I said:
Cabbage said:
Isn’t this the same position as our contestant was in with Monty Hall? The number of doors Monty has left closed diminishes, but the chance that he’ll win isn’t affected. Are you claiming that because the car was hidden ahead of time, that makes a difference?
Al, I was just wondering how you respond to me slight reworking of the problem above. If Monty gives you a choice of opening one door, or two, which would you take? This has the (completely irrelevant) caveat that, if you choose to open two doors, Monty will be the guy who turns the knob and yanks on one of them, and if the car is behind Monty’s door, you still get it. How is this different from the original Monty-always-opens-a-losing-door puzzle?
It’s not different at all. Switching after Monty adds info doubles your chances.
Yes, Tominator2, the fact that the car was placed behind a specific door ahead of time does affect the results. For your raffle analogy to be applied to the Monty Hall problem, it would be akin to placing the car behind one of the two doors left, after Monty opens one of the empty doors. That results in the chance of the contenstant winning being 1/2, regardless of switching doors or not. Does that make sense to you now?
Boris, your variation is quite different from the original premise. If Monty offers you both doors instead of the one you picked you would be deranged not to take it. It is irrelevant that Monty opens one of the doors. A scantily clad bimbo would be the most desired door opener. There’s a huge difference between (a) Monty opening one losing door and you having the choice of keeping your door or switching to the other (which is a clear 50/50 coin toss when the smoke and mirrors are done away with and you are the poor car-lusting chap on the spot), and (b) Monty offering you the chance to switch to the other two doors to win whatever is behind both (2 out of three times the car is going to be behind one of the other two doors).
In scenario (a) half the time you switch you will lose and in scenario (b) 2 out of three times you switch you will win.
As for you, my little Cabbage (I call my wife “my little cabbage” sometimes) you missed something. In three out of the nine possible combinations (of the prize door and the initial door selection by the contestant) the door picked and the prize door coincide. In those three scenarios, and only those three, Monty has the freedom and intellectual licence to actually pick between two possible doors to open. In the other six there is only one door that meets Monty’s conditions - a losing door that was not picked by the contestant. That’s why Monty has twelve possible door opening combinations rather than the basic nine scenarios, because, in the three cases where the winner was picked correctly Monty has two options. In all six cases where the contestant picked right and Monty opens either losing door A or losing door B, the contestant will lose if switching. Look at it again. The three scenarios where Monty has free choice of two doors count as two each because Monty can create two possible combinations in each scenario. If you figure this out you will reach 50/50 and you will agree that there is no benefit to switching.
pick pick pick
zip zip car zip zip car zip zip car
zip car zip zip car zip zip car zip
car zip zip car zip zip car zip zip
These are the nine possible combinations. In the first group, third row, Monty can choose door zip or door zip, and similarly in the second group, second row, and third group, first row. In the other six there is only one door Monty can open. The choice of two increases the number of possibilities to twelve and, whether you switch or stay you will win six times and lose six times.
How about on “Who wants to be a Millionaire” when a contestant is completely stumped, takes his fifty-fifty, and two wrong answers are deleted from the four choices. Then he makes his wild-assed guess and Regis says “final answer?” The odds went from 3:1 to 1:1 and now the contestant has a chance to switch. Is he going to improve his odds by switching his WAG? No, even though two answers have been revealed as incorrect.
I hope my little table posts OK.
Shucks, the table didn’t post. You’ll have to take pencil and paper and draw it yourself. Sooo twentieth century.
Al Zheimers, I will try to explain it another way.
There are three doors, A, B, C.
Assumption: the prize is behind door A.
You (the viewer) pick at random.
therefore:
1/3 probability you pick door A.
1/3 probability you pick door B.
1/3 probability you pick door C.
Case 1:
My strategy is : don’t switch after Monty opens a door with no prize.
Possibilities:
I pick A, Monty opens B or C, I stay with A. I win (probability 1/3).
I pick B, Monty opens C, I stay with B. I lose (probability 1/3).
I pick C, Monty picks B, I stay with C. I lose (probability 1/3).
Therefore in 2/3 of the cases above, I lost.
Case 2:
My strategy is : switch after Monty opens a door with no prize.
Possibilities:
I pick A, Monty opens B or C, I switch to C or B. I lose (probability 1/3).
I pick B, Monty opens C, I switch to A. I win (probability 1/3).
I pick C, Monty picks B, I switch to A. I win (probability 1/3).
Therefore in 2/3 of the cases above, I win.
The example above makes it clear (in my mind at least) why you should switch.
Or are you saying that Cecil Adams is wrong? 
Mr. Zeimers, pardon me if I’m flogging a dead horse, but I just can’t see how the two are different.
Situation A (as originally conceived):
You pick a door and stand if front of it. Monty opens one of the remaining two doors, but only a losing one. Then he asks you if you want to switch.
Situation B (my rewording):
You pick a door and stand in front of it. Monty asks you if you’d rather open the other two doors. If you say yes, he opens one of them; if either of you reveal the car, you win it.
The only difference is, if your initial guess is wrong, in Situation A you are guaranteed to reveal the car; in Situation B, either you or Monty might reveal the car, but you win it in either case (it would be kind of weird for a show employee to drive home with a prize…).
The reason that the odds increase in Who Wants to be a Millioniare is because the two choices removed are completely independent of the original WAG. If the computer removed two choices, and those two choices could not be the contestant’s first choice, the contestant should switch. And, if the contestant is smart, will select the least likely answer, since if he is right that his first choice is wrong, the remaining choice will be correct.
Now back to the Monty Hall problem. If and only if you choose the correct door at first it will be to your benefit to stay. To expand on what Arnold said:
Assume that you choose door #1.
There is a 1/3 chance that the car is behind this door. In this case, Monty will open door #2 half the time and door #3 the other half. Each of these occurences has a 1/6 chance of happening. Just because there are extra occurences that can happen in this situation, doesn’t change the probablility of being in this situation in the first place.
If you were to play six times, and choose door #1 each time, this is what would happen:
Out of six tries, the car will be behind each door twice. Now that this has been explained from every possible angle, can we please just put it to rest. At least read Cecil’s take on it. If you persist in thinking that the chances are 50/50, I will be forced to think that you haven’t read the various explanations, and will not pay any attention to your claims of “fighting ignorance”.
So, if I’ve picked door 1, the possible outcomes are:
(1) The car is behind door 1, Monty opens door 2, if I stay I win, if I switch I lose.
(2) The car is behind door 1, Monty opens door 3, if I stay I win, if I switch I lose.
(3) The car is behind door 2, Monty opens door 3, if I stay I lose, if I switch I win.
(4) The car is behind door 3, Monty opens door 2, if I stay I lose, if I switch I win.
Are there any other possible outcomes?
Now, looking back at my representative sample above, I see that in two of the four times I exercise my option to stay I lose and in two I win. I also see that in two of the four times I exercise my option to switch I lose and two I win.
Must I write out the other eight possible scenarios based on the selection of door two or door 3? Okay.
If I’ve picked door 2, the possible outcomes are:
(1) The car is behind door 1, Monty opens door 3, if I stay I lose, if I switch I win.
(2) The car is behind door 2, Monty opens door 1, if I stay I win, if I switch I lose.
(3) The car is behind door 2, Monty opens door 3, if I stay I win, if I switch I lose.
(4) The car is behind door 3, Monty opens door 1, if I stay I lose, if I switch I win.
If I’ve picked door 3, the possible outcomes are:
(1) The car is behind door 1, Monty opens door 2, if I stay I lose, if I switch I win.
(2) The car is behind door 2, Monty opens door 1, if I stay I lose, if I switch I win.
(3) The car is behind door 3, Monty opens door 1, if I stay I win, if I switch I lose.
(4) The car is behind door 3, Monty opens door 2, if I stay I win, if I switch I lose.
As per my previous post, it is preordained that Monty is limited to twelve, and only twelve, possible different door opening combinations. They are now all listed here. Reviewing all twelve I see that, if I switch, I win six and lose six. If I stay I win six and lose six. This, to me, clearly indicates that there is no advantage to switching.
I do study your posts. If this thread frustrates or annoys you then there are plenty of other threads to go to. The fight against human ignorance includes fighting my own ignorance and I am willing to concede defeat as soon as I see a definitive and convincing explanation. I expect the same from you.
I maintain my position that there is no advantage to switching, based on the complete lists in this post of all possible outcomes. I await a clear explanation of the flaw(s) in my analysis.
Al Zheimers:
OK, let me try it from this angle. You are correct, there are exactly 12 different possible outcomes, 6 where you win, 6 where you lose. That’s not enough, however, to conclude that your odds of winning are 50%–you must also consider the probability of each of those 12 events occuring. For an (unrelated) example, consider a six-sided die. You win if you roll an even number, lose on odd. 6 outcomes–3 you win, 3 you lose. 50% chance of winning, right? Not if it’s loaded.
I’ll limit it to these 4 cases where you pick door 1 (picking door 2 or door 3 will work out the same way).
In the first two cases, the car is behind door 1, so Monty has a choice of opening door 2 or door 3. What will dictate his choice? For simplicity’s sake, let’s say it’s the flip of a fair coin. Heads–open door 2, Tails–open door 3.
We’ll calculate the probability of winning assuming you stick with your original door.
(1) The car is behind door 1 AND Monty flips heads (opens door 2). The probability of the car being behind door 1 is 1/3. Exactly half of that time, Monty will flip heads. Half of 1/3 is 1/6, so the probability of this event happening is 1/6. If you stay you win, so this event gives a 1/6 chance of winning.
(2) The car is behind door 1 AND Monty flips tails (opens door 3). The probability of the car being behind door 1 is 1/3. Exactly half of that time, Monty will flip tails. Half of 1/3 is 1/6, so the probability of this event happening is 1/6. If you stay you win, so this event gives a 1/6 chance of winning.
(3) The car is behind door 2. The probability of this is 1/3. If you stay you lose, so this event gives 0 chance of winning.
(4) The car is behind door 3. The probability of this is 1/3. If you stay you lose, so this event gives 0 chance of winning.
To sum up, if you stay you have a 1/6 + 1/6 = 1/3 chance of winning. (Therefore, if you switch you have a 2/3 chance of winning).
The point that’s been missed is that each of (1) and (2) are only half as likely as (3) or (4).
Al, I can relate to what you are saying. I also once tried to work out all the possible outcomes as you did, and like you mistakenly concluded the odds were the same. The error we made in our logic is that while the number of possibilities is equal the likelihood of the possibilities is not. You outline twelve possible scenarios and correctly count that in six of them you should stay and in six of them you should switch. But the reality is that the six switching scenarios are twice as likely to occur as the six staying scenarios, so your odds are twice as good if you switch.
Trust me on this one. Or if you don’t trust me, try the actual experiment.
Al Zheimers, I don’t want you to think that we’re ganging up on you. But we’re trying to make Cecil’s answer clear.
What you said above is correct, there are four possible outcomes. But where you’re mistaken is in assuming that all four outcomes have the same probability!
What are the odds that the car is behind door number 1? 1/3, because there are three doors.
Therefore, the outcomes (1) and (2) combined have a probability of 1/3, because both of those outcomes start out with “car is behind door 1.” The probabilities for all the outcomes starting out with “car is behind door 1” need to add up to 1/3.
Outcome (1) has probability 1/6.
Outcome (2) has probability 1/6.
Outcome (2) has a probability 1/3.
Outcome (3) has a probability 1/3.
So therefore “if I switch I win” beats “if I switch I lose” 2 to 1.
P.S. I just realized that I duplicated Cabbage’s post. Read that post instead, Cabbage goes into more details.
Regarding the Millionare question:
Assume you have no idea what the answer is; all four answers are equally probable. Assume your random decision is ‘A’, and then computer removes C and D. You gain no benefit from switching because, as waterj2 pointed out, the computer’s decision is independent of your guess. It is equally likely that it will remove your guess as not.
Actually, based on my observations, it seems that the retained answers are pre-defined as the correct and most plausible alternative; the choice does not seem random. Thus this lifeline is only useful when you have no idea whatsoever what the answer might be and can’t determine the two most plausible answers without the lifeline.
Now, let’s consider an alternative: You get to guess an answer, and the computer keeps that answer and removes two incorrect answers, keeping either the correct answer and a random (or most plausible) wrong answer. In that case, the Monty Hall theorem holds, and the probability of the other answer being correct becomes 3/4 and you must switch.
This would of course render the 50/50 option a “gimme” if you have any knowlege of the question at all: You merely pick an answer that seems totally implausible, and switch after the 50/50. Regardless, your odds of winning cannot be less than 3/4 (unless you fool yourself, as I did on the Exorcist question).
No matter where you go, there you are.
Excellent Smithers!
Cabbage, you are good. Your analysis is excellent. I find no fault with it.
You have selected door 1. Monty opens door 2 and offers you the chance to switch.
You think:
(1) The car is behind door 1, Monty opened door 2, if I stay I win, if I switch I lose.
Or
(2) The car is behind door 3, Monty opened door 2, if I stay I lose, if I switch I win.
Since option (1) is only half as likely to occur as option (2) the wise player will play the odds and switch, improving his odds of winning from 1:1 to 2:1.
Arnie, I never felt ganged up on. I realized I was arguing a minority opinion but I truly was in search of illumination. I’m outta this thread, a wiser man.
Cool. Thanks, glad to be of help! 
Oh My God!!! Is it over? Well, it was fun while it lasted… 
I could be wrong…it happend once before…
Wait a second! In the game, isn’t there actually only one loser? Then there is a middle prize and the big prize. What do you do in that case? And at the end of the show, isn’t there two people both doing some sort of chosing? What should they do?
Oh nevermind.
PeeQueue
Zor said:
Darn, I was afraid that might be the answer. It just seems so counter-intuitive, like the contestant could discover something they shouldn’t be able to.
Al Zheimers, does this mean we’ve brainwashed you into the Cecil-Monty Hall cult?