Tuesday, April 04, 2006

The Monty Hall Problem, Part Two

I'm afraid I'm very pressed for time today, so only brief blogging.

The answer to Variation One of the Monty Hall Problem, given at the end of yesterday's post, is that it now makes no difference whether or not you switch. Really.

The answer to Variation Two is that your best strategy is to stick with your initial choice until there are only two doors remaining. No other strategy will give you as high a probability of winning.

I'll go into some of the mathematical aspects of these problems in later blog entries, though some of my commenters to the previous post have already done an excellent job of it (note particularly Micahel Ralston's excellent comment explaining the difference between the situations where Monty Hall does, or does not know where the car is.)

Richard Wein points out that the Times version of the problem, quoted in yesterday's entry, is rather vague, since it is not stated explicitly whether Monty Hall knows where the car is. He's right about this, though I think it's stronlgy implied in the Times version that Monty Hall does know where the car is and therefore always reveals a goat.

If this is the point Somerby had in mind in saying that in the Times version it makes no difference whether or not you switch, then I will be very impressed. I don't believe that to be the case, however. Somerby's reply was so snide and harsh towards the Times reviewer that I think he was treating it as obvious that after Monty Hall opens one of the doors there is a 50-50 chance of winning. I don't think he was making subtle points about what Monty Hall does and does not know.

As of this writing Somerby has not revisited this problem.

Anyway, there's lots more to say about this problem, and I'll try to get around to it in the near future. Let me thank all the commenters to the previous post for their thought-provoking replies.

In the meantime, you can have a look at Wikipedia's entry on this subject. It contains several other variations in addition to the ones I mentioned.

133 Comments:

At 10:28 PM, Anonymous Jeff Chamberlain said...

The Monty Hall problem has baffled and frustrated me for years. I don't challenge the mathematics; I'm "convinced" that it makes sense to switch. But I don't understand it. At all. The explanations don't make sense to me (including 100 doors and 99 goats -- so no smiling at me, either).

I'm not stupid. And while I'm not mathematically sophisticated I'm neither mathophobic nor innumerate.

So what's wrong with me? (And, undoubtedly more arrogantly, what's wrong with people who do understand this that they can't seem to explain it so I do, too?)

 
At 11:25 PM, Anonymous Anonymous said...

I was wrestling with the same issue for hours. It is just counter-intuitive that, faced with two choices you don't have a 50-50 chance. This is what helped me (if your mind doesn't work like mine - perish the thought - it might not help you :)

Reading the Wikipedia helped - the venn diagrams and the 100 door example. When you choose your door from 100, you have a 1% chance of getting the car. There is a 99% chance the car is behind the pool of 99 doors. Now, think of them as a group with a 99% chance of having the car. Doors start to get opened in that group, revealing goats, but the odds don't change for the group. Your original choice is will ALWAYS only have a 1% chance of being correct. The other group will ALWAYS have a 99% chance of being correct (99 doors). As more and more doors get opened, there are fewer and fewer closed doors, each with an increasing probability of having a car (YOUR doors chances don't change - it was one of a hundred). Finally, 98 doors have been opened, eliminating all options but one. Remember, your door was picked before you knew those 98 were goats - so it still has only 1% chance of car. But here's the thing, the rest of the doors had a combined chance of 99%, and all of the false choices have been removed, all but one - so there is a 99% chance that the last door is a car.

That's about the best I can do, hope it helps - I love this stuff!

 
At 7:56 AM, Blogger Lord Runolfr said...

Jeff,

The Opposing Player explanation was one of the most helpful to me.

"Consider the game as a two-player game in which Player A chooses and opens a door. The game host then opens a goat door. Player B then gets what is behind the remaining door. Since the first player will choose the car door only 1 in 3 times, the second player will win the car 2 out of 3 times. Thus, the car is behind the remaining door 2 out of 3 times."

 
At 9:10 AM, Anonymous Anonymous said...

Definitely not intuitive. But, the Wikpedia entry helps understand at least the original problem. Thx.

 
At 11:42 AM, Blogger Karen said...

I found your blog through Daryl at cobranchi.com. Reading through your posts and the comments has been fun, although I find myself in exactly the same position as Jeff in his 10:28 comment here.!
I would like to ask a question. Today, the game show is not Let's Make a Deal, but rather Deal or No Deal. Does the Monty Hall problem shed any light on when you should take the bank's offer versus when you should keep opening suitcases?

 
At 11:47 AM, Anonymous Anonymous said...

Jeff,

I understand the problem and din't get the 99 goat thing either. Think of it like this:

-Monty MUST open one of the two goat doors.
-Monty CANNOT open the door you pick first.

Neither of these two things alone affects your chances, but combined they make a difference.

Here's why:
You have a 2/3 chance of selecting a goat door on your first pick. Since Monty can't open your own door, he must reveal the other goat. (Which therefore reveals the car.)

You have a 1/3 chance of selecting the car on your first pick. At which point Monty will open either of the other two doors.

So 2/3 of the time Monty is REVEALING the correct door, and 1/3 of the time the last door is a false hope.

----Easy Summary---
If you pick goat first, the final door is guaranteed to be the car.

If you pick car first, the final door is guaranteed to be goat.

Since you are more likely to pick the goat first, you should always pick the final door.

 
At 1:48 PM, Anonymous Jeff Chamberlain said...

I appreciate the attempts to educate me, but my question really was a little different. I've tried for years to understand the MH problem, to no avail. I've pretty much given up on that part of it (sigh). However I'm not used to being unable to understand things that I put my mind to, and so I'm really curious to learn what it is about me that seems to have prevented me from understanding this.

That said, hope springs eternal. Remember that I don't know what I'm talking about -- that is, after all, the underlying point -- and so it's difficult for me to identify just what it is that I'm not "getting." One possibility -- maybe -- is that I don't see why the opened door remains relevant once it's been opened. There are still 3 doors, of course. But once that third one has been opened (to reveal a goat), what does it now contribute? Why hasn't the problem changed from a "3 unknown doors" problem to a "3 doors only 2 of which are unknown" problem? It may still be a "3 door" problem, but that opened door does not seem relevant any more.

Maybe an alternative way to express this is that I don't understand why the specific history influences the current probabilities. I'll choose Door A. Monty opens Door B (revealing a goat). Now a third party enters. She's given the opportunity to choose any of the doors. She obviously won't choose Door B. Door B might just as well not be there as far as her choices are concerned. She'll choose Door A or Door C. Are her odds 50-50? If not, why not? And what's the difference between this and if I am offered the same choice after Door B has been opened?

Please permit me to reiterate that my point is not that I challenge the solution to the MH problem. I like cars a lot better than goats, so I'll switch. Every time. I promise.

 
At 2:50 PM, Anonymous JY said...

Hi Jeff,

How about this for yet another way of looking at it. Suppose we're watching two guys play a game of cards. Each one is trying to pick the ace of spades out of a deck. The first guy has to pick one at random. The second guy then takes the rest of the cards, and looks at all of them, searching for the ace. If he finds the ace, of course, he takes it. Otherwise, he just picks one at random.


Who do you think has the ace? Would you bet that guy #1 has it, or guy #2?

This is all the Monty Hall problem is.

 
At 7:47 PM, Blogger Michael "Sotek" Ralston said...

Jeff: Her odds are the same as yours, if she knows what door you selected. If she doesn't, then all she knows is one door has a car, and one door has a goat - which gives her 50-50 odds.

This is because she makes her first pick after a door has been removed.

You, however, don't - you made your pick BEFORE the door was removed.

If you picked wrongly (then), which is what you were more likely to do, then Monty Hall didn't have a choice as to which door to reveal - and you should switch.

