That’s why I showed the text of the program, so people could check my work. I’m actually barely a programmer at all, but I’m trying to learn. I invite you to look at my code and check it out. It’s not very complicated, and the syntax is close enough to a lot of languages. Heck, if you want, I can give you the Visual Basic version, which is pretty close to a flat out BASIC version.
As for scholarly articles, it’s so ubiquitous that I probably am best off just giving you the [Google Scholar link](Google Scholar"monty+hall+ problem). But I can also throw in the study I was talking about, both the full PDF form, and the abstract available on PubMed.
As for the information–Monty’s picking a door from two choices does not give you as much information as him being forced to pick a certain one. Unless you think Monty can pick the door you already have. Two not random choices and one random one is not the same as three random choices between two.
No, as qazwart explained in post #84 with Martin Gardner’s alien example, your brother’s odds would be 50/50 if he came into the game at the last decision point, without knowing how you got there.
But saying that a coin flip gets you the answer of 50/50 does not tell you what the odds are if you choose a door based on the information you have.
Here’s a simple example that I hope will help. Take a well-shuffled deck of cards. Cut the deck into two unequal piles, where one pile has about five cards and the other has the remainder. Now guess which pile is likely to have the ace of spades. I assume you agree that the large pile has approximately a 90% chance of containing it.
Now decide which pile to pick by flipping a coin. You’ll win 50% of the time. But that doesn’t tell you what the odds are if you decide with your knowledge! The coin flip is a red herring that’s throwing you off.
One other point - you’ve said that you know from the outset that Monty will reveal a door, and that door will not have the prize, so when he does this, there’s no information that you didn’t have from the beginning. But you’re mistaken here - at the beginning, you didn’t know which door he would be revealing. That makes all the difference.
dzero, it really isn’t that hard. Get a deck of cards and a friend to play Monty. Take the Queen of Diamonds as the prize (the chick with the bling), and two jokers as the goats (or pick whatever set you want to designate).
Now you look away and have Monty shuffle the three cards, check which is which, and lay them out as 1 2 3. Then you pick the first card. Now have Monty reveal a goat. Now pick a second time - switch or stay. Record the results of whether you switched or stayed and whether you won the prize or the goat.
Now repeat this for a couple dozen times, and then tally the results.
What, you don’t have a friend? Okay, version 2:
Take your three cards, shuffle them, and then lay them out. Pick your card. Now tally whether the prize was in the card you selected or the other 2 cards.
Those are equivalent scenarios.
And I think that is the root of your problem. I suspect you will argue that is not equivalent. Understanding why that is equivalent should clarify why the odds are 1/3:2/3.
The problem here is that dzero has “never included the concept of switching or not” in his analysis, so its not that he doesn’t get the odds difference between switching and staying, it’s that he’s simply not considering that as a variable.
Correct. My point is that if he happened to pick the “switch” door, those who’d seen what had gone before would know that his odds of winning would be 2/3.
It’s clear that acting without foreknowledge makes the situation similar to a coin flip, which means 50-50 odds. It’s (slightly) interesting to note that this is entirely independent of what the actual distribution up to the coin flip happened to be.
This seems to be at the root of dzero’s failure to understand - he thinks that the situation pretty much requires a coin flip, and (correctly) concludes that a coin flip means 50-50 odds.
The simplest explanation I have ever heard is this:
what if Monty didn’t revela but gave you the opportunity to switch to both doors versus your one door, i.e. he sais “Ok now you can keep what is behinnd Door 1 or you can have what is behind Doors 2 and 3” clearly the odds are favorable for switching. How does this relate to the actual problem at hand? Becasue Monty won’t reveal the good door, he has effectively only shown that one of the doors is a dud, which wew already know. And we know at least one of the doors is a dud. So by revealing one of the doors as a dud, NO NEW RELEVANT INFORMATION has been learned, therefore no change to the odds.
Right, he’s trying to analyze the situation but using an improper tool/method. I’m trying to skip the formal analysis and instead offer an easy way for him to collect data for himself. Analysis, schmanalysis, the data will give the answer, then he can figure out why that answer is the way it is.
It’s a great explanation, except it’s difficult to grasp why the two cases are equivalent. The trick is understanding what happens with the info you have and don’t have.
The odds started 1/3 for each door. 1/3 - 1/3 - 1/3
Monty now reveals a dud door. He always reveals a dud door. What happens to that 1/3 possibility?
Intuition suggests that 1/3 chance splits between the remaining two doors, and that you make a second guess at 50:50.
The Monty Hall problem was devised to demonstrate that that intuition is false. The odds on the first door were selected at 1/3 and do not change. The odds on the other two doors that are Monty’s doors sum to 2/3. Monty shows you either of his doors that is a dud, which means you get 2/3 possibility if you swap to the remaining Monty door.
Probabilities have nothing to do with right or wrong. They simply are the chance that something will happen. Let’s take the two son problem. Someone says “I have two children and one is a boy”. What are the odds they’re both boys?
Certainly, the person who had the two children knows whether they’re boys or not. But we can still talk about the odds. What are the odds that out of thirty people in a room, two will share a birthday? Everybody knows their birthday, and I might even know the other people who shares my birthday. But, I can still talk about the odds.
I covered this in an earlier post. If Monty didn’t know and was simply guessing, switching would only be 1:2 odds. It is because Monty knows, and Monty reveals this information by purposefully opening the wrong door that my odds in switching increases to 2:3.
My youngest son explained it this way: Monty’s choices of which door to choose are limited. He can’t choose the door we picked, and he can’t choose the door with the car. He is constrained to pick one and only one door. Thus, revealing where the car is located. If you look at the truth table I published in post 121, you can see he’s right. When Monty only has a single door to choose from, switching is a winning strategy. When Monty can choose multiple doors, winning is a losing strategy.
If Monty doesn’t know which door has the car, his choice of doors isn’t constrained: He could pick either one. If he doesn’t reveal which door has the car, switching in this case is only a 1:2 odds of winning.
I don’t knwo much about mathematical probability, but this one is simple.
You have a choice of taking what is behind 1 door, or taking what is behind BOTH of the other doors. It’s NOT BOTH you say. No, but one of the 2 doors you didn’t pick is empty, and you are shown which one, so switching is essentially the same as taking what is behind BOTH of the other doors. 1 in 3 times your initial pick wins. 2 out of 3 times, the winner is one of the other doors, and you have been shown which one of those is not the winner.
I don’t know how it’s modeled mathematically, but people seem to be confused about a key point. When Monty shows you what is behind one of the doors you didn’t pick, it isn’t random. At least one of those two doors is not the big prize, and he has to show you that one, never the one with the big prize. So the second part of the situation, where Monty picks a door to show you, is not random.
**Hasn’t Marilyn Vos Savant already definitively answered this question? **
Her answer seemed logical and made sense; as I thought the same when I read the question myself–because…by increasing your choice, you then increase your chances of a win. Simple math…see below for ENTIRE debate from Marilyns website:
:smack: Old School “Scholars”–[COLOR=“Black”]and after all those people wrote in PROVING that Marilyn’s theory was accurate over and over again, the men still need to shake their heads.
What a shame so many “intellects” are still so close-minded…alas, the earth is still flat for them. Pity…**[/COLOR]
Welcome to the Straight Dope Message Boards, ecco477, we’re glad to have you with us.
You might want to take a look at our rules on quoting from copyrighted material: FAQ: What’s the policy on copyrights?. Basically, you provided a link to the material (that’s good!) and so there is no need to cut-and-paste the whole thing (which is bad, and a violation of our rules.) I understand you’re new, so no problem, I’ve edited it out – those who want to read it, go to the link – and no worries.
I’ve been having arguments about this one since I took prob and stat in college, and I think bringing it muddies up the issue…because it’s wrong!
You’re getting the 1/3 in the first example by taking this chart:
G G
G B
B G
B B
And on hearing that there’s one boy, you cross off GG and have:
G B
B G
B B
Ah, so there’s a 1/3 chance they are both boys, and a 2/3 chance there’s one boy and one girl. So there’s a 2/3 chance the other child is a girl.
This should mean you can, on hearing the gender of one random child, predict the gender of the other with 2/3 accuracy.
You can’t.
Go ahead, get a friend and go test it. (There’s a nifty iOS dice app that includes boy/girl dice in case you don’t actually want to do lots of reproducing and wait 18 months or more for results).
You’ll find that if try to predict the gender of the other child by picking “opposite gender to the one I know about”, you’ll be right 1/2 of the time, instead of the 2/3 the probabilities tell you.
Keep that in your head for a moment.
Now you get to your second part. You are told “The older one is a boy”. Ah ha, you say. Now you started with:
O Y (O = Older, Y = Younger)
G G
B G
G B
B B
and can cross off G G and G B, leaving you:
B G
B B
A 50/50 chance. You can now predict with actual statistical accuracy.
“Well, I got more information this way”, you say.
But actually, in the first case, you got the same amount of information.
Because we can just make that chart “The one you first learned the gender of is a boy”.
1 2
B G
B B
And we’re back to our 50/50.
There is always an order, even if it was not explicitly stated in the problem.
If you really want to ignore order, you don’t have four options. You have three.
Two Girls.
One of each.
Two Boys.
B G and G B are the same when ignoring order. You can’t set up your sets as ordered and then ignore order when considering them.
Back on page 1, a poster comments that “Sandy is a girl and Chris is a boy” is not the same as “Chris is a girl and Sandy is a boy” are not the same. But as far as probabalities are concerned, they are. Adding the names is a distraction that sounds logical, but isn’t mathematically correct. It’s like saying that using one sparkly die and 4 white dice in Yahtzee changes the odds.
To use the logic that gets you to a 1/3 chance, you SHOULD have grouped them as just 3 options (2B, one of each, 2G) instead of 4.
Since my point was that saying that logic doesn’t actually work, we’re not really disagreeing here.
We start with four equally likely possible outcomes:
[ol]
[li]The oldest child is a boy and the youngest child is a boy.[/li][li]The oldest child is a boy and the youngest child is a girl.[/li][li]The oldest child is a girl and the youngest child is a boy.[/li][li]The oldest child is a girl and the youngest child is a girl.[/li][/ol]
If you’re told that the oldest child is a boy, that rules out outcomes 3 and 4, leaving behind two equally likely possible outcomes. The probability that both children are boys is 1/2.
On the other hand, if you’re only told that at least one child is a boy, that only rules out outcome 4, and you’re left with three equally likely outcomes. The probability that both children are boys is 1/3.
No, either way works to give two-thirds:one-third logically.
I’m not sure how you’re coming up with 50% as the correct answer using your iOS dice app. It sounds to me like you’re saying that given a boy, the next child has a 50-50 chance of being a girl. That’s absolutely correct (assuming the odds are 50-50 for any given birth, which isn’t quite true in reality). But then you’re the one who has imposed order by implicitly assuming the boy is born first. That is, you’ve actually eliminated two of of the initial four possibilities: GG and GB - not just the GG case.
Am I misunderstanding what you were trying to do there?
I always thought this problem was silly because the boy must obviously be either the oldest or youngest. If he’s the oldest, the odds of the other child being a girl are 1/2. If he’s the youngest, the odds of the other child being a girl are 1/2. What difference does it make if I know whether he’s older or younger than his sibling?