Computer Forums Probability question

 04-22-2011, 04:41 PM #1 Site Team     Join Date: Jul 2009 Location: England, UK Posts: 3,428 Probability question Some may have heard of this before, but it's a good one nonetheless! You're playing a game - it's pretty simple, there are 3 doors, one with a prize behind and the others with nothing behind. You pick a door to open, however before you open it the game host arbitrarily picks another empty door (not the one with the prize behind) and opens it. This will therefore leave you with your original choice and one other door - one containing the prize, one containing nothing. He then gives you the option to choose either remaining door. Question is, should you stick with your original door or switch to the other? Or doesn't it matter? __________________ __________________ Save the whales, feed the hungry, free the mallocs.
 04-22-2011, 04:53 PM #2 Guru     Join Date: Jan 2008 Location: U.S. Posts: 7,841 Re: Probability question You switch. This is the famous Monty Hall problem. __________________ __________________
 04-22-2011, 04:54 PM #3 Site Team     Join Date: Jul 2009 Location: England, UK Posts: 3,428 Re: Probability question It is - I was trying to fox a few people first but thought you might get there before I did __________________ Save the whales, feed the hungry, free the mallocs.
 04-22-2011, 04:55 PM #4 Guru     Join Date: Jan 2008 Location: U.S. Posts: 7,841 Re: Probability question If you want I can edit my answer __________________
 04-22-2011, 04:59 PM #5 ~~~~~~~~   Join Date: Jan 2005 Location: Wales Posts: 5,556 Re: Probability question Yeah, I am aware of this question! It's a bit of a mind-bender __________________
04-22-2011, 05:00 PM   #6
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Re: Probability question

Quote:
 If you want I can edit my answer
It's cool, I'm sure someone else would post the same thing in a bit anyway! It is a bit of a mind bender but once you've got it round your head it does actually make perfect sense.

Anyone who's struggling and thinks it's 50/50 ask away!
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 04-22-2011, 05:41 PM #7 Guru     Join Date: Jan 2008 Location: U.S. Posts: 7,841 Re: Probability question The key bit of information here is that the host picks and opens his door after you've chosen your door but before you open it. __________________
04-22-2011, 07:28 PM   #8
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Re: Probability question

for this question, (the way it's worded above) you must stay as you have a 100% chance of winning if you stay.

Quote:
 however before you open it the game host arbitrarily picks another empty door (not the one with the prize behind)
if you pick a door, and the the host is able to decide at random what door to pick (i.e it doesn't matter what door they pick), then you must have gotten the correct door to start with. (otherwise they would have to make an informed choice and could not pick at random).

Even if you say that the host doesn't pick arbitrarily, (i.e the host knows when the car is and know that he'll open one of two doors (whichever you don't pick).
I've read the statistical proof, and i (like many others) disagree.

in the first question, the change of you picking the prize is 1/3.

now the host has opened a door, you have a new choice, that bears no ground on the original question.

you now have a 1/2 chance of winning, so it doesn't matter.

there is a simillar probability question.
you see a woman walking down the street with a young girl, it's her daughter, she says that she has two children and one is at home, what are the odds that the other is a girl.

with two kids the possibly combinations are

BB 1/2 * 1/2 = 1/4
BG 1/4
GB 1/4
GG 1/4

you know that one of the kids is a girl so that wipes out the first combination
you're now left with
BG = 1/3
GB = 1/3
GG = 1/3

the idea is that there is a 2/3 possibility that the other child is a boy (2/3rd possibility of boy compared to 1/3rd possibility).

but it ignores the fact that it's a mutually exclusive event. (and random)

I just flipped two coins, one of the coins was a head, what's the other most likely to be?
the answer is that it's a 50/50 chance, the events are mutually exclusive, two independent choices.

in the original question, odds of getting the door right the first time are 1 in three.
now the host reduced the amount of doors to 2, forget your original choice and pick again (which is what the question do you want to swap really means)
the odds are now 1/2 it makes no difference if you change...

it's a different independent choice, so you don't compound the probabilities from the original question.

by the time the host has taken away a door that they know to be incorrect, then you're basically picking one of two, with an equal chance of the prize being behind either door.

so yes,
if you take the view that the host is asking you if you want to stick with your original choice 1/3
or now that he's opened the other door you can effectively have both doors (2/3) then I can see why they say switch...

on the other hand, you know that one of those doors does not contain the prize, you're not seeing 2 doors for the price of 1 by switching, one of them is empty, the odds aren't 2/3 one is definitely not the answer it's still 1/3 (you've seen the third door. if you stay you don't un-see it, if you stay you get to see what's behind your door and the open door, so you get 2/3 odds the same as if you switch). with 1/3 odds on the unpicked unseen door.

but now we've taken the third door away it's 1/2 staying, or 1/2 switching.
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04-23-2011, 12:18 AM   #9
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Re: Probability question

^ Disagree. Let's increase the number of doors.

Quote:
 Originally Posted by Wikipedia It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially. To extend the above, it's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens the first 999,998 of them for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one. This example can also be used to illustrate the opposite situation in which the host does not know where the prize is and opens doors randomly. There is a 999,999/1,000,000 probability that the contestant selects wrong initially, and the prize is behind one of the other doors. If the host goes about randomly opening doors not knowing where the prize is, the probability is likely that the host will reveal the prize before two doors are left (the contestant's choice and one other) to switch between. This is analogous to the game play on another game show, Deal or No Deal; in that game, the contestant chooses a numbered briefcase and then randomly opens the other cases one at a time. Stibel et al. (2008) propose working memory demand is taxed during the Monty Hall problem and that this forces people to "collapse" their choices into two equally probable options. They report that when increasing the number of options to over 7 choices (7 doors) people tend to switch more often; however most still incorrectly judge the probability of success at 50/50.
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04-23-2011, 04:01 AM   #10
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Re: Probability question

Quote:
 however before you open it the game host arbitrarily picks another empty door (not the one with the prize behind)
Perhaps I shot myself in the foot slightly with the ambiguous wording - I meant the host picks any unopened door without the prize behind. I was trying to cover all bases because it's possible that the host uses psychological / other statistical techniques to bias which door is opened and reduce the chances of winning overall (though this is the grounds of much more complex proof and subjectiveness.)

Quote:
 if you take the view that the host is asking you if you want to stick with your original choice 1/3 or now that he's opened the other door you can effectively have both doors (2/3) then I can see why they say switch...
Thing is, the choices here aren't independent. Let's put it another way - there's 3 possible routes:

You pick the prize to start with and switching always gives you nothing
You pick nothing to start with and switching always gives you the prize
You pick nothing to start with and switching always gives you the prize

Each of those have an equal likelihood, saying we should win in 2/3 of the cases if we switch.

Another way of looking at it is the chances of you picking the prize to start with is 1/3 - that doesn't change. As said above, if there's a thousand doors and the host opens 998 others, the chances of you picking the right one to start with were still less than 1%. And unless you picked the right one to start with, switching will always lead you to win.
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