How's the Weather in Nerdville?
In Nerdville during the rainy season if today is clear there is only a 10% chance of rain tomorrow, but if it rains one day there is a 80% chance that it also will rain on the following day. Assuming that this is the rainy season, what is the probability that it is raining right now in Nerdville?
This is a "you must show your work" question.
DorothyGale only, please.
Quote: statmanSince DorothyGale joined the thread in "A Most Unusual Puzzle," I see that she's not through with me yet. Since she likes to play Whack-a-Mole, here is a question from the mole:
In Nerdville during the rainy season if today is clear there is only a 10% chance of rain tomorrow, but if it rains one day there is a 80% chance that it also will rain on the following day. Assuming that this is the rainy season, what is the probability that it is raining right now in Nerdville?
This is a "you must show your work" question.
DorothyGale only, please.
Seems like a good puzzle (at first glance, it seems that there is some kind of a boundary condition - e.g., when the rainy season starts, and the chance that it rains on the first day - missing here, but still...) Saving it just in case ;)
Quote: statmanSince DorothyGale joined the thread in "A Most Unusual Puzzle," I see that she's not through with me yet. Since she likes to play Whack-a-Mole, here is a question from the mole:
In Nerdville during the rainy season if today is clear there is only a 10% chance of rain tomorrow, but if it rains one day there is a 80% chance that it also will rain on the following day. Assuming that this is the rainy season, what is the probability that it is raining right now in Nerdville?
This is a "you must show your work" question.
DorothyGale only, please.
Its a forum. If you want to talk to Dorothy on her own, there's private messages....
Besides, you haven't finished on the most unusual puzzle. Weaselman and I both posted solutions.I was hoping my work could get marked....
thecesspit: DorothyGale does not accept private messages. As for "A Most Unusual Puzzle," you and some of the others have created a very unpleasant working environment. I have submitted the puzzle to Marilyn Vos Savant, who probably is the best in the world at this kind of puzzle.
Technically, you don't provide enough information. We can't assume that it is raining 24 hours a day on the rainy days and none at all on the dry days. If you re-word the question to ask "what is the probability there was or will be rain in Nerdville today?" then we can come up with an answer.
Assuming that's what you meant, I know the answer, it was very easy.
Quote: statmanthecesspit: DorothyGale does not accept private messages. As for "A Most Unusual Puzzle," you and some of the others have created a very unpleasant working environment. I have submitted the puzzle to Marilyn Vos Savant, who probably is the best in the world at this kind of puzzle.
Unpleasant? You mean answering the question posed, giving a working of the answer and discussing it? Or pointing out someone's mistakes? I don't mind being shown I am wrong. When that happens, I don't start crying that someone is foolish and the problem is wrong. I work at fixing it. In fact, your very insistence that the problem was unsolvable was why I went and worked on it again, and I -think- found at least two solutions, and weaselman showed a much more generic set of answers.
"Unpleasant working environment" came from the person trying to impose their own rules and attitudes with clod boots on an establish culture with out a by your leave. Maybe that works in academia (actually, or maybe in your world of academia... not where I've studied and worked), but in the wild world of the internet, that won't wash, buddy.
And if DGale doesn't want to accept private messages... well, there we go, she doesn't want to answer questions individually. You can't in a public forum expect to have a one-on-one conversation and not be interrupted.
As for this problem... it's not one I've seen before, but I suspect by some thought I'd learn something with the solution. I'm suspecting either Bayesian probabilities or Markov chains are the tools needed.
Cheers.
Quote: thecesspitAs for this problem... it's not one I've seen before, but I suspect by some thought I'd learn something with the solution. I'm suspecting either Bayesian probabilities or Markov chains are the tools needed.
Good instincts -- you can do it with both.
And Statman, this isn't a working environment at all. If you view the reactions from forum members as unpleasant, it's likely because you keep talking down to them as if you're a wise master and they're ignorant initiates. You haven't demonstrated the aptitude to justify that haughtiness, and even if you had, it's still not warranted in a forum such as this.
Edit: probably the easiest way is to use the Law of Total Probability and the fact that p(rainy) = 1 - p(sunny), thereby allowing you to solve one equation with one variable.
I like easy pie.
