That’s why it’s unintuitive. The problem as stated is perfectly clear.
I’ve been convinced for years that Marilyn vos Savant and everyone else who claimed it was better to switch doors was just, well, wrong. Your chances are 50/50! One out of two! What’s wrong with you people?
Well, TriPolar finally explained it so it made sense to me.
Now I get it. By opening that empty door, the 33% chance associated with that door has now been transferred to the other door you didn’t pick, giving it a 66% probability. Finally … Ignorance has been fought! Thank you, TriPolar!
If these examples don’t sink in, consider a restructuring of the problem:
Your friend has a deck of cards and offers you a “totally fair” game of chance. Whoever has the Ace of Spades wins. The catch? You get one draw and he gets the rest of the deck. You’d call bullshit pretty fast, right? Well, what if after you drew your card and looked at it, he said “WAIT, how about we trade piles?” You’d absolutely take that. You’d have to be loony not to.
It’s exactly the same problem, you know 51 of those cards are losers, and your blind draw is a 1/52 chance it’s the right one. If you take the deck you have a hell of a better chance.
So I wrote a little program to calculate your winnings if you always switch. Writing it reveals the huge reason that this problem holds – Monty’s choice is dependent on yours… and he’s omniscient.
int NUM_DOORS = 3;
int ITERATIONS = 1000000;
int winCounter = 0;
for(int i = 0; i < ITERATIONS; i++){
Random rand = new Random();
int winningDoor = rand.nextInt(NUM_DOORS);
int yourDoor = rand.nextInt(NUM_DOORS);
int montysDoor;
if(yourDoor == winningDoor){
montysDoor = rand.nextInt(NUM_DOORS);
while(montysDoor == yourDoor)
montysDoor = rand.nextInt(NUM_DOORS);
} else {
montysDoor = winningDoor
}
// Because we switched doors, we win if monty won
if(montysDoor == winningDoor){
++winCounter;
}
}
Here’s the thing, you have a 1/NUM_DOORS chance of picking the winning door. Monty, being omniscient, always picks the winner if you haven’t. That’s a NUM_DOORS-1/NUM_DOORS chance (i.e. 2/3 in a 3 door game). What Monty is basically asking you is if you want to trust omniscience or blind luck. He ALWAYS gets the winning door if you didn’t trip on it.
Fun fact about the trials – the more doors, the better it is to switch. A game with two doors causes you to win about 50% of the time by switching (you pick a door, Monty picks the other, you switch). 3 doors is always 65-67%, but usually 66.x% (remember, I’m doing a million iterations here). If you have about 100 doors, results are usually in the 99% range. Once you get one million doors, it’s not uncommon for switching to win 100 percent of the time, though it’s usually in the upper 99 percent range.
The exact probability of winning by switching is 1 - 1/NUM_DOORS.
Which is also NUM_DOORS-1/NUM_DOORS, isn’t it? Maybe I should have used parens, I meant (NUM_DOORS-1)/NUM_DOORS
Your choice is 50/50. Joe’s choice is 2/3 vs 1/3. And there’s no problem.
Consider an alternate arrangement. I’m thinking of a number. I tell Bob to choose a number between 1 and 1,000. He chooses 538. I tell him that the number is either 538 or 48; which would he like to choose? If he switches, he’s got a 99.9% chance of being right; if he stays with his guess, he’s got a 0.1% chance of being right.
Later, knowing that you didn’t overhear my conversation with Bob, I give you a different choice: 48 or 538? You’ve got a 50% chance of being right with either guess.
If you don’t believe me, here’s the experiment to try with a friend:
Step 1) Visit this website and generate 2 numbers between 1 and 1,000. Call them “target” and “backup.”
Step 2) Ask a friend to guess a number between 1 and 1,000.
Step 3) If they guess correctly, tell them, “Your number is either what you guessed, or it’s [backup]. Which do you think?”
Step 4) If they guess incorrectly, tell them, “Your number is either what you guessed, or it’s [target]. Which do you think?”
Step 5) Record their guess.
Step 6) Put the friend’s guess and either [target] or [backup] (depending on whichever was the other number you told to the first friend) in envelopes and ask another friend to choose an envelope.
Step 7) Record their guess.
If your first friend switches their guess, of course, 99.9% of the time they’ll be right. But the second friend doesn’t have that information, so they’ll only be right 50% of the time.
Dammit! :mad:
Actually, that’s pretty interesting. I was vaguely worried when I made that dataset that its lack of true randomness would cause a problem, but I didn’t really think about what the problem would look like.
No, it’s intentionally misleading. That’s the entire point of the problem, to be misleading and trick people into giving the “wrong” answer.
The problem, clearly stated, would be “There are 3 doors. You can either pick 1 door or 2”. In that case it’s blindingly obvious what the correct strategy is. But that’s NOT how people present the problem, so they’re being intentionally misleading.
Calling it the Monty Hall problem or otherwise presenting it in terms of the game show adds a layer of obfuscation that is clearly designed to mislead people. It implies that there is an intelligent actor who’s making decisions to hold the odds relatively even (or at least appear even). But once someone gives the “wrong” answer, they pull the rug out and say no, haha, the problem’s actually different, there is no intelligent actor and the player always gets a chance to switch, so you’re wrong and stupid. That is entirely the point of presenting the problem this way. If it wasn’t, they’d state it clearly and no one would ever give the “wrong” answer or find it confusing.
So kenetic, can I assume you’re also not a fan of the “two people are dead in a cabin in the woods. What happened?” puzzles?
Well, if one of them was Monty Hall…
No, you’ve misunderstood the whole exercise. If you explain the problem completely (as you are supposed to), outlining that MH always opens an empty door, people still make the wrong choice. It’s because of that behavior that this is an interesting problem. If everyone made the right choice no one would be talking about it. Common sense is misleading people, that’s what makes this problem so interesting. The probability math is trivial once people understand the implications.
First, I’d make sure that they were dead.
I agree with kenetic. I don’t know how to exactly describe what I’m saying about it being misleading, but it’s presented more like a riddle than a math problem. Many of the components have nothing to do with the actual problem.
I forgot to mention before the part where every description leaves out the key factor that Monty always shows you an empty door (an important factor, one that common sense could explain, but usually isn’t explicit in the description). The term ‘switch’ is constantly used, even though there is no switch. The choice hasn’t been made, so there is no switch. The first door pick is not a choice, it’s just a means to seperate the doors into two sets, one with 1 door, the other with 2 doors. The contestant doesn’t have to name that door either. The problem is the same if Monty names that door, or if it’s chosen at random. After all, the contestant has no idea what door the prize is behind. The reveal of an empty door after the designation of the first door is also misleading. This could be done after the actual choice (the ‘switch’ as it’s called), without changing the problem either.
But the bigger problem is in the explanations. These are full of the concept of a change in probability following the ‘switch’. Nothing changes, from the beginning you have better chances by picking two doors instead of one. And I’ve seen a couple of explanations that are flat out wrong, claiming the chances go from 1 in 3, to 1 in 2. The chances of picking the winning door are always 2 in 3, unless you foolishly choose the 1 door option with a 1 in 3 chance of winning.
So this is a great ‘word problem’ for probability students, forcing them to deconstruct the problem to understand it. But if you’re explaining this to people, it’s just misleading, and I doubt many people who didn’t understand it before are educated by most of the ‘solutions’.
I used a similar analogy to get my nephew to understand it. I told him the Ace of Spades was the winner and spread the deck out on the coffee table. I told him to select a card but don’t look at it yet. Then, playing the host, I went through the rest of the deck and turned over 50 of the 51 remaining cards. There were now two face down cards left, mine and his. I then offered him a choice of sticking or switching and the light bulb of understanding came on pretty quick.
And now you join the ranks of those who can’t see how people don’t get it no matter how you try to explain it.
A true story:
At a job interview, a friend of mine was asked a version of the Monty Hall problem. When the interviewer asked if she would stay with her original choice or switch, she said she would stay with her original choice and she refused to change her mind.
She thought it meant to test of her loyalty, not her analytical skill!
The reason why pigeons have an easier time with this, and the reason why every time the Monty Hall problem pops up the thread gets hundreds of argumentative replies is because of the difference in how a pigeon and a human think.
The pigeon just wants the prize. That’s the goal. When it learns that what it’s doing doesn’t get the prize it’ll switch because it wants the prize.
Humans don’t care about the prize. What they want is to outsmart the host. And they think that because the host is trying to trick them, there’s a real need to outthink the host as well.
Look at some of the replies given in this thread to prove my case.
What I think helps to clarify this problem (not all the time, mind you) is to state upfront that Monty Hall will ALWAYS offer you a second choice, he will always open a door, regardless of what door you originally picked.
What happens when you set up the problem as a singular test is that people start to think that the only reason Monty opened another door is because you managed to pick the right door and he’s trying to get you to lose.
And, honestly, given what we know of Let’s Make a Deal and given just that one hypothetical in a vacuum, it’s not an unreasonable assumption to make.
What if that door’s father is my father’s son?
Almost. Remove the part I bolded, and it becomes the same problem. The person who chooses the door doesn’t get to look behind it before deciding whether or not to switch.
If you get to look at your card first, you have a 100% certainty of getting the ace of spaces. If you don’t, then your chances of winning are 1 in 52 if you stay, and 51 in 52 if you switch.
Yeah, I meant to write “but before you looked at it.” :smack: