Fuzzy Math…

[Slick91]

Grab a Calculator
1.key in the first three digits of your (US only) phone number (not the area code)

2. Multiply by 80.

3. Add 1

4. Multiply by 250

5. Add the last four digits of your phone number

6. Add the last four digits of your phone number again

7. Subtract 250

8. Divide by 2.


Check out the number.

[riverbravo]

6096553.....and?

[IanAM]

8700272

...close, but no cigar...

[Slick91]

Well, I can't say what you're doing but I tried it with four different telephone numbers and it worked out fine.

It is supposed to calculate out to the phone number you picked to solve the equation.

It's not fun if I have to explain it. [;)]

Oh, and a disclaimer, I'm not sure if it works for phone numbers outside the USA.

[IanAM]

Guessed that...my number's 870272...get an extra zero in the middle...

[IanAM]

Sorry...just read the "(US only)" - assume you have more digits than us!

[Makoto]

phone nerds [8|]

I remember one guy actually calculated out what the White House's secure line # was just by looking for a gap in numbers in a Washington DC phonebook.

[cnjpratt]

Interesting at first glance, but not exactly rocket science:

If x = first 3 #'s
y = last 4 #'s

Then the instructions become: ((((x*80)+1)*250)+2*y-250)/2

Simplify: x*40*250+y = x*10,000 +y

Therefore, whatever 3 digit number you put in for x ends up a 7-digit # ex. 756*10,000 = 7560000. Then you add the last 4 digits Ex 4156 = 7564156.

They just made it look complicated.

[Cheeks]

NOW HEAR THIS:

ORIGINAL: cnjpratt

Interesting at first glance, but not exactly rocket science:

If x = first 3 #'s
y = last 4 #'s

Then the instructions become: ((((x*80)+1)*250)+2*y-250)/2

Simplify: x*40*250+y = x*10,000 +y

Therefore, whatever 3 digit number you put in for x ends up a 7-digit # ex. 756*10,000 = 7560000. Then you add the last 4 digits Ex 4156 = 7564156.

They just made it look complicated.

I'm just guessing here.....but,...... did you get beat up alot when you were in school? [;)]

THAT IS ALL:

[Svennemir]

The Car and the Goats
You are a contestant on a television game show. Before you are three closed doors. One of them hides a car, which you want to win; the other two hide goats (which you do not want to win).

You get to pick one of the doors, and you will win what is behind it.

However, the way the game works is that the door you pick does not get opened immediately. Instead, the host (Monty Hall) will open one of the other doors to reveal a goat. He will then give you a chance to change your mind: you can switch and pick the other closed door instead, or stay with your original choice.

Which of these two strategies gives you the better chance of winning the car? This simple question recently caused quite a storm of mathematical controversy!

(From:
link)

[Belisarius]

Classic probability example Svennemir.

Answer: Switch doors.

[dinsdale]

Well when first picking the door the chance of winning the car is 1:3

After that, regardless of which door you pick the chance is 1:2. The probablility has improved after the first door was revealed, but picking either after that results in the same.

What was the controversy?

[Svennemir]

After that, regardless of which door you pick the chance is 1:2

No

Belisarius' answer is actually correct. I'll let you guys who probably read this continue thinking for a while (or you can click the link and find the solution somewhere).

[mavraam]

You should always switch.

[Rainbow7]

This is a great example that I always use in discussing probability and Bayesian statistics with students. It stumped many professional statisticians at the time (and continues to, from my experience). The problem becomes clearer (only to some) if you start with 1000 doors instead of three. Would you stick with your first door (only 1/1000 chance) or would you switch to the other door remaining after Monty opens 998 doors?

[mavraam]

ORIGINAL: Rainbow

This is a great example that I always use in discussing probability and Bayesian statistics with students. It stumped many professional statisticians at the time (and continues to, from my experience). The problem becomes clearer (only to some) if you start with 1000 doors instead of three. Would you stick with your first door (only 1/1000 chance) or would you switch to the other door remaining after Monty opens 998 doors?

Yes but that still doesn't clear it up for most. Its the fact that at the decision point you APPEAR to be effectively choosing 1 of 2 doors so it shouldn't matter.

But in fact are choosing 2 of 3 if you switch instead of 1 of 3.

I'm not sure that makes sense, but one thing I've found that will convince some is to get 3 playing cards, 1 red, 2 black and simulate the situation 20 times. The first ten, never switch. The second ten always switch. The second 10 should almost always win over the first barring some statistical anomoly.

The fact that its so difficult to convince someone of the correct answer despite the simplicity of it points to how brilliant of a puzzle this is.

I remember a puzzle about 3 people splitting a check that ended up shorting the waiter $1 even though it appeared that everyone had paid their fair portion but I can't remember the specifics. [:(]

[Belisarius]

ORIGINAL: mavraam

I remember a puzzle about 3 people splitting a check that ended up shorting the waiter $1 even though it appeared that everyone had paid their fair portion but I can't remember the specifics. [:(]

Yeah that's also a classic, but in that case it really is fuzzy math due to rounding errors.

Sorry, couldn't dig up a link either.

[Svennemir]

I'm not sure that makes sense, but one thing I've found that will convince some is to get 3 playing cards, 1 red, 2 black and simulate the situation 20 times. The first ten, never switch. The second ten always switch. The second 10 should almost always win over the first barring some statistical anomoly.

It's funny, I also like to use playing cards to convince people. When they've tried to not switch 10 times, it becomes clear that they could as well select the door/card right away without the host revealing anything, and then the 1/3 follows.

Lo and behold! The power of google enabled me to find the one about the waiter you mentioned. (and in the first try!)

Here it is: Three men go to stay at a motel and the clerk
charges them $30.00 for the room. They split the cost ten
dollars each. Later the manager tells the clerk that he over-
charged the men and that the actual cost should have been
$25.00. He gives the clerk $5.00 and tells him to give it to the
men. But he decides to cheat them and pockets $2.00. He then
gives each man a dollar. Now each man has paid $9.00 to stay in
the room and 3 X $9.00 = $27.00. The clerk pocketed $2.00.
$27.00 + $2.00 = $29.00. So where is the other $1.00?

Link

[Svennemir]

By the way, it can be tricky! But I've figured it out (and it's not about rounding errors - come to think of it, maybe you thought about another one, Belisarius?)

[Svennemir]

Of course I shouldn't forget this little gem:

Did you hear about the mathematician who was looking all over for the eigenvalues of a matrix, but couldn't find a trace?

(you have to be a nerd to get this one)