Afterward, I created a small probability table, much like a Heads/Tails possibility listing. I totaled probabilities of all combinations ending with "R", and ending with "N", then did a ratio. Even at 4 days worth of weather, the ratio approached the correct answer.
Finally, I did some Markov Chain research, and found the limiting matrix approached the answers I got from the two above methods.
How can you expect to post a juicy question like this for one person and not expect people to respond? :)
Quote: waltomealIt sounds to me like the weather in Nerdville is... stormy.
YEEEAAAAAHHHHHHHHHHHHHHHHH
If A is the probability of Clear, 1-A is the probability of Rain.
The probability of Clear on the next day is always: .9A + .2(1-A). = .7A + .2
The probability of Clear on the following day is always: .9 (.7A + .2) + .2 (.8 - .7A) = .34 + 49A
The probability of Clear on the next day is always: .9(.34 + .49A) + .2(.66 - .49A) = .438 + .329A
The probability of clear on the following day is always = .9(.438 + .339A) + .2(.562 - .631A) = .5066 + .2401A
....
After n days, the probability of clear is 2/3 + (.7A)^n. Therefore, the initial probabilty doesn't matter.
I'll have to be honest with you and tell you that I can't express the first constant as a function of n trials.
When n=1, c = .2
When n=2, c = c(n=1)*.9 + (1-c(n=1))*.2 = .34
When n=3, c = c(n=2)*.9 + (1-c(n=2))*.2 = .438
When n=4, c = c(n=3)*.9 + (1-c(n=3))*.2 = .5066
So, c(n) = c(n-1)*.9 + (1-c(n-1))*.2.
Limit of c(n) as n approaches infinity?
Editing:
So, the Markov Chain would look like:
| C | R | |
|---|---|---|
| C | .9 | .1 |
| R | .2 | .8 |
And to solve we get the following equations:
(a) C + R = 1
(b) .9C + .2R = C
(c) .1C + .8R = R
(b) and (c) yield .1C = .2R and therefore C = 2R
Subbing into (a), 2R + R = 1 and therefore R = 1/3.
Woot!
Assuming the probability it was raining yesterday is the same as today, this is really simple: p = 0.8p + 0.1(1-p) ; p = 1/3, you don't need Markov chains for that.
But it is not a given, is it?
For example, what if the rainy season starts today? If so, it was probably not raining yesterday, and then the probability of a rain today is only 0.1. And the probability of a rain tomorrow is 0.17 etc., and those are all valid answers.
It will converge to 1/3 eventually, if the rainy season is long enough, but without a condition to that effect, we cannot just assume it did not start yesterday, can we?
If R and C are the probabilities of being Rainy and Clear, respectively, you don't need to make the matrix. You just have:
p(Rainy today) = p(Clear yesterday) * p(Rainy today | Clear yesterday) + p(Rainy yesterday) * p(Rainy today | Rainy yesterday)
which turns into
R = C * 0.1 + R * 0.8
and since C = 1-R
R = 0.1(1-R) + 0.8R
R = 1/3
Also, your other function above 2/3 + (0.7A)^n, as n goes to infinity, is 2/3. That's p(Clear), so p(Rainy) is 1/3 -- same answer.
Quote: weaselmanBut it is not a given, is it?
No, it's not given. It was simply stated that you are in the rainy season. It is only possible to calculate the long run probability of a rainy day since we don't know the distribution of rainy days while not in the rainy season and we don't know which day of the rainy season we are on.
I'm doubt that the original question was ambiguous.
Quote: CrystalMathNo, it's not given. It was simply stated that you are in the rainy season. It is only possible to calculate the long run probability of a rainy day since we don't know the distribution of rainy days while not in the rainy season and we don't know which day of the rainy season we are on.
I'm doubt that the original question was ambiguous.
No, the original question (from the Internet a few months back, at least) actually asked it a different way: how many rainy days are expected be this year if the first day is sunny? The probabilities were different but the solution methodologies are the same.
…so immediately after the great flood, Noah released all of the animal pairs with the command “go forth and multiply”. But one particular pair of snakes was giving him loads of trouble. You see, they were Adders. So after considerable study he came up with a plan. He went into the forest, cut down some trees and built them a playground and he presented the Adders with log tables.
The question is why can the Adders now multiply?
Quote: FrGambleSo here is a little brainteaser/joke for those who live in Nerdville, especially if it starts raining a lot you might need this info:
…so immediately after the great flood, Noah released all of the animal pairs with the command “go forth and multiply”. But one particular pair of snakes was giving him loads of trouble. You see, they were Adders. So after considerable study he came up with a plan. He went into the forest, cut down some trees and built them a playground and he presented the Adders with log tables.
The question is why can the Adders now multiply?
Because their zygotes divide?
Quote: MathExtremistBecause their zygotes divide?
I was working on something with addition (Adders) and logorithms (logs) and tables, as in multiplication tables. But my math is about as good as your ability to tolerate falsities, so maybe someone can finish my idea for me.
Quote: FrGambleSo here is a little brainteaser/joke for those who live in Nerdville, especially if it starts raining a lot you might need this info:
…so immediately after the great flood, Noah released all of the animal pairs with the command “go forth and multiply”. But one particular pair of snakes was giving him loads of trouble. You see, they were Adders. So after considerable study he came up with a plan. He went into the forest, cut down some trees and built them a playground and he presented the Adders with log tables.
The question is why can the Adders now multiply?
Very funny. This is the best post in the thread.
Quote:Dween: How can you expect to post a juicy question like this for one person and not expect people to respond?
I welcome responses: just don't give it away. Since DorothyGale does not accept private messages, this is the only means I have of communicating with her. One of the ways I may be able to win respect in this forum is to pull the rug out from under my most vicious assailants. I have done this once, and now I want to see if I can do it to Dorothy Gale. In the thread "New Probability Tables for Roulette IV" she remarked that I didn't know who I was dealing with. Well, I would like to find out, and most especially I would like to find out what kind of person she is. It is a Markov chain problem, but just a little algebra is all that is required to solve it.
Quote: statmanOne of the ways I may be able to win respect in this forum is to pull the rug out from under my most vicious assailants.
You really believe that, don't you? Making a clown of yourself can win you admiration sometimes, for being funny, but respect? ... Not really.
Quote:It is a Markov chain problem, but just a little algebra is all that is required to solve it.
As it has been shown above, the problem (1) is underdefined, and (2) does not need any algebra.
Quote:MathExtremist:And Statman, this isn't a working environment at all. If you view the reactions from forum members as unpleasant, it's likely because you keep talking down to them as if you're a wise master and they're ignorant initiates. You haven't demonstrated the aptitude to justify that haughtiness, and even if you had, it's still not warranted in a forum such as this.
Although you have criticized me on many points, you never have been rude. Most of the others who have taken shots at me seem like clones of the three stooges. They should use you as an example.
I think the Wizard's response concludes the matter. He must have passed his actuarial exam. I am grateful to him for giving me 15 years to live on his life expectancy calculator. I went to a local doctor who gave me six months to live and charged me $100, so I went to another doctor who gave me two years to live and charged me only $50. I'm going to him from now on.
Since you guys are intent on running interference for DorothyGale, I suppose I shall have to find a problem even the Wizard can't solve.
Quote: statmanQuote:MathExtremist:And Statman, this isn't a working environment at all. If you view the reactions from forum members as unpleasant, it's likely because you keep talking down to them as if you're a wise master and they're ignorant initiates. You haven't demonstrated the aptitude to justify that haughtiness, and even if you had, it's still not warranted in a forum such as this.
Although you have criticized me on many points, you never have been rude. Most of the others who have taken shots at me seem like clones of the three stooges. They should use you as an example.
Stay classy there statman, stay classy.
Three stooges who can do mathematics, though. I see you've made zero comments on the many solutions provided.
Quote: thecesspitQuote: statmanQuote:MathExtremist:And Statman, this isn't a working environment at all. If you view the reactions from forum members as unpleasant, it's likely because you keep talking down to them as if you're a wise master and they're ignorant initiates. You haven't demonstrated the aptitude to justify that haughtiness, and even if you had, it's still not warranted in a forum such as this.
Although you have criticized me on many points, you never have been rude. Most of the others who have taken shots at me seem like clones of the three stooges. They should use you as an example.
Stay classy there statman, stay classy.
Three stooges who can do mathematics, though. I see you've made zero comments on the many solutions provided.
Isn't that what happens in school when the professor has no idea how to solve the problem at hand?
