This is the foo bar.

### Since a few people said they liked it

I put the prime numbers via dice thing on my website.

Yeah, it's a copy/paste, but it suffices.

Let me know if there are other things similar you'd like to see. I'll do what I can.
Aaron F Stanton
August 23rd, 2005
I was always a fan of the Monty Hall problem as a good demonstration of how "intuitive" probability can get you in trouble...
Mat Hall
August 23rd, 2005
Your last step you say "Hey maybe 5 has the same property as 2 and 3". Question - will you then look back at you "8" dice and say "Hay maybe 4 has the same property as 2, 3 and 5"? (which it does for the property you stated, that every 4 more dice you'll be able to make a 4 by x rectangle)
just to be annoying
August 23rd, 2005
Except that you didn't follow the problem at all.

8 dice can make an 8x1 rectangle or a 4x2 rectangle. 2, 3, and 5 dice only have one configuration.
muppet
August 23rd, 2005
In other news, prime numbers still aren't divisible except by one and themselves.

August 23rd, 2005
Whoa - what?! When did that happen?

God. Next thing you know P.Diddy is gonna drop the "P"

Philo
Philo
August 23rd, 2005
It's a physical representation of primes, not an abstract one. May I should explain that in the post.
Aaron F Stanton
August 23rd, 2005
Not to go all McKinstry, but I noticed a day or two ago that someone with IE 5.17 hit my site. 5.17??? What browser reports itself as that? (And no, I don't care enough to try and track down their IP address, country, and shoe size. It's just something Mambo reported, and this is idle curiousity.)
Aaron F Stanton
August 23rd, 2005
IE 5.17 was one of the MacOS 9 versions of IE.
Mat Hall
August 23rd, 2005
Ah...thank you. That explains it.

By Monty Hall, you mean the "Pick one out of three, I show you one of the other two is bad, do you switch?" problem?
Aaron F Stanton
August 23rd, 2005
Aaron : Yep, that's the one. Probably the classic counterintuitive probability problem ... though, for a "purer" counterintuitive probability problem, I like the "test for cancer with a 5% false positive" problem.
Snark
August 23rd, 2005
Not familiar with that one.
Aaron F Stanton
August 23rd, 2005
That's the chap. It's totally counter-intuitive and some people get quite angry when you explain it to them. On the face of it it makes no sense whatsoever that swapping improves your chances, and there's often a complete refusal to accept evidence to the contrary. Hours of fun. :)
Mat Hall
August 23rd, 2005
Aaron :

It's a Bayesian probability problem, like the Monty Hall one. I like it slightly better, since the explanation is somewhat easier to grasp.

The setup is :
1. A certain disease has a .1% rate of incidence in the population at large -- that is to say, of the entire population, one out of every thousand people will have this disease.
2. There is a simple blood test for this disease. The blood test has a 5% rate of giving a false positive result. For simplicity's sake, we'll say that there is a 0% rate of false negatives (the test always signals "positive" if the person tested has the disease).
3. You are tested for the disease as part of a routine medical checkup, and the test comes back positive. Should you trust the test results, or is it more likely that you -don't- have the disease?

The answer, of course, is that -even though the test result is positive for you-, you are in fact very unlikely to have the disease. For most people, this result is counterintuitive, since we don't usually think in terms of conditional probability. The explanation (pretty easy to sketch out on a cocktail napkin, which is the merit of this particular setup) is as follows :

Consider what happens when we apply this test to 1000 people. Out of those 1000 people, 1 will actually have the disease (since its rate of occurrence in the population is .1%, or 1 out of 1000). This person will be detected by the test ... let's set them aside into our group of "positive results."

Now out of those 999 remaining people who don't have the disease, the test will return 50 false positives (5% rate of false positives, but I rounded up the fraction for clarity's sake). So let's add those 50 people to our group of "positive results."

The "positive results" group now contains 50 disease-free individuals and 1 person who actually has the disease. This means that any person with a positive test result really only has a (1/51 = ~.0196) less than 2% chance of actually having the disease.

Interestingly, when I first encountered this problem, it was part of a study to determine whether doctors had a good grasp of what test results and rates of incidence actually mean. In the study, a sample group of doctors was presented with this problem as a real-world example -- IIRC, something like 75% of the doctors answered -incorrectly-.
Snark
August 23rd, 2005
That's an interesting one. By the way, you might as well round up in this case - a non-integer number of people is rather odd to think about. (Even though most here would have understood.)
Aaron F Stanton
August 23rd, 2005
+++That's an interesting one. By the way, you might as well round up in this case - a non-integer number of people is rather odd to think about. (Even though most here would have understood.)+++

Especially Mat Hall, since he's barely half a man to begin with.
muppet
August 23rd, 2005
Yeah, it's a simple way to introduce conditional probability. There is one important point to emphasize, though : the test is applied randomly in the population (I like "as part of a routine checkup") -not- just if you're suspected of having the disease, since that imposes an unquantified condition and skews your results.

I also like to add a clause about how the disease is treatable, but the treatment has unpleasant side effects. This way, people actually think about "well, -should- I trust the test or not?" and it really sticks the point home if they do trust the test (as most people do).

<nerd>
If you want another amusing example for Bayesian inference, you can use the real-life 1968 search for the USS Scorpion nuclear submarine. Research on that one is left to the discretion of the reader. ;)
</nerd>
Snark
August 23rd, 2005
I'm *all* man, baby!
Mat Hall
August 23rd, 2005
A simple one:

There are three cards in a hat. One is painted red on both sides, one is painted black on both sides and another is painted red on one side and black on the other. I pick a card and we see that the side facing us is red. I reason that since there are equal chances that it can be either a red or a black, I bet my money that it is red. Is it a fair bet?

The problem I find is that if there's a convincing explanation one way, it is very difficult to think how it would've been some other way. I understood the Monte Carlo problem well. I was able to explain it once. But there are chances that it will again grow hazy. Is there any way to understand it better? Or may be it's because of the IQ level being low..:-)
Senthilnathan N.S.
August 23rd, 2005
I guess I'd explain that by saying that as you know it's not the black-on-both-sides card it's either the red-on-both-sides or red-black card. Of the two possible cards, there's a one in three chance of pulling one out with the red facing up with two of those belonging to the red-on-both-sides card, meaning that it's more than likely it's the red card -- two out of three "red face up" events will be the red-on-both-sides card...
Mat Hall
August 23rd, 2005
Re. the three cards problem :

If you pick a card with the red face up, you should always bet that the other face will also be red. Let's identify the faces of the cards :

1. Face = red, reverse = red.
2. Face = red, reverse = red.
3. Face = red, reverse = black.
4. Face = black, reverse = red.
5. Face = black, reverse = black.
6. Face = black, reverse = black.

Faces 1 and 2 belong to the card with both faces red, faces 5 and 6 belong to the card with both faces black, and faces 3 and 4 belong to the card with one black and one red.

You must have chosen either 1 or 2 or 3 ... which gives you a 2/3 probability of the reverse being red. This isn't really a conditional probability problem at all, right?

Or am I completely wrong?
Snark
August 23rd, 2005
Both Mat and Snark are perfectly correct. And as Snark says it's not conditional probability at all. It's a simple one, atleast for people here.

Kahneman and Tversky explain how we aren't wired to consider Bayes's rule when we ought to do it.
http://history.behaviouralfinance.net/TvKa74.pdf
Senthilnathan N.S.
August 23rd, 2005
> Let me know if there are other things similar you'd like to see.

Pictures of the stripper?