I meant "one of the children is a girl, and it doesn't matter which one, and the gender of the other child doesn't matter either."
I meant "one of the children is a girl, and it doesn't matter which one, and the gender of the other child doesn't matter either."
The the probability that your neighbor has at least one girl is 3/4.
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Attachment 566
Last edited by JivinJeffJones; 02-16-2008 at 07:49 AM.
I think I get the door thing..
I think it's explained better this way:
There is a 3/3 chance that you will pick a door that Monty is not going to open, and a 3/3 chance that he is going to open an incorrect door. The door he is going to open will have a goat in it.
It doesn't matter which door you pick at first. You have a 1/3 chance of it being the correct door, and as you know for certain that Monty will pick an incorrect door anyways, the fact that he will do that means it will not change your 1/3 chance.
You most likely wouldn't have gotten the car on your first try, since it's a 1/3 chance. There's 1 car and two goats. This is why your first choice remains a 1/3 choice.
Monty eliminates another one. So between your 1/3 chance, and an elimination, that gives the unchosen door as the winning one a 2/3 chance.
Alright, it has been a busy couple of days; I got home real late yesterday, and just fell asleep.
Maybe by now, everyone understands the probability stuff, but I will use most of my dinner break to post what I said I would. I’ll do the teacher thing and give another puzzle to check for understanding. You already know I am arrogant enough to do that.
For reference:
Monty Hall Problem
Cab Problem
Neighbor Problem
The basic idea is to mentally run a bunch of identical experiments in parallel to see what the answer is. Because probability is the percentage of times you get a particular outcome when you run a large number of identical experiments with random variables involved
The tree notion that ptgatsby mentioned is a good way to keep track.
I will try to avoid using math except for the terms I define.
For the probability related problems posted here, IMO, there are some key concepts that help a lot.
Please prepare to be patient. Solving small but tricky problems in a slow and deliberate manner is like mental Tai Chi. (Though I am not sure I would be qualified to correct thinking)
Sample spaces:
A sample space is the set of all possible outcomes for an experiment. There is some leeway here to consider all sorts of unlikely outcomes, if you want, but usually we can pick the simplest ones that fit. You could go back and assume more complications as a sanity check and see if you get approximately the same answer (a double check on your choice of abstraction, if you will).
- The relevant experiments and sample spaces I chose in the Monty hall problem:
- You choosing a door. The possible outcomes are: you chose a goat, or you chose a car.
- Monty Hall Revealing a door after you choosing a door. The possible outcomes are: The revealed door has a goat behind it. The revealed door has a car behind it.
- The final revelation after running through out the whole problem scenario. The possible outcomes are: It is better to have switched, It is better to have stayed.
- The relevant experiments and sample spaces I chose in the cab problem:
- An accident happens. Possible outcomes are: The cab involved was green, The cab involved was blue
- A witness reporting color of car when accident happened. Possible outcomes are: Accurate, Inaccurate
- Checking the actual color of the car when a witness repots it to be blue. Possible outcomes are: The cab involved was green, The cab involved was blue.
- The relevant experiments and sample spaces I chose in the neighbor's child problem:
- The family had a child. Possible Outcomes are: The child is male. The child is female.
- The family has two children. Possible outcomes are: The children are male. The children are female. One child is male, the other is female.
- Meeting a family w/ two children with a male child. Possible outcomes are that both male children or that they have one child of each gender.
- Meeting a family w/ two children with a male child and checking to see if they also have a girl. Possible outcomes are that they do or they don’t.
Probability Distributions:
We assign probabilities to all the possible outcomes based on what we know. This part is also what makes probability theory work. The better we assign probabilities to outcomes, the more accurate the results. There are many stipulations on how we can assign probabilities, but the important ones are that in a sample space, the sum of all the probabilities should add up to 1, and all the probabilities are non-negative.
Usually, we are forced to use intuition to assign probabilities to particular outcomes of events, but what is stated in the problem, or the evidence, supersedes our intuition (that is part of what makes things hard).
- Probability distributions I chose in the Monty hall problem:
- You choosing a door. The probability you chose a goat is 2/3 and the probability you chose a car is 1/3.
- Monty Hall Revealing a door after you choose a door. The probability the revealed door has a goat behind it is 1. The probability the revealed door has a car behind it is 0. (This is the assumption that The_Liquid_Laser mentioned).
- The final revelation. The probabilities need to be derived.
- Probability distributions I chose in the cab problem:
- An accident happens. The probability the cab involved was green is 0.85, and the probability the cab involved was blue is 0.15.
- The witness reports the color of a cab after an accident. The probability the witness was accurate is 0.8, inaccurate 0.2.
- The true color of the car is revealed after a witness reports it to be blue. These probabilities need to be derived.
- Probability distributions I chose in the neighbor problem:
- The family had a child. The probability the child is male and the probability the child is female is 1/2.
- The family had two children. The probabilities need to be derived.
- Meeting a family w/ two children with a male child. The probabilities need to be derived.
- Meeting a family w/ two children with a male child and checking to see if they also have a girl. The probabilities need to be derived.
Rules of Probability:
There are various rules for deriving probably distributions from other probability distributions based on how the sample spaces/experiments are related. These are not arbitrary rules or technicalities, but rules similar to those of arithmetic (for getting how many things there are), or geometry (for calculations of volumes). They have an uncanny ability to reflect what is observed in real life. There are certainly many rules of probability, but I will state the two I find most useful. I believe, if you try using these rules (perhaps generating more intermediates experiments), you will arrive at the answers given in the spoilers.
- The combination of two experiments yield a joint probability distribution. Two experiments are independent if and only if the probabilities for the joint distribution can be derived by multiplying the probabilities associated with the two events.
- If an experiment, A, is the result of restricting another experiment, B, the desired probability distribution is called a conditional or Marginal distribution. You get this distribution by dividing the probability of getting a particular outcomes of A, by the probability of getting any of the outcomes in A.
- Probability derivations in the Monty hall problem:
- You choosing a door. The probability you chose a goat is 2/3 and the probability you chose a car is 1/3.
- Monty Hall Revealing a door after you choose a door. The probability the revealed door has a goat behind it is 1. The probability the revealed door has a car behind it is 0. (This is the assumption that The_Liquid_Laser meant, I believe).
- The final revelation.
Derivation:
Note, any experiment for which only one outcome has a non-zero probability (and therefore a probability of 1) is independent of all experiments.
So, you can now combine the two experiments to get that the probability that you chose the car and Monty revealed a goat is 1/3, while the probability that you chose a goat and Monty revealed a goat is 2/3.
A pure probability problem is rare, so to get the distribution for the final experiment by simply mapping the combination we found to the desired outcomes. If you chose a car, then you shouldn't switch. If you chose a goat, then you should. So 1/3 of the time it is better to stay. 2/3 of the time it is better to switch.
- Probability derivations the cab problem:
- An accident happens. The probability the cab involved was green is 0.85, and the probability the cab involved was blue is 0.15.
- The witness reports the color of a cab after an accident. The probability the witness was accurate is 0.8, inaccurate 0.2.
- The true color of the car is revealed after a witness reports it to be blue.
Derivation:
The probability that the witness is accurate is the same no matter what the color of the cab (a convenient fabrication, making it an independent experiment). So combine the accident and the witness... The probability that the cab was green and the witness was accurate is 0.68. The probability that the cab was green and the witness was inaccurate is 0.17. The probability that the cab was blue and the witness was accurate is 0.12. The probability that the cab was blue and the witness was inaccurate is 0.03. Now we want to confine this experiment to when the witness saw blue (accurate witnessing of blue, inaccurate witnessing of green). The probability the witness saw a blue cab and the cab was blue is 0.12/(0.12+0.17)=12/29. The probability the witness saw a blue cab and the cab was green is 0.17/(0.12+0.17)=17/29.- Probability distributions, I chose in the neighbor problem:
- The family had a child. The probability the child is male and the probability The child is female is 1/2.
- The family had two children.
Derivation:
It is reasonable to assume the genders of children are independent, so the probability that both children are male is 1/4, so is the probability that are both female. The probability of having one child of each gender is 1/2.- Meeting a family w/ two children with a male child.
Derivation:
This is a constraint of the previous example to families with at least one male child. The probability that both children are male is 0.25/(0.25+0.5)=1/3. The probability that the family has one child of each gender is 0.5/(0.25+0.5)=2/3.- Meeting a family w/ two children with a male child and checking to see if they also have a girl.
Derivation:
This is a simple translation based on the sub-type of family you met. The probability that the family also has a girl is 2/3.
Frankly, I find avoiding math annoying. A lot of what I described isn't really accurate (but I believe it is close), and I was way more verbose than I would have been if I hadn't been avoiding math to talk about math. It would be so easy to draw trees with numbers assigned to them or to use probability notation.
But, I hope this made sense and was some help. Hopefully, it has also helped to build intuition, instead of being a bunch of rules and technicalities as Athenian said.
Sorry, for being so long winded.
Last edited by ygolo; 09-21-2007 at 01:46 AM. Reason: DAMN QUOTES!
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Try this modification:
The problem also assumes that when a woman gives birth there is an equal chance of the child being either a boy or a girl.
My neighbor has two children. I met one of them. He is a boy. What is the probability that the other child is a girl?
I would appreciate if someone humored me and used the concepts in my previous post. I am hoping people will get the correct answer. I want to see how good I am at explaining things.
Accept the past. Live for the present. Look forward to the future.
Robot Fusion
"As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
"[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
"[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield
As much as I'd like to join you and give credit to people not knowing mathematics, it's impossible to do this way. The situation described earlier already reflected reality accurately. With you presenting entirely other framework, one with your premises - of course the results may change. It's ok to present premises, ok, of course. I can't agree that the multi-round Monty hall problem would "naturally" and realistically extend to the rules and dynamics you said. As I've read from the subject, the host (and the game show) really actually keeps the rules the same. Now why is that? Now opening such possibilities we go very far outside mathematics, so this is not certain anymore. But, in short, producers may just decide on a program format by some criteria and to change that format by other criteria. What's been observed (as I understand from reading some sources) that the host does not try to decieve the guest, and there is not a meta-game involved in trying to interpret the everchanging rules.
Apparently the program has been presented with great enough similarity each time, so that it has gained the audience's trust that the rules will stay the same.
Now that in my interpretation the facts dont support your postulation, I welcome you to speculate. However such non-supported speculation may not be entirely relevant to the scenario we discussed before, not relevant enough as to disprove it.
SOrry for long post.
Remember that you postulated the whole thing in trying to make something that you wanted?