And if she knows which one you selected, she can do precisely what you would do - which is select the other one.

 
At 4:10 PM, Blogger trrll said...

Jeff, try this. No matter what door you choose, Monty can always (and in the usual version of the riddle, always does) open a door and show you the goat. Since he does this regardless of whether you are right or wrong, it cannot change the probability that you picked the correct door at the outset. If it was 1 in 3 before, it still is after he opens the door. In other words, the history does not affect the probability of your choice being correct. Neither can it affect the probability that the car is behind one of the unchosen doors, which is 1 minus the probability that you chose correctly at the outset--i.e. 2 in 3.

What it does affect is the probability that the car is behind the door that he opens--now that you see the goat, that probability is necessarily 0. But the probability that the car is behind one of Monty's doors is still unaltered: 2 in 3. Since he has only one closed door left, all of that probability must reside in that door.

 
At 9:37 PM, Anonymous Jeff Chamberlain said...

I understand the explanations. However, I don't understand why opening the third door doesn't change the game, and thus the probabilities. (I undedrstand that it doesn't, in the sense that I have comprehended this as a fact because so many people who know more about this than I do have said so. But I don't understand why.)

However, again, what I am really curious about is why so many people, myself included, can't seem to "get" this.

 
At 9:59 PM, Anonymous Kevin from NYC said...

I don't believe it!

I went to the web site. I chose the middle door all the time.

I never changed.

I was 10 out of 21...50%

 
At 9:32 AM, Blogger bcarson said...

Curious.

Kevin went to the website and chose the middle door all the time.

He won 10 out of 21 ... roughly 50%.

I did the same thing and won 5 out of 21 ... roughly 25%.

I conclude that if your name is bcarson you should switch but that if your name is Kevin you should not.

Seriously, if you find that a "random" choice from among three options prefers a certain option 50 percent of the time, you might conclude either that the choice is not truly random or that your sample is not representative.

 
At 12:43 PM, Blogger Michael "Sotek" Ralston said...

Jeff: It doesn't change it because your decision was already made.



That said ... I suspect most people don't get it, because they don't think in terms of the history.

They say "There's two doors - so it must be 50/50". But when they picked, there were three - and the fact that there was a goat behind at least one door they didn't pick doesn't ACTUALLY tell you anything, because you already knew that before you even picked.

Knowing *which* door it is is informative, but only in the sense that it moves the probability of the car somewhere else, but not to where you already fixed it.

 
At 9:51 PM, Anonymous Kevin from NYC said...

Ha....I did some other posts but they got eaten..

I went to the test page and choose the middle again and got like 11/20. THen I chose the middle and alway switched and got 13/20 and 14/20. and did the don't switch and got like 9/20 so it seems 50% ie 1 out of 2 doors for hold and 66% if you switch.(the predicted outcome) .wierd.

 
At 9:56 PM, Anonymous Kevin from NYC said...

so don't switch got a worse 7/20 and switch clocked in again at 13/20....

 
At 7:21 AM, Anonymous Anonymous said...

Jeff,

When you pick your door it remains relevant to Monty, which is what is important. You basically only have 3 choices, to pick the car, to pick Goat A, or to pick Goat B.

If you pick Goat A, Monty must open the door containing Goat B. In that case you win the car if you switch. If you pick Goat B, Monty must open the door containing Goat A. Again you win the car if you switch. The only remaining possibility is if you pick the car. Then Monty can show you either Goat A or Goat B, and you'd be wrong to switch. But doing the math, you see that switching every time will win twice and only lose once.

So, think of the problem from Monty's perspective and not your own, and that might help.

A similar problem I like goes like this:

You have 3 cards. One is red on both sides. One is blue on both sides. The remaining card is red on one side and blue on the other. You draw a card unseen and place it on a table...it shows a blue side up. What is the probability that the other side of the card is also blue?

Dave S.

 
At 9:11 AM, Anonymous Jeff Chamberlain said...

I'm sorry, but unfortunately not surprised, that I still don't get it. (Like I've said, this is not the first time I've tried.)

Mr. Ralston (and others): I understand that on my first pick the odds were 1 of 3. I also understand that my odds remain 1 in 3 as related to my initial pick no matter what happens later. I don't understand why, when 1 of the remaining doors is opened (showing a goat) and I'm offered the chance to switch my initial choice, that does not constitute a new game (with only 2 relevant alternatives). Since I'm offered the chance to "start over," and choose anew, this time between 2 relevant doors (isn't that what the opportunity to "switch" means?), why does my initial choice among 3 (or more) doors "fix" anything?

Dave S: There are 2 relevant cards, B-B and B-R. Mine could be either of these, but it can't be the R-R card. (Since I know this, isn't the R-R card irrelevant?) So 1 out of 2?

 
At 1:03 PM, Blogger Michael "Sotek" Ralston said...

Jeff: It doesn't start things over, because knowing that a door you didn't pick has a goat doesn't give you any information about the door you ALREADY picked. (As, after all, there is no way for it not to have a goat.)

It does give you information about the second (goat) and third doors, however.



Here, let me try a reformulation that may make some sense:

There's three doors, two with goats, one with a car, same as before.
You pick one door, then Monty Hall tells you that there's at least one goat behind the other two doors, and offers you the chance to take what's behind *both* of the ones you didn't pick, discarding any goats.

I'm hoping that it's sufficiently clear what the correct choice is in that situation, and why.
I think the equivalence between it and the original is also clear, but ever since I first understood the problem, I had a bit of trouble figuring out how people *don't* get it after doing the math, so... I may be off.

 
At 4:52 PM, Anonymous Anonymous said...

Jeff -

You must keep in mind that Monty, unlike you, does not make his pick at random. Monty will always pick a goat. So if you pick a goat, and Monty picks a goat, then switching will be the correct decision as only the car is left. Since you pick a goat 2/3 of the time, you win 2/3 of the time by switching.

Your card solution is incorrect. The answer is, the other side has a 2/3 probability of being blue also.

Dave S.

 
At 5:03 PM, Anonymous Jeff Chamberlain said...

I'm hopeless.

 
At 2:07 AM, Blogger Michael "Sotek" Ralston said...

Dave: Actually, I'm pretty sure yours is wrong.

The key to the Monty Hall paradox is that you don't learn anything about *your* door when he tells you there's a goat not behind it.

In the card one, though, you know you don't have the RR card if you see blue on your side.

Okay, let's go through the three possibilities.

1) RR.
2) BR.
3) BB.

All are 1/3rd initially.
But RR is not the case - so we can discard that.

There's nothing stacking BR over BB, or the other way around, so the odds are 50/50.

This is the "Monty doesn't know where the goat is, and if he finds the car, you start over" variation.

 
At 9:12 AM, Anonymous Anonymous said...

Michael -

The key to the Monty Hall paradox is that you don't learn anything about *your* door when he tells you there's a goat not behind it.

Hmmm, I'm sorry, but I don't understand this.

There's nothing stacking BR over BB, or the other way around, so the odds are 50/50.

The problem here is that you are dealing with the cards when you should be dealing with the sides. To see what I mean, label each side on the double blue card, B1 and B2. The blue side of the blue/red card is B3. Similarly, we can call the red sides R1, R2 and R3 (which is opposite B3). Now there are 6 "sides" which can be selected initially. We know the selected side was blue, so R1 and R2 are eliminated as possibilities for the other side of that card. That leaves B1 (if we're looking at B2), B2 (if we're looking at B1), and R3 (if we're looking at B3) as the only possibilities. That 2:1 blue, so the probability of getting a blue on the other side is 2/3.

No?

Dave S.

 
At 12:53 AM, Blogger Michael "Sotek" Ralston said...

Dave: Hmm. Yes, I see what you mean about the cards. Point taken.


As for Monty Hall...

I select a door. I know there is a 2/3rds chance it will have a goat, and 1/3rds chance it will have the car.

I also know there is a 1/1 chance that when Monty opens a door, there will be a goat - independant of what my door was.

Thus, I gain NO information (about my door) when Monty opens that door.

Thus, him showing me a goat behind a door has no effect on the probability of the door I selected having a goat, any more than if he'd rolled dice or done nothing at all... or said "There is at least one goat behind a door you did not pick". I already KNEW that... so how can it affect my probability?

Meanwhile, it has a very strong effect on the probability of the car being behind the door he opened (went from 1/3 to 0), and thus increased the probability of the car being behined the third door.

 
At 9:52 AM, Anonymous Anonymous said...

Michael -

I think we are in agreement about the Monty Hall problem. I was trying to approach it from a slighly different perspective to help Jeff "get it". Looks like I may have confused things still further instead. :)

Cheers,
Dave S.

 
At 1:57 PM, Anonymous Anonymous said...

Jeff,

Hopefully you're stil coming back here. A few things for you.

"One possibility -- maybe -- is that I don't see why the opened door remains relevant once it's been opened."

The opening and closing of the doors is irrelevant. The issue is what probabilities are. They are about how things can be. Which is quite different from how things are.

How they CAN be is determined by the initial conditions.

I mentioned in the first of these that people are bad at probabilities. You've asked about the difficulty understanding. What's wrong with you?!? Nothing. You just notice the world around you. And EXPECT certain things. Like the sun coming up tomorrow. And generally you are justified in that.

Look at coin tossing. A string of heads will often lead people to believe taht a Tail is more likely... since it is "due". This comes from a simplistic understanding of what's going on.

The fact that things tend to balance out in the long run is not the same as predicting the NEXT toss.

In fact, I can make a more compelling argument for the opposite.

Theoretical probability is a calculation of what "should" or "can" happen based on the number of outcomes and such.

Experimental probability is what "should" happen based on what we know.

If I toss a coin once and get heads. The theoretical probability of getting a T on my next toss is, effectively 0.5. The experimental probability of getting T is zero. It has never happened.

Of course tossing a coin once is hardly a good sample size.

With the Monty problem, people ignore the initial conditions that are determining the probabilities. Those initial conditions have not changed by opening a door. Let's get really silly here... would opening and closing a "goat" door a thousand times change how you perceive the question? A hundred times? 10? Then why should opening it once do so?

Fact of the matter is and this is going to sound absurd... when you open a goat door... technically, the chance of that door having been right is STILL 1/3. You know it's not. But the chances of what it could have been have not been altered.

Opening and closing doors doesn't change where the car can be. And it doesn't change the number of doors.

;-)

Cheers

 
At 8:36 PM, Blogger M.E. Bachofen said...

I think the problem of understanding lies in thinking that picking the prize is 1:3 at the start. Remember, it's about switching. So, think of it as the odds of picking a goat at the beginning. You have a 2:3 chance of picking a goat from the start every time . Every time you do pick a goat and switch, you win! Pretty simple really.
meb

 
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The Monty Hall problem has baffled and frustrated me for years. I don't challenge the mathematics; I'm "convinced" that it makes sense to switch. But I don't understand it. At all. The explanations don't make sense to me (including 100 doors and 99 goats -- so no smiling at me, either).


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At 12:09 AM, Blogger blogcar said...

So what's wrong with me? (And, undoubtedly more arrogantly, what's wrong with people who do understand this that they can't seem to explain it so I do, too?)
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eading the Wikipedia helped - the venn diagrams and the 100 door example. When you choose your door from 100, you have a 1% chance of getting the car. There is a 99% chance the car is behind the pool of 99 doors. Now,


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The period of his life was marked by a split between the Trịnh Lords of the north and the Nguyễn Lords of the south, followed by the Tây Sơn rebellion.
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At 10:34 PM, Blogger smalltk said...

The Monty Hall problem has baffled and frustrated me for years. I don't challenge the mathematics; I'm "convinced" that it makes sense to switch. But I don't understand it. At all. The explanations don't make sense to me (including 100 doors and 99 goats -- so no smiling at me, either).
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For plants, this usually requires some form of irrigation, although there are methods of dryland farming; pastoral herding on rangeland is still the most common means of raising livestock. In the developed world, industrial agriculture based on large-scale monoculture has become the dominant system of modern farming, although there is growing support for sustainable agriculture (e.g. permaculture or organic agriculture).
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Hạ Long Bay is located in northeastern Vietnam, from E106°56' to E107°37' and from N20°43' to N21°09'. The bay stretches from Yên Hưng district, past Hạ Long city, Cẩm Phả town to Vân Đồn district, bordered on the south and southeast by the Gulf of Tonkin, on the north by China, and on the west and southwest by Cát Bà island. The bay has a 120 km long coastline and is approximately 1,553 km² in size with about 2,000 islets. The area designated by UNESCO as the World Natural Heritage Site incorporates 434 km² with 775 islets, of which the core zone is delimited by 69 points: Đầu Gỗ island on the west, Ba Hầm lake on the south and Cống Tây island on the east. The protected area is from the Cái Dăm petrol store to Quang Hanh commune, Cẩm Phả town and the surrounding zone.
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Hai Phong Railway Station, established in 1902, is the eastern terminus of the Hanoi–Hai Phong railway line, also known as the Yunnan–Vietnam Railway. Built at the initiative of the French during their occupation, the railway once connected Haiphong to the city of Kunming in China's Yunnan province, although service along the Chinese portion of the line is currently suspended. Rail travel from Haiphong is still possible, with connections to the rest of the Vietnamese railway network possible via Hanoi. Besides this, there are lots of railway roads in the whole city to served the requirement of moving goods out of and into ports.
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At 10:10 PM, Blogger GameforYou said...

The length of canal from Rhoswiel, Shropshire to the Horseshoe Falls including the main Pontcysyllte Aqueduct structure as well as the older Chirk Aqueduct, were visited by assessors from UNESCO during October 2008, to analyse and confirm the site management and authenticity. The aqueduct was inscribed by UNESCO on the World Heritage List on 27 June 2009, alongside previously inscribed sites such as the Taj Mahal, Great Wall of China and Stonehenge.
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At 11:32 PM, Blogger GameforYou said...

The NT-based versions of Windows, which are programmed in C, C++, and assembly,[9] are known for their improved stability and efficiency over the 9x versions of Microsoft Windows.[10][11] Windows XP presented a significantly redesigned graphical user interface, a change Microsoft promoted as more user-friendly than previous versions of Windows. A new software management facility called Side-by-Side Assembly was introduced to ameliorate the "DLL hell" that plagues 9x versions of Windows.[12][13] It is also the first version of Windows to use product activation to combat illegal copying.
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At 11:33 PM, Blogger GameforYou said...

During development, the project was codenamed "Whistler", after Whistler, British Columbia, as many Microsoft employees skied at the Whistler-Blackcomb ski resort.[14]
According to web analytics data generated by W3Schools, from September 2003 to July 2011, Windows XP was the most widely used operating system for accessing the internet. As of October 2011, Windows XP market share is at 33.4% after having peaked at 76.1% in January 2007.
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At 12:07 AM, Blogger GameforYou said...

Unlike Windows Vista, which introduced a large number of new features, Windows 7 was intended to be a more focused, incremental upgrade to the Windows line, with the goal of being compatible with applications and hardware with which Windows Vista was already compatible.[8] Presentations given by Microsoft in 2008 focused on multi-touch support, a redesigned Windows shell with a new taskbar, referred to as the Superbar, a home networking system called HomeGroup,[9] and performance improvements. Some standard applications that have been included with prior releases of Microsoft Windows, including Windows Calendar, Windows Mail, Windows Movie Maker, and Windows Photo Gallery, are not included in Windows 7;[10][11] most are instead offered separately at no charge as part of the Windows Live Essentials suite.
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At 11:11 AM, Blogger GameforYou said...

The coins struck between 1859 and 1864 contained 88% copper and 12% nickel. During this time, prior to the issuance of the Five-Cent nickel coin, the cent was commonly referred to as a "Nickel" or "Nick," for short. Due to the hoarding of all coinage during the Civil War, the nickel cents disappeared from daily use and were replaced in many Northern cities by private tokens. The success of these copper tokens prompted the change of the cent to a similar metal. In 1864, the alloy changed to Bronze (95% copper and 5% tin and zinc), and the weight of the coins was reduced from 72 grains to 48 grains. This weight continued for copper-alloy U.S. cents until the 1982 introduction of the current copper-plated zinc cent.
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At 9:49 AM, Blogger GameforYou said...

Ashikaga Takauji establishes the shogunate in Muromachi, Kyoto. It is a start of Muromachi Period (1336–1573). The Ashikaga shogunate receives glory in the age of Ashikaga Yoshimitsu, and the culture based on Zen Buddhism (art of Miyabi) has prospered. It evolves to Higashiyama Culture, and has prospered until the 16th century. On the other hand, the succeeding Ashikaga shogunate failed to control the feudal warlords (daimyo), and a civil war (the Ōnin War) began in 1467, opening the century-long Sengoku period ("Warring States").
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Well constructed (and humorous) camera test for the hot new Canon EOS C300.
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it is unclear if the majority of students really need to learn software engineering or would benefit from it. Force-feeding coding skills into students who may not have the aptitude or proclivity to want to learn them seems unwise to me and is likely to slow down the students who might actually have a desire to learn the subject."
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William John McGee reasoned in 1890 that the East and Gulf coasts of the United States were undergoing submergence, as evidenced by the many drowned river valleys that occur along these coasts, including Raritan, Delaware and Chesapeake Bays. He believed that during submergence coastal ridges were separated from the mainland, forming lagoons behind the ridges.[9] He used the Mississippi-Alabama barrier islands (consists of Cat, Ship, Horn, Petit Bois and Dauphin Islands) as an example where coastal submergence formed barrier islands, but his interpretation was later shown to be incorrect as the coastal stratigraphy and sediment ages were more accurately determined.
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At 8:18 AM, Blogger Let'go said...

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Saturn has an intrinsic magnetic field that has a simple, symmetric shape—a magnetic dipole. Its strength at the equator—0.2 gauss (20 µT)—is approximately one twentieth than that of the field around Jupiter and slightly weaker than Earth's magnetic field.[18] As a result Saturn's magnetosphere is much smaller than Jupiter's.[54] When Voyager 2 entered the magnetosphere, the solar wind pressure was high and the magnetosphere extended only 19 Saturn radii, or 1.1 million km (712,000 mi),[55] although it enlarged within several hours, and remained so for about three days.
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I'm currently working my buttocks off on my new book so I didn't have time to create a Christmas comic this year, but here's an oldie from 2010:

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At 8:04 PM, Anonymous Anonymous said...

This is a post on the "Monty Hall" problem. James Hein, in his 2009 book "Prolog experiments in Discrete Mathematics" (free download from internet), page 36, provides a Prolog program you can run to show that you should switch rather than say with your original choice.

I have a major in theory of mathematical statistics. The Prolog program he provides does not correctly model the problem. His solution of the problem is incorrect.

Richard Mullins
bulsitim@gmail.com

 
At 9:24 PM, Anonymous Anonymous said...

mea culpa. I still don't understand Hein's Prolog code. But
Wikipedia says: Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999).


Richard Mullins
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