What if the goat is on a treadmill?
Does it have wings?
Yep. I think it’s as simple as that. With your original choice you get one door.
Monty’s offering you the other two. Not a choice between the other two but actually both of them.
Damn, this thing is slippery! I think I’ve finally got my head round it by inflating the numbers. (I don’t think that changes the basics but please tell me if I’m wrong.)
Monty starts by asking you to choose one of 100 million doors. You don’t like the odds on this at all but you choose one. He then opens all but one of the remaining 999, 999, 999 doors and then asks if you’d like to switch.
Of course, you’d be crazy not to, it’s a near certainty that you’ll win by switching. Now, reducing the initial number of doors lessens each time the advantage of switching but that advantage never disappears until you get to just 2 initial doors.
With the original problem and its 3 initial doors the advantage has come right down, but it’s still indisputably there - 2/3 as opposed to 1/2.
Someone else suggested this in another thread somewhere (or was it Cecil’s column?) and it really helps me to nail it down. But did I remember it aright, or is the problem changed by inflating numbers?
Wow, this brings back memories! I don’t understand why Monty knowing that he will reveal a goat or randomly selecting one of the two remaining doors to reveal a goat changes anything. If the friggin’ wind blows one of the remaining doors open to reveal a goat, I should switch. Yes, a random choice could well reveal a car, but that’s not what we’re seeing. In a single instance, if a goat is revealed, regardless of how or why, the odds with switching are the same, right? vos Savant is wrong, I think.
There’s a 2/3 chance I selected wrong. I should always switch if I know which of the two remaining doors NOT to pick. In the population of permutations where a goat is revealed, 2/3 of the time the other remaining door has a car–regardless of how it is revealed. What’s wrong with that logic?
You’re correct: inflating the numbers doesn’t change the logic of the problem, it just makes the differences more intuitively obvious. Cecil suggested running the problem with a deck of cards (using the ace of spades as the “prize” and everything else as goats), which has the advantage of being actually do-able.
The difference is between the Monty Hall problem (host always reveals only goats) and “Deal or No Deal” (where what is revealed is random).
This might be more obvious if we change the problem setup a bit. Suppose we have three doors, behind one is a prize, and behind the other two are goats. Now: you, aldiboronti, and I are each randomly assigned one of the doors. Each of us has 1/3 of a chance of winning, right?
Now Monty says: “the rule of this game is, I always open aldiboronti’s door.” So he does, and there’s a goat behind it.
Now, is it in your advantage to switch? After all, one of the other two doors has been revealed. BUT, if you think it’s in your advantage to switch…is it in my advantage to switch?
Let’s see.
Option 1: probability 1/3: You choose the door that has a car. It doesn’t matter which of the other two doors Monty randomly opens, because neither has the car. Switching is bad in this case.
Option 2: probability 2/3: You choose one of the doors that does not have a car. Within this scenario, Monty has a 50/50 shot of opening the door that does have a car. If he opens the door with the car, the question is moot. If he doesn’t, switching is good.
So:
Option 1: probability 1/3: switching is bad.
Option 2a: probability 1/3 (2/3 * 1/2): switching is moot.
Option 2b: probability 1/3: switching is good.
You can’t ignore the possibility that Monty might open the door with a car, because that’s 1/3 of the possible outcomes. If he opens a door randomly and doesn’t see the car, he only eliminates option 2a from consideration, leaving two equal-probability options, and you don’t know which one is right.
It’s completely counterintuitive, but the math doesn’t lie.
Think of it this way: In the original scenario, you have 2 chances out of 3 to win if you switch. But if there’s a 1/3 chance Monty will open the door with the car behind it, then that “sucks up” one of those two original chances out of three, leaving just 1/3 chance of winning by switching, instead of the original 2/3.
Powers &8^]
I don’t think any of the examples of “what if there were x doors” (with x being some large number) help clarify anything. People who think the odds are equal when there are only two doors left in the three game version are going to think the odds are equal when there are only two doors left in a 1000 door version.
Sounds like someone could make a killing playing 100-Card Monty.
Powers &8^]
Lots of people do. My point was that, to the extent you want other people to understand, it is helpful to address the point they are missing, not just multiply what appears to them to be irrelevent.
But several people have said they understood better by inflating the numbers – and that’s just here in this thread.
Powers &8^]
Yes, that is probably the simplest and clearest way to describe the situation.
That’s where the 52 card variety can really help demonstrate. Here’s how one person can run it alone. Play the game as Monty.
Take a standard deck, assign the Ace of spades as the car, all others are goats.
Shuffle well. Decide that the first card laid down will be the player’s random choice. Lay out all 52 cards face up.
The player’s card probability of being the Ace is 1/52. Now acting as Monty, look at all the other cards, and select the Ace of Spaces, and flip it over to cover it up. If the Ace is the first card (i.e. the player won), pick a card at random and cover it up.
Now is it to the player’s advantage to stay with the first card, or pick the flipped over card? Remember that the player doesn’t know what their card is. If it helps, lay the first card face down, so at the end the two face down cards are what the player doesn’t know.
In the world of statistics what is happening is that revealing the goat gives NO NEW INFORMATION with regard to the door you have picked, but it does give information about the unrevealed door you haven’t picked. Beofe the reveal the doors all have 1/3, after the reveal all the probablity of the revealed door is placed on the unrevealed, unpicked door
It was suggested in this thread. 
The basic problem isn’t changed by inflating the numbers, it just makes the odds of having guessed correctly the first time even smaller.
I don’t think there’s anything wrong with it. What Monty knows doesn’t matter once we’re at the point where the initial door has been chosen and one of the two remaining doors has been opened to reveal a goat. There’s still only a 1/3 chance that the original guess was correct, and since the already-opened door did not in fact contain the prize then there’s a 2/3 chance that the remaining door does.
As long as the game is fair it doesn’t matter if Monty knows where the prize is, as long as the actual situation is that either through chance or design the door he opens does not contain the prize. The only way Monty’s knowledge of where the prize is makes a difference is if the game is NOT fair in some way. If Monty is a trickster and only offers the chance to switch to people who guessed right the first time, then obviously switching would be a bad idea. But if we accept that the game is fair then once the goat has been revealed it’s best to switch.
That’s not true.
Suppose the game is that you choose a door, and Monty will randomly open one of the two remaining doors.
Let’s suppose you choose door A (the following logic will apply no matter which door you choose). The possibilities are:
Prize behind Monty chooses
Door A Door B
Door A Door C
[del]Door B Door B[/del]
Door B Door C
Door C Door B
[del]Door C Door C[/del]
Before Monty opens a door, all six of these possibilities are equally likely. Once Monty makes his choice and reveals a goat, you know the two struck through are no longer possible, so it must be one of the other four. Of these four, two have the prize behind your door, and two behind the “other” door…so you have no advantage to switching.
If Monty doesn’t know where the prize is, this is really the same thing as if there were two contestants playing the game. Player 1 gets to choose a door but doesn’t open it yet. Player 2 (Monty), not knowing where the prize is, then chooses and opens one of the other remaining doors. If Player 2 doesn’t win the car, Player 1 may then choose to stick with his original door or switch to the remaining unopened door.
Player 1 has a 1/3 chance of guessing correctly when he first picks a door.
Does Player 2 then have a 50/50 chance of winning the car when he opens a door? No, because there’s that 1/3 chance the car was behind the door that Player 1 picked, a door that Player 2 isn’t allowed to open now. Player 2 also has a 1/3 chance of choosing the car this round, and a 2/3 chance of getting a goat.
If Player 2 doesn’t win, Player 1 is in the same position as in the original problem. He thus has a big advantage over Player 2. There’s still a 1/3 chance that Player 1 was right in the first place and a 2/3 chance that the car is behind one of the other doors. Since Player 2 has already opened one of the other doors and revealed a goat we know the car wasn’t there.
The odds are now 1/3 that Player 1 was right about the car being behind the first door he chose, 0/3 that Player 2 was right about the car being behind the door he chose (we’ve already seen that he was wrong), and 2/3 that the car is behind the remaining door. Player 1 is better off switching.
[quote=“Lamia, post:56, topic:479220”]
If Monty doesn’t know where the prize is, this is really the same thing as if there were two contestants playing the game. Player 1 gets to choose a door but doesn’t open it yet. Player 2 (Monty), not knowing where the prize is, then chooses and opens one of the other remaining doors. If Player 2 doesn’t win the car, Player 1 may then choose to stick with his original door or switch to the remaining unopened door.
Player 1 has a 1/3 chance of guessing correctly when he first picks a door.
Does Player 2 then have a 50/50 chance of winning the car when he opens a door? No, because there’s that 1/3 chance the car was behind the door that Player 1 picked, a door that Player 2 isn’t allowed to open now. Player 2 also has a 1/3 chance of choosing the car this round, and a 2/3 chance of getting a goat.
[quote]
You are correct until here.
And here’s where you go wrong.
Think about this a second: You’re saying that in a two-person game, revealing what’s behind the third door will help one player, but not the other. The order that the players choose in–indeed, whether they choose, or are randomly assigned-- doesn’t matter At the point where each player has a door, and the third one is unopened, everyone has a 1/3 chance of winning, and it’s a parallel game. Revealing additional information can only help or hurt each player equally.
Run through the possibilities like I did in the previous post.
Not if only one player is offered the opportunity to switch doors, which is the scenario I described. Player 1 chooses but does not open a door, Player 2/Monty chooses and does open a door, and then Player 1 gets to either keep his first choice or switch.
If, after opening his first choice and finding a goat, Player 2 were offered the chance to go ahead and open the final door too then obviously he’d be better off doing so than sticking with just the door he’s already opened and knows was wrong. If Player 2 gets to open two doors then he has a 2/3 chance of winning.
This is just the same as the player in the original problem, or Player 1 in my two player scenario. He can either stick with his first choice (1/3 chance of being right), or he can get to see what’s behind BOTH other doors (2/3 chance of being right). As other posters have pointed out, it doesn’t matter that Monty is the one who opens one of the doors for him. The only way it could make a difference is if Monty/Player 2 opened the door with the car, but as the original problem is set up then this isn’t the situation the player is faced with.
Ah, I misunderstood your rephrasing.
No matter, my argument still remains–by randomly opening a door, you’re creating a situation where the two unopened doors still have equal probability.
Again, run through all the possibilities like I did above.
No, the only situation where the two unopened doors have equal probability is if Monty opens the door with the car. (This will happen 1/3 of the time if Monty chooses at random.) The two remaining doors would then have an equal chance of 0%. But in that case the game is over and we have nothing to discuss.
If this doesn’t happen, if the door Monty opens has a goat, then it tells the player nothing about what is behind the door he chose. It doesn’t magically increase his odds of having guessed right from 1/3 to 1/2. Whether he was right or wrong then he knew at least one of the other doors had to have a goat behind it, so seeing the goat doesn’t give him any new information about his own door. It only tells him that IF he was wrong with his first choice (as he would be 2/3 of the time), then the car is definitely behind the other closed door. In other words, he has a 2/3 chance of being right if he switches.
Once we’re at the point where a door has been opened and the goat has been seen, it’s exactly the same as the original problem where Monty deliberately revealed a goat. It doesn’t matter how we got there as long as the game is fair.