It's that time of year again. (Gindin's annual tomfoolery. (2015 Update)

[Ben E Lou]

Check out the Solecismic site. :D

[Ben E Lou]

Solecismic Software to Cease Publication



Solecismic Software has announced that it will no longer publish Front Office Football games.

"We have decided to move toward titles less suited to hardcore gamers," said Ethan Potter, director of Customer Relations for Solecismic Software. "We have an exciting new Hello Kitty screen saver in the pipeline that we think can sell in the hundreds of thousands."

Potter went on to say that the sports simulation genre is out of favor in publishing circles. He scoffed at today's announcement of Front Office Football 2006 (http://www.solecismic.com/fof/index.php) as too little, too late.

"Who is Gindin kidding?" Potter laughed. "Terrell Owens for the end-zone celebrations? That is so 2001. Talk to me when you get Joe Horn. Talk to me when you replace those boring statistics with floating 3-D numbers that sparkle and dance." Jim Gindin, founder and lead developer of the Front Office series of games, could not be reached for comment. But a reporter stationed outside his home in rural Amherst, New Hampshire heard the distinct sound of a sledgehammer smashing a big-screen television. Gindin is rumored to be considering a career change. He was offered the head coach position on the University of New Hampshire varsity women's curling team. <SMALL>

</SMALL>

[Ben E Lou]

Front Office Football 2006


Be the first on your block to own a 2006 football game. Forget about all those products that merely mention the year 2005, Front Office Football 2006 is truly a next-generation product.

FOF 2006 takes advantage of rules changes even the NFL hasn't yet implemented. With scores down during the 2003 season, drastic action was necessary. It won't get any better in 2004 and 2005. So, in 2006, the league will experiment by reducing the yardage required for a first down from ten yards to nine.

FOF 2006 has trademarked the phrase "first and nine, time to move the line." Take that, Madden.

The NFL will also significantly change defensive pass interference penalties. Twice a game, each wide receiver will be allowed to throw his own flag. The referee will then review the play, and if the defender initiated any contact past the five-yard bump zone, it's an automatic penalty. Receivers successfully using two challenges during the game will be granted a third challenge.

FOF 2006 will incorporate this change with a new technology never before found in a video game. DirectX 11.0, which is scheduled for release in 2005, will include a new "touch and feel" texture-pad interface on the mouse. So, when you're using the Microsoft IntelliTouchyFeelyMouse, you'll be able to sense when your receiver has been unfairly abused, and then you can instruct him to throw that flag.

Front Office Football 2006 will also incorporate the following new features into the game:

Interactive Combine Test Results - see your potential draftees take the Solecismic Intelligence Test (SIT) during the annual scouting combine. You will be allowed to interfere with the test by giving certain players the correct answers, thus overinflating their scores and making them more interesting to opposing teams during the draft. But cheat too often, and risk losing your own draft picks.
NFL Afghanistan - forget about NFL Europe, it's failing. By 2005, all the hot offseason NFL action will be in Afghanistan and western Asia. In FOF 2006, you can train your young players by sending them to play in NFL Afghanistan. Be careful, as it's still a violent world, and some of them may not come home.
Peytronizing - When it comes to negotiating a contract, all players feel they need more money. The NFL stops this horrible example of greed by allowing each team one franchise or transition player. That player automatically receives a one-year contract at a high rate. But that's not enough in 2006. You'll be allowed to Peytronize an additional star player. For a one-time fee of 40% of the salary cap, you'll be able to keep your favorite player. But there's a cost. He'll get to choose five of his teammates for immediate release. And one of them might well be your liquored-up idiot all-pro kicker.
End-Zone Celebrations - Yeah, we all heard how celebrating was evil. In 2004, the NFL is supposed to end touchdown routines, sack dances and first-down pointing. Well, attendance will suffer, and by 2006, everyone will be begging Terrell Owens to come up with a new way to humiliate his opponents. Here at Solecismic Software, we've heard the clamor for improved graphics and yummy little eye-candy routines. So, we rented some time at a motion-capture studio in Philadelphia, traveled down there and convinced Owens to try out 100 or so original routines. We then hired a studio experienced in Playstation 3 development to process these routines and incorporate them into FOF 2006. While the famiar scoreboard interface will still be used to display the play-by-play, when you score a touchdown, Terrell will dance for you in his full 24-bit color, 3D, 1280x1024 screen glory.
New Injuries - Fans of the Front Office series are already familiar with the remote possibility that half your team will come down with food poisoning right before a key matchup. Bo-ring. Now, each week, in addition to the weather and the attendance for each game, you'll have access to a description of each team's pre-game meal. Now, if your team suffers food poisoning, you'll know what caused it, and you'll be able to avoid the Chicken Kiev on your next road trip to Buffalo.
Leonardo da Vinci Mania! - We've heard the complaints that Front Office Football looks too much like a spreadsheet. Well, we've made a special deal with Microsoft founder Bill Gates' Corbis company to access his library of fine art. Now, if you're tired of looking at statistics, with one button-press you can change that ugly graph of numbers into the Mona Lisa. Or that salary chart can become Jean Baptiste Camille Corot's Civita Castellana. Or those infuriatingly unattractive league leader lists can become a pretty kitty cat. The possibilities are endless.
There's just so much more in FOF 2006, we can't even begin to tell you about it. You'll just have to download the demo and try it out. Because of piracy concerns, we've limited this full-function demo to 11 seconds of game play. But it's still worth the 2 terabyte download just to see what we've done with the installation routine.

Front Office Football 2006 is scheduled for release on April 1, 2004 at your neighborhood Wal-Mart, Target and Lane Bryant stores.

[Easy Mac]

Thank God, now we can retitle the board HKC (Hello Kitty Central).

[JAG]

LOL :)

[Tekneek]

Because of piracy concerns, we've limited this full-function demo to 11 seconds of game play. But it's still worth the 2 terabyte download just to see what we've done with the installation routine.

:D That was...AWESOME.

[fantastic flying froggies]

Hey, I thought the WR's own penalty flags were a pretty cool idea, imagine what Owens or Moss would do with those ?

:D

[condors]

i love the peytronizing thing :)

[mckerney]

A great April Fools Day Prank would be for OOTP Developments to release OOTP6. I'd get a kick out of that one.

[Ksyrup]

:D That was...AWESOME.
Yep. You've got to respect a man who can poke fun at himself. Well done!

[Flame Eater]

MEEEOOOOOOWWWW!!

[Young Drachma]

hehehehe...

great stuff

[Celeval]

Rock on! I'm heading to Lane Bryant right now!

[Subby]

Pick me up some size-16 mom jeans while you are there!!!

[Tekneek]

Rock on! I'm heading to Lane Bryant right now!
:(

[WSUCougar]

LOL

Love it.

Now, what's the over/under on newbs posting "I can't find the FOF2006 download" threads? :D

[SirFozzie]

Cool, I'll have to tell my friend Sheila about this one, she can get the Hello Kitty Screensaver to go with the Hello Kitty Vibrator she already owns.

(yes, there really IS a Hello Kitty Vibrator)

[JeeberD]

Outstanding stuff. I just wish Jim had hired George Teague for the endzone celebrations as well...

http://mywebpage.netscape.com/RICE%20vs%20SUSHI/TeagueOwens.gif

:D

[Darkiller]

Excellent stuff from Jim, very creative.;)

[Passacaglia]

Bi Fny F Tts F2b F Icm Nio Pbeyl Eahsrwi Npaa Lswce Tpd F

I don't get it.

[JAG]

Jeeber, thanks for the trip down good memory lane. :) My only regret is that he didn't get an even better shot on him.

[BigJohn&TheLions]

I read on and the site also mentioned a "FOF2004" What is that? Like we'd fall for that! Jim is such a kidder!

<img src="http://people.ucsc.edu/~awu/Hello%20Kitty%20Castle.jpg">

[Ben E Lou]

Bump for 2005...


Solecismic Software is Hiring!

We have a new position available supporting our fall 2005 game release. If you're interested in becoming part of our vibrant company, please complete an application (http://www.solecismic.com/application.php) and tell us all about yourself.

[TLK]

"Football 101".... can't wait.... :)

[Ragone]

hehe, i like homestarrunners.com 's april fools day prank :)

[Tekneek]

hehe, i like homestarrunