If a goat “HAS to be revealed” then that implies Monty must know where the car is, otherwise there’s a chance the car will be revealed instead. That makes a difference.
Monty Hall II Part II
More simply:
1/3 of the time the goat is revealed because C has the car.
1/3 of the time a goat is revealed because H has the car
1/3 of the time the car is revealed.
The first two reasons for the goat being revealed are equally likely therefore there is no gain from switching in Monty Hall II.
“If a goat “HAS to be revealed” then that implies Monty must know where the car is, otherwise there’s a chance the car will be revealed instead. That makes a difference.”
I stated that the goat must be revealed for the switch to be an advantage, knowledge of where the car or goat are doesn’t change that FACT. I’ve shown you this multiple times now. Monty doesn’t even have to be the one to open the door, as long as the result is the goat, the switch is an advantage, you’ve stated this yourself.
Mike stated @872: "knowledge of where the car or goat are doesn’t change that FACT [that the goat must be revealed to be an advantage] "
It’s not the knowledge that matters, Mike, it is how the first door came to be opened. Purely by chance.
But you’re not one of those people who only absorb posts that accord with their view, are you?
@871 I explained in fairly simple terms how the chances of any door containing the goat were 33% before the reveal and 50% for the each of the remaining doors after the reveal.
I also explained that unlike in Monty Hall, in this random version of the game, aka Monty Hall II, the contestant is eliminated, upfront, in 33% of games by the revelation of the car, something that cannot occur in the standard version of the game. In other words, this leaves the contestant with a 1/3 chance of winning the car, or 50% of the 2/3 of games they still have a chance in of winning.
These are actually facts that should give anyone who thinks a reveal is a reveal is a reveal considerable pause.
Perhaps considering the situation with many more doors, which has helped people understand MHP in the past, will also assist here:
You choose one door from 100.
At random the host opens 98 other doors.
There is a 98% chance the car door will be among the doors randomly opened by the host.
Ergo, in the 100 door game, you are eliminated upfront 98% of the time and win 50% of the 2 times when the car is not revealed upfront, ie 1/100…ie chances of winning in Monty Hall 2 = 1/n, where n = number doors.
Of course in the traditional version you’re “in the game” 100 times, not just twice, and you win 99% of the time when you switch. In the random version, you win once whether you stick or switch. It makes no difference what strategy you adopt. QED.
Capisci?
“It’s not the knowledge that matters, Mike, it is how the first door came to be opened. Purely by chance.”
This was my original statement, knowledge of the reveal doesn’t matter, revealing the goat is what matters to get the 2/3 advantage on the switch, and I’ve been told I was wrong multiple times.
Just answer me this one question, Palmer won’t. Monte tells you to pick one door and you choose door #1, he then asks someone in the audience without knowledge of the contents of both remaining doors to open one and they reveal a goat, do you switch to the last door?
I switch because the probability that the car is behind the 3rd door is 2/3, the probability that it is behind my original door is 1/3. That’s it.
I do appreciate the banter with you guys, Jonathan and Palmer, where did all the 50/50 people go, did they all see the light?
Mike, in relation to your question @874, I refer you to my explanation @873 and, in particular the 1/100 analogy therein.
Specifically, however, and as stated @871:
- 33% of the time the audience member will kill the contest by revealing the car;
- 33% of the time you will have the car already, so switching will be a losing strategy; and
- 33% of the time you will not have the car, so switching will be a winning strategy;
The likelihoods of your door or the other closed door concealing the car being no different to one another, there is no benefit or disadvantage to switching.
Therefore, I recommend stitching.
@Mike
There IS a difference between saying “a goat door HAS to be opened” and “a goat door IS opened”. The 1st implies certainty the 2nd doesn’t. As I said earlier the 33.3% of games that you lose immediately when Monty reveals the car are games you would normally win by switching.
If a member of the audience opens Door3 (say) then it is equally likely for the car to be behind Door1 or Door2
If you pick Door1 300 times the audience member will open Door2 150 times and Door3 150 times. Of the 150 times he opens Door2: 50 times the car is behind Door1, 50 times behind Door2 and 50 times behind Door3. The same for the 150 games he opens Door3. Pick ANY game at random where a goat is revealed and there’s a 50% chance you’ve already picked the car.
Just answer me this one question. Monte tells you to pick one door and you choose door #1, he then asks someone in the audience without knowledge of the contents of both remaining doors to open one and they reveal a goat, do you switch to the last door?
This is the only scenario I am asking about. You guys are adding the reveal of the car and I’m not. I don’t know how more simple I can make this, when a goat is revealed, knowledge or no knowledge, then the decision to switch to the other door is always a better decision. That’s it!
“There IS a difference between saying “a goat door HAS to be opened” and “a goat door IS opened”
I’m saying both, the goat door HAS to be opened for the decision to gain the 2/3 advantage.
If a goat door IS opened then the decision to switch gains the 2/3 advantage.
I contend both are true regardless of knowing the contents of the doors before they are opened.
When Monty reveals a goat because there is a100% chance of one of his two doors concealing a goat.
When an audience member randomly selects a door and reveals a goat, it is because there was a 67% chance they would choose a goat door. It is unsurprising in this single game of yours that a goat, and not a car, is revealed as this outcome is twice as likely as any other possible outcome.
However, the chances of the other two doors concealing a car remain 33%.
67% of the time Monty chooses a particular door because he had no other choice and that is what makes his choice so instructive.
In the computer simulator above, I know that when Monty chooses the right-most of two available doors, there is a 100% chance switching will win.
When Monty chooses the left most door, there is a 67% chance the car is behind the door he did not choose.
This is valuable information.
When I door is chosen without an such criteria, the revelation of a goat is quite meaningless.
If Door 2 is chosen it does not mean that door 3 has a car, and vice versa.
Therefore, the same conclusions cannot be drawn about a selection unconstrained by the rules of the game and a selection made by Monty which has regard for what is behind each door.
You’re looking at goat #2, Ben. Now what?
But to summarise that interesting explanation:
If the Monty Hall problem was a 50/50 proposition, the contestant, on average, would win the car 50% of the time by “doing nothing”. But you can only win a one-in-three guessing game 33.33% of the time so doing nothing cannot win 50% of the time, only 33.33% of the time. Therefore the probability of the only other door concealing the car must be 1 - 33.33%, or 67.67%.
you’re confusing the aspect of “winning” with the probability of choosing between two doors. The probability will always remain at 50% because you only have two choices. The “winning” percentage is 67% because you accounted for the 3 doors and by eliminating one the remaining percentage is 67 %.
I refer to certain claims above that a single door cannot have 2/3 probability of the car and the switch is to two doors, not one. This is untrue, and the probability of the single unaddressed door having the car changes from 1/3 to 2/3 when Monty has revealed a goat.
899 "I presume you are quoting yourself."
Sorry, I just clicked ‘post’ without filling my name in so my comment 898 came up ‘Anonymous’. I’m not actually the other ‘Anonymous’, whose comment 893 I was responding to by asking what their odds are of making the right decision (they are of course 1/3 by sticking, 2/3 by switching). I used the same term as Anonymous, odds, but could equally have called it probability, chances etc.
Incidentally, I’m not sure if Anonymous is being ironic or falling for gambler’s fallacy in 896 ‘pick the door that last won then switch since the odds of it being there multiple times in a row is statistically less’.
Happy 93rd birthday Monty Hall 8/25/15
Yes hindsight is a wonderful thing. And this has precisely what to do with the Monty Hall Problem? No one said you’re guaranteed to win the car by switching doors
To say that playing the odds is “a great way to make a wrong decision with confidence” is pure sophistry. The more accurate, and honest, statement is that playing the odds is a great way to make a decision with 67% confidence that it will yield a car.
As any frequentist would, Anonymous seems to argue against the utility of probability where only one event, or game, as in this case, is in prospect. Seemingly swayed by the closeness of the probabilities of 33.33% and 50%, he/she reasons that only the outcome matters, not the probabilities, as you are not engaging in enough instances of the game to reflect the probability. So, they reason, if you play the odds and switch and lose, the case is proved.
However, if you were addressing a room of 100 prospective “Let’s make a Deal” TV show contestants who each had their one shot in the locker to win the car, would advising them that it matters not whether they stick or switch, on their one opportunity to do so, be wise, fair and helpful advice?
The laws of probability tell us that if no contestant switched, we would expect 33 of those 100 contestants to win a car and if everyone switched, we would expect 67 of those 100 contestants to win a car.
It does not matter that each contestant only gets to play once; the more contestants that switch the more contestants will win a car.
Accordingly, the wiser, fairer and more helpful way to advise each of the 100 contestants who will each play the game only once is to consistently tell each one of them that they should switch when provided the opportunity to do so.
In all probability, 33 will hate you and 67 will thank you for that advice, statistically speaking.
Reduction ad absurdum:
Let’s assume that you have an INFINITE number of doors to choose from. It should be pretty obvious that you have NO (unless you want to nitpick) chance of picking the RIGHT door.
You pick a door.
Monty Hall picks another door.
All of the other doors (an infinite number less 2) now swing open, and you become desperate to escape the infinite-2 goats.
You now know that the door with a car MUST be either yours or Monty’s, and you already know that your door, well, can’t possibly (unless you want to split hairs) be the One True Door. So… do you switch or not? Do I really need to ask?
Even when knowing the reasons why we sould switch, the mind still thinks the odds are 50-50 for both doors.