you guys probably know all about this so please excuse the newbie.
but there’s something I’d like someone to explain to me if possible, because I still can’t really get my head around it.
Using the scenario of 3 doors: 2 with goats behind and 1 with a car… let’s just imagine a game where you didn’t select a door at the start. The host carries on and opens a door with a goat behind it. So then two doors remain: one with a goat behind, one with a car. You didn’t select a door at the start, so you can’t “switch”, therefore the odds of now choosing the ‘car’ door from the remaining two doors are 1/2, regardless of which one you pick.
And yet somehow, if you’d selected a door at the start, the host is still obliged to open a door with a goat behind it, and yet you apparently now have the advantage that switching your original choice of door will give you a 2/3 chance of winning the car.
so my question is: how does a hypothetical door selection - which can’t influence the host’s decision to open a door with a goat behind it, because the host HAS to open a door with a goat behind it - somehow increase your chances of winning the car to 2/3 if you “switch”?
I’ve read all the blurb and examined all the diagrams and charts, but I still don’t understand this part, so where am I going wrong, and what am I not understanding? :smack:
In your example, the host may open your selected door, in which case… I think I’d agree that you have a 50/50 chance at that point.
But not otherwise. My favourite explanation is to imagine 100 doors with 99 goats, and after you select a door, the host opens 98 of them. You’d feel pretty strongly that you should switch at that point. Then just reduce the problem size.
^ thanks for the link… probably what I should’ve searched for first.
yeah I get that, but only when there are more than 4 doors to start with… I won’t type anything else till I’ve read that link though and hopefully resolved my confusion.
Unfortunately for me - even though I think I am smart - that explanation of opening 98 doors didn’t actually register with me - it still seemed to me to be 50/50
It took me a good day or so when the whole Marilyn/Cecil thing happened before I could convince myself you should always switch. I actually had to resort to the whole simulation thing before I got it.
While the 98 doors is a great explanation - some of us are dumb enough where even THAT didn’t make the light bulb go on…
Now it’s easy for me to look down my nose at other people that don’t get it
It comes down to this. Call the door you choose Door 1, and assume that the car has an equal chance of being behind Door 1, 2 or 3, and Monty knows where the car is, and must open a door which can be neither your door nor the car. Then:
Car is behind Door 3: Monty has no choice but to open Door 2
Car is behind Door 2: Monty has no choice but to open Door 3
Car is behind Door 1: Monty chooses randomly between Door 2 and Door 3.
As you can see, two times out of three Monty opens the only door he can, and the car is behind the door he could not open. One time out of three, he had a choice. If you therefore act on the assumption that Monty opened the only door he could, you will be winning two times out of three.
Or look again at those three choices above: Half the time, Monty will be opening Door 2. It will either be because he had to (one-third of all possible trials) or because he randomly chose to (he has a random choice one third of the time, and will randomly choose Door 2 half of these times; that is, one-sixth of all possible trials). When he opens Door 2, the odds are two to one in your favour that he is opening it because he has to.
ok read it… not really seen my specific problem addressed.
how can I put it more simply…
(and I’m only talking about the 3 door scenario here)
regardless of what door you select at the start, you already know 100% that whether you choose the door with a car, or one of the doors with a goat, the host is going to open one of the doors that has a goat behind it.
Therefore surely it would make no difference if he opened a door with a goat behind it before you’ve selected a door? leaving you with two doors and a 1/2 chance of winning. and yet apparently if he does it after you’ve selected, you can change your mind and increase odds of winning?
I’ve been told that a computer program demonstrates that the 3 Door “switch” technique works, and I’m not about to argue with a computer, but I seriously cannot get this thing to make sense in my head!
But the two scenarios are different. If you pick a door with a goat behind it at first, it limits Monty to one possibility—he will open the door with the other goat. If you didn’t pick, he might open either door with a goat. That doesn’t answer your question as to why it matters, but surely you can see that the potential outcomes are at least different. When you select a door and pick a goat, Monty has no choice as to what to do. When you don’t select a door, he always has a choice.
Now, as to why it matters, look at it this way. If you switch doors, you will always get the prize that is the opposite of what you selected at first. If you picked a goat, and he reveals the other goat, switching will get you the car because that’s all that’s left. But if you picked a car, by switching you will always get a goat. It’s impossible to switch from a goat to another goat, or a car to another car. So, since you’re going to initially pick a goat two times out of three, by switching you flip the result so you instead get the car two times out of three.
ah…thanks. that’s how I needed someone to spell it out it for me. I understand this explanation better than anything in that wiki page to be honest. I could understand the concept as far as 1000 doors or 50 doors went but not 3. that makes it a lot clearer, thank you.
The hypothetical choice isn’t the help , on its own.
Its what Monty DOES that helps you.
When Monty opens a door BEFORE you choose a door, he’s leaving you with a 50/50 chance. Simple.
When you select one, and then Monty removes the goat door, and then offers you the remaining door, he is offering you the best of the remaining doors. The best of two doors is worth TWO doors when there is only one prize. That is also explaining the “information” question… the information Monty provides is “this door is the best of the two!”.
Exactly. When Monty picks first, he has an unrestricted choice. When you pick first, Monty has a restricted choice two times out of three if the rules are as we’ve stated them.
Interesting to note that one of Cecil’s correspondents on this question is now “a Director of Credit Suisse in the Private Banking & Wealth Management division, based in New York. He is the Head of Research of Alternative Index Replication.”
Pay attention to Cecil. It might help your career.
It occurs to me that there is a slight variant on the game that might help the explanation as well.
Image there are a two of contestants in the game. (Perhaps there has been a draw in a preceding selection game.) You are one of the two, and I am other other.
Monty asks you to select a door. He then turns to me, and asks me to select anther door, but just as I am about to do so, he says, hang on, “I like you, so I’ll make your choice easier.” He opens a door with a goat behind, leaving me with no useful choice of the doors to choose - but since you have already chosen your door, you can’t change. You had three to choose from, I have none. Do you feel pissed off? Would you like to swap places with me?
Francis, that’s an interesting twist and a good way to look at it Not sure if it would help someone who didn’t already grok the solution, but it is yet another good way of thinking about why it’s better to switch.
What are chances of choosing a “goat” door. Two out of three, right? If you succeed in choosing a “goat” door and always switch you automatically win. If you happened to choose the “car” door and always switch, you automatically lose. So playing the “always switch” strategy means you win 2/3 of the time. Obviously, playing the “never switch” strategy means you win only the 1/3 of the time you guessed right the first time.
Suppose you play a mixed strategy of rolling a die and switching when you roll a 1 or 2 and standing pat otherwise? In that case, a similar analysis will show that you win 2/9 + 2/9 = 4/9 of the time. For this game, no mixed strategy can beat the pure strategy of always switching.
I think that at least some of the difficulty of this problem is due to the fact that you feel psychologically disappointed if you guessed right the first time and then switched to lose. You should redefine the problem as that of choosing a goat the first time (yum).
Yeah, and that’s why my original point was basically: if his unrestricted choice = a goat, and his restricted choice also = a goat, how can be of any use to you, because you don’t know where any of the boxes are anyway. He’d be bound to pick a goat in either scenario.
But the way Mr. Shark explained it has made me see the light… hallelujah! (now where are my glasses?)
ETA: Yet another way to look at it is that Monty offers you either the door you first picked or both the other two. You’d swap in a heartbeat - even when Monty shows you that one of the doors he has just given you is useless; you already knew one of them had to be.
You have more information; it is the added information (that he didn’t pick your door) which ultimately increases your odds.
The computer will show that out of 100 plays, you would win approximately 66 times by switching in the original game. Under your scenario, it would likely show a winning rate of 50 times as you predict.
