Binary person, your fallacy is in multiplying the scenarios where keeping the door wins by two. Just because Monty can has the option of giving you either of the two losing doors when you have picked the winning one does not make each of those occurrences equally likely!
Which one of the two losing doors Monty gives you is irrelevant, and should not be included in your enumeration.
1010011010, since you went to all the trouble of showing so many cases, I feel a little embarassed to correct you. However, this is The Straight Dope, and truth rules…
The key to this problem is that your Cases 5, 6, 7, and 8 are each twice as likely as 1, 2, 3, or 4. The assumption is that MH will ALWAYS choose to open a door with a goat. When you have chosen a goat, and he has only one such door available, this situation has the same total probability as do MH’s 2 options when you have chosen the door with a car.
E.g., Case 5 corresponds to Case 1 plus Case 3. In Case 5, you chose Door 2 and MH opened Door 3. (The car is behind Door 1.) Now,imagine that Case 5 also includes Case 5a, where the contestent selected Door 2, and MH would have opened door #1, but there was a car behind it, so he had to switch his choice and open Door #3.
Regarding the Monty Hall problem, Marilyn vos Savant was wrong and your original answer was correct. The key is in the statement that the host knows what is behind each door, consequently, he can always show a door with no prize behind it. Assuming your original choice is door #1, there are three four possiblities as to which door can be opened and which contains the prize. These possibilities are as follows:
Door #1 has prize and door #2 is opened
Door #1 has prize and door #3 is opened
Door #2 has prize and door #3 is opened
Door #3 has prize and door #2 is opened
As you can see, of the four possibilities, two have Door #1 having the prize and two have door #1 not having the prize. Therefore, the odds are even. The argument about choosing one out of a million or one card out of a deck of 52, then revealing all except the the original choice and one other is not a valid comparison. It only appears that it is because revealling one out of three leaves two.
Also, concerning the two children problem, you state that the possibilities are:
A (female) B (female)
A (female) B (male)
A (male) B (female)
A (male) B (male)
More correctly they should be stated by which child enters first, that adds four more possibilities:
B (female) A (female)
B (female) A (male)
B (male) A (female)
B (male) A (male)
Now, you can see that if the first child is female, there are four possiblities, in two of which the second child is female and in two of which the second child is male. 50-50.
Or, conversely, using your four possibilities, in both of the last two the first child is male, and you should eliminate both of them, again leaving a 50-50 probability as to the gender of the second child.
Robert D Schader
Green Bay, WIRobert D Schader certainly displays self-confidence, contradicting the (self-proclaimed) smartest man in the world and smartest women in the world!
I do agree with Robert’s 4 possible cases. However, the the 1st and 2nd cases each have probability 1/6 and the 3rd and 4th cases each have probability 1/3.
The simplest way I can put it:
The probability of you picking a loser door initially is 2/3 (clearly). If you pick a loser, you will always win by switching because Monty will have helpfully eliminated the other loser. So if you follow the strategy of always switching you will win 2/3 of the time.
The always-switch strategy only loses when you picked the winning door with your first choice, and clearly that will happen only 1/3 of the time.
jen and dec, admirable effort. You might have had bad arguments but you were fighting for the right side. Mandrake has convinced me.
I have seen the err of my ways
Possible scenarios:
Winning scenarios: 1, 4, 6 or 1:2
Winning scenarios where you switch: 4,6 or 1:3
Winning scenarios where you stay: 1 or 1:6
IE- you’re twice as likely to win when you switch!
<<switches>>
Aufwiederlesen
1010011010 ¦¬)
“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
Note that my intention was NOT to convince you. Sufficient materials for that, including previous incarnations of Manduck’s explanation, had already been posted.
My intention was to point out the fallacy in your particular argument, which seemed on first sight to be logically phrased. Usually people who bother to enumerate come up with the correct answer, which made your enumeration interesting to me.
Manduck2 – Must congratulate you. Your explanation is the simplest and clearest I’ve seen.
Let me try to explain this one other way. First I’ll list all of the possible outcomes.
The first number is the door that the prize is behind, and the second number is the choice made by the contestant.
PC—PC—PC
11—12—13
21—22—23
31—32—33
Each of these possiblities have the same probability, 1 in 9.
Now the host, knowing where the prize is, selects a door that does not contain the prize and shows it to the contestant.
The contestant chooses a door, removing 6 of the possiblities, then the host opens a door(not containing the prize), removing 1 of the remaining possibilities. That leaves 2 possibilities, both of which have the same probability, one where the original choice was correct and one where it was not.
RD --You wrote:
The contestant chooses a door, removing 6 of the possiblities, then the host opens a door(not containing the prize), removing 1 of the remaining possibilities. That leaves 2 possibilities, both of which have the same probability, one where the original choice was correct and one where it was not.
The trick is that these remaining cases do NOT have the same probability. You have to also factor in the probabiklity that MH opened a particular door.
E.g., assume that you chose door #1 and MH opened Door #3.
If the car was behind Door #1, then MH had a 50% chance of opening #2 and 50% chance of opening #3. If the car was behind Door #2, then MH was 100% certain to open Door #3. For this reason, when MH opens Door #3, it is twice as likely that the car is behind Door #2.
RD - Look at the outcomes you listed:
PC—PC—PC
11—12—13
21—22—23
31—32—33
Note that in 3 of those 9 cases (11, 22 and 33) your first choice was correct. In the other 6 cases, you would have to switch doors to win the car. So in 6 of the 9 cases the switching strategy wins, while not switching only wins in 3 of the cases.
Let’s assume that the contestant chooses door #1. Now there is a 1/2 probability that MH will open door #2 and a 1/2 probability that he will open door #3.
MH opens door #2 - probability 1/2
Prize behind Door #1-probability (1/2)(1/3)
Prize behind Door #3-probability (1/2)(2/3)
MH opens door #3 - probability 1/2
Prize behind Door #1-probability (1/2)(1/3)
Prize behind Door #3-probability (1/2)(2/3)
Probability of Door #1 being correct
(1/2)(1/3)+(1/2)(1/3)=1/3
Probability of Door #2 being correct
(1/2)*(2/3)=1/3
Probability of Door #3 being correct
(1/2)*(2/3)=1/3
Opening Door #2, for example, removes 1/3 of the possibilities, but the probabilities for Door #1 and Door #3 being correct remain in a one to one ratio.
One of the arguments I read was that showing a non-prize door was equivalent to offering both of the other doors for your one. If this were what was happening then it would indeed be to your advantage to switch. This,
however is not the case. The host is just removing from consideration a door that he knows does not have the prize. He can do this 100% of the time. If your choice was correct, then he has a choice as to which door he opens, if your choice was not correct then there is only one door he can open.
Hypothetical situation:
You consider choosing door #3, but at the last momment change to door #1. Now the host reveals door #2, showing that it does not contain a prize.
The probability of the host opening door #2 is the same had your choice been door #3 or door #1.
What you are saying is that it is now to your advantage to switch back to door #3, but if you had not change your mind, and originally choosen door #3, it would now be to your advantage to switch to door #1. Clearly these can not both be true.
RD - You state that if you choose door #1, there is a 50% chance that MH will open door #3. This is not necessarily true - if the prize is behind door #2, MH is forced to open door #3, i.e. the probability is 100%.
Let’s break it down again. Assume the prize is behind door #3. The possibilities now are:
You pick door #1 (probability = 1/3), MH opens door #2 (probability = 1): total probability = 1/3
You pick door #2 (P = 1/3), MH opens door #1 (P = 1): total P = 1/3
You pick door #3 (P = 1/3), MH opens door #1 (P = 1/2): total P = 1/6
You pick door #3 (P = 1/3), MH opens door #2 (P = 1/2): total P = 1/6.
The first two cases above are the ones that you would win if you switched, and their probabilities add up to 2/3.
Manduck, after careful reconsideration, I am forced to admit that you are indeed correct.
It doesn’t matter whether the game show host always opens the door or not. Switching doors provides no advantage. The way the question is worded though should indicate to you that the game is that the host opens a door every time. If it was optional for the host to do that, then they would have to say that the host “opened” another door, not that he “opens” another door. Even if the game WAS that it is optional for the host to open a door, the question is about a particular instance where the host does open the door, so you don’t even need to consider all that. You guys are way off. Anyway, it doesn’t matter, here’s why:
Without a door being opened, you have 3 choices, in 1 of which you win, so your chances of winning are 1/3 for each choice.
With a door opened, you have 2 choices, in 1 of which you win, so your chances of winning are 1/2 for each choice.
If the host opens a non-winning door, you’re not learning any more about the door you didn’t choose than you are about the door you did choose. Someone said that having the
host open a door and opening the other door you didn’t originally choose is like opening 2 out of 3 doors, whereas if you keep your original door, you’re only opening one door. Take a closer look, in the latter case 2 out of 3 doors are also open. People who claim
to have proven an advantage one way or another by doing experiments or by writing
programs are either totally incompetant or are lying.
Marilyn is wrong quite often (see http://www.wiskit.com/marilyn),,) and I suspect she does it on purpose to generate responses and hence publicity.
I believe there is still some specious logic afoot in the analyses of this problem, Cecil’s responses included. I refer specifically to the notion that Monty Hall’s pre-knowledge of what’s behind the doors is a factor that distinguishes this from the lottery ticket quandary. While it adds a certain Machiavellian intrigue to the proceedings, I’m not sure it’s relevant.
It’s just not clear to me why this should matter at all (I’ll avoid the smug certainty that has come back to haunt virtually every person who has commented on this important issue). Stated more emphatically, if this fact is apropos, I am completely misunderstanding the logical conclusion, which is why I bring it up at all (since the conclusions are otherwise “accurate”).
Cecil himself asserts that this dilemma is tantemount to being offered all the unselected doors in return for your original choice. If this is true, why would that fact change because this option emerged randomly and not from the schemings of the evil Hall?
I believe that even if Monty randomly selected one of the unchosen doors, faced with an infinite series of the subset of trials where the knucklehead hasn’t selected the actual correct choice–that’s what we have here–I’m still better off switching. The same is true for the lottery ticket if I calculate the benefit derived from this strategy over the course of an infinite series of trials.
Am I loopy?
thatsmrberns2u said:
Them’s fightin’ words, Mr. Berns. And on your first post, too - tsk, tsk.
I was going to stay out of this, and just read this one. I’ve had arguments about this subject before. But about seven years ago, when this subject was a little less stale, I did, in fact, write a computer program to play this game a few thousand times, and test the always-switch and never-switch strategies. So your offhand remark becomes a personal insult! I must respond!
The program did show that the always-switch strategy wins 2/3 of the time and the never-switch strategy 1/3 of the time. And I wrote the program as a skeptic, mind you - I was rooting for a 50/50 result. But I’m convinced now.
It was a very simple program, so even if I am “incompetant” (which is usually spelled “incompetent”, by the way - what an unfortunate word to misspell) I’m sure I managed to get an accurate result, as have many others. And why in the world would anyone lie about such a thing?
If you’re sure that ALL of the experiments and computer programs testing the hypothesis are flawed, why not do one yourself? If you turn out to be right, you’ll turn the worlds of math and physics upside down.
The mistake in your reasoning is a common one; you assume that when the number of closed doors changes from 3 to 2, the 1/3 probability of your original choice being the correct one auto-magically changes to 1/2. It doesn’t. Nuclear war wouldn’t change it. Monty going postal wouldn’t change it. The contestant suddenly getting abducted by aliens wouldn’t change it. And certainly, opening some silly door isn’t going to change it. That’s really the heart of the argument; the door you initially pick is probably the wrong door, and opening more doors does NOT change that. No matter how many doors get opened later, your first pick was probably wrong. It’s simple to see that if the door you picked is probably wrong, then the doors you didn’t pick probably include the prize. Opening doors doesn’t change that, either, which is why switching usually wins.
To believe anything else is pretty close to believing in time travel. If the odds of your first pick being right change (from 1/3 to 1/2), then the events of the present (opening doors) are effecting the events of the past (your first pick)! In this universe, most of us like our causes to come BEFORE our effects. Opening doors after you make your choice doesn’t effect the odds that your choice was right or wrong. Making a new choice, however, can have rewards aplenty.
OK, i know that this question has been completely beaten to death, but i have a very simple way of looking at it.(IN 5 SIMPLE POINTS)
How can those of us at the Always Switch School convince you? People from the Information Has No Value School have lined up one by one and argued with Manduck and the rest, and eventually given up the argument, only to be replaced by someone else.
You have a two-thirds chance of picking a losing door. Monty always opens a losing door, which is not the same as the one you are standing in front of. Monty does not open a door at random (in which case it would have a 33% chance of being the winner, allowing you to just pick the car … not a very great challenge). Thus Monty has added information about which doors are losers. If you know one door is a winner, knowing that another is a loser adds information, which is highly prized by those of us outside the Information Has No Value School. So assuming you’re wrong to begin with, you will win if you switch.
It’s a 67% likely assumption. Assuming that you are right, upon guessing, is a 33% likely assumption.
Or are you guys seriously arguing that you have a 50-50 chance of picking the winning door out of the three closed doors?
If door #3 is eliminated, that means it has a 0% of being the winner, not a 33% chance. Your first guess has a 33% chance of being right, not a 50% chance of being right. Switching to the door Monty did not open gives an overall 67% chance of winning.