole assumption is based on a random element of choice by the host to open the door or not, and it doesn't matter which prize the contestant picked. The two constants are: If a door is opened, the host will open a door with a goat (not much point in asking you if you want to switch if the car is revealed, is there?) - and your odds are now 1 in 2. If the door remains closed, your odds remain 1 in 3.
This problem becomes easier to understand if you imagine that the two goat doors can be distinguished. Let's say that there is a billy goat behind one door, a nanny goat behind the other. Also note that if I happen to choose the car door, the host can mentally flip a coin to decide which door he opens, while if I pick a goat the host has no choice. Now there are four possible sequences of events:
I pick the car door, host opens the billy door - probability 1/3 * 1/2 = 1/6
I pick the car door, host opens the nanny door - probability 1/3 * 1/2 = 1/6
I pick the billy door, host opens the nanny door - probability 1/3 * 1 = 1/3
I pick the nanny door, host opens the billy door - probability 1/3 * 1 = 1/3
The last two cases are the ones where I would win if I switched, and their probabilities total 2/3.
jens did say:
I will happily play this game for even stakes with anybody who will promise not to switch.
Come on, be sporting. I'd pay out $66 dollars when our friend wins, and take just $34 when I do.
using some cards
Waaaay too slow, let's do it by computer. 3,000,000 rounds minimum...
This discussion shows how powerfully a faulty premise can influence a discussion. Let me restate the rules of the game.
"You get to pick one of three doors. Behind one door is a good prize. Behind the other two doors are Dud prizes. After you make your selection, Monty will reveal a Dud from behind one of the doors that you did not select. After revealing the non-selected Dud, Monty will allow you to change your selection, if you wish."
The question is "Should you change your selection after the first Dud prize is revealed?"
The answer is, it doesn't matter whether you switch or not. Your odds were 50/50 from the start! The game has two parts. The first part ends with Monty revealing the first Dud prize. This part of the game is irrelevant. You do not win or lose in the first part and it has no bearing on the second part. In the first part of the game (up to revealing the first Dud prize), the possibilities are summarized in the table below:
You select | Monty reveals
Good Prize | Dud 1
Good Prize | Dud 2
Dud 1 | Dud 2
Dud 2 | Dud 1
Monty opening one door is part of the initial setup of the game. It is irrelevant because you do not win or lose in part one. It is only the second part of the game, where you must select one of the two remaining doors that has a win or loss associated with it. Perhaps it would clarify the situation if the rules allowed you to not select a door in the first part and Monty just opened one of the Dud doors at the start. Your probability of winning is the same before and after the door is opened (50%). You do not gain any new information when the door is opened.
If you're having trouble with the concept up to this point try thinking about this one. You are going to flip a coin twice. What is the probability of correctly calling "heads or tails" for the second coin flip before you flip the first time?
The fact that your probability of winning goes from 1/3 to 1/2 between the first and second parts of the game is an apparent paradox rather than a real one. It is irrelevant. Only the second part of the game counts - selecting one of two remaining doors.
To anybody who still believes in 50-50 as opposed to 1/3-2/3:
Play the game, using some cards. Convince yourself. Better yet, I will happily play this game for even stakes with anybody who will promise not to switch. High stakes.
To those who insist on the 50/50, you're working on the same logical fallacy that you have a 50/50 chance of winning the lottery, either you'll win or you won't. It's just that this fallacy is cleverly disguised in the statement of the problem.
Kocher says << You select | Monty reveals
Good Prize | Dud 1
Good Prize | Dud 2
Dud 1 | Dud 2
Dud 2 | Dud 1 >>
Yes. But just because there are four outcomes doesn't mean they have equal probability. That's the fallacy. The odds of Good Prize is 1/3, the odds of Dud1 is 1/2, so the probability of your first alternative is 1/3*1/2 = 1/6.
Similar for Good Prize/Dud2, chance is 1/6.
The odds of Dud1/Dud2 are 1/3, and so are the odds of Dud2/Dud1.
You win if you switch in the latter two cases, sum 1/3 + 1/3 = 2/3. Manduck has spelled this out in more detail using Nanny and Billy instead of Dud1 and Dud2, but same idea.
Kocher goes on: << Perhaps it would clarify the situation if the rules allowed you to not select a door in the first part and Monty just opened one of the Dud doors at the start. Your probability of winning is the same before and after the door is opened (50%). You do not gain any new information when the door is opened. >>
This is dead wrong. Your probablility of choosing the right door at start is 1/3. After one door is opened, your odds are improved to 1/2. How can you possibly think that opening the door doesn't give new information?
And, as noted, if you had already picked a door at 1/3 odds, you will improve your chances if you change your selection when the odds are changed to 1/2. The initial door still had 1/3 odds, the new door has had it's status improved to 1/2.
This is why people give the example, suppose there are 100 doors. You select one (odds: 1/100). Monte opens 98 of the others to show duds. Do you switch? You bet, because the odds backing your door are still 1/100, the other door has odds of 99/100 (NOT of 1/2)
Well boys, i wrapped my brain around this one for a whole day and came to realize this. It really doesn't matter any more. Let's Make a Deal went off the air before i even realized that cartoons weren't real people. But i did enjoy tormenting myself for a day. This is the kind of mental exercise one needs to keep from going insane.
This question was rather conclusively settled, IMO, by the Car Talk guys from National Public Radio. Their websight, www.cartalk.com, ran a simulation of this problem. You had three doors with two wrong doors and a right door. You chose one, the program opened a known wrong door, you decided whether to switch or stay, and then your door was open. The program tabulated the results from all the people who played it online. Out of thousands of games, twice as many people won by switching as won by staying. I played several times myself, and my chances fit the 2/3-1/3 pattern as well. The secret is that your original choice has a 1/3 chance of being right, the first door opened has zero chance of being right (because Monty knows what's behind the doors), therefore the third door must have 2/3 chance of being right, and tests have born this out.
"I had a feeling that in Hell there would be mushrooms." -The Secret of Monkey Island
Does anyone really want a simulation of this problem? If so, I'll code one up in C, put it on the web, and post a link here. In case someone suspects that I'm biased, I'll include source code, so you can inspect or recompile it yourself.
We had this argument in the office last year, and one guy did simulate this on one of the big mainframes. After something like 30 million iterations, he finally agreed with Cecil and the Car Talk guys.
Here are your options:
1) You chose the Correct door and stayed.
2) You chose the Correct door and switched.
3) You chose the 1st Wrong door and stayed.
4) You chose the 1st Wrong door and switched.
5) You chose the 2nd Wrong door and stayed.
6) You chose the 2nd Wrong door and switched.
Games where you win: 1, 4, 6.
Total probability of winning: 3:6
Games where you win by staying: 1
Total probability of winning by staying: 1:6
Games where you win by switching: 4,6
Total probability of winning by switching: 2:6
Switch, you'll win twice as often.
"All I say here is by way of discourse and nothing by the way of advice. I should not speak so boldly if it were my due to be believed." ~ Montaigne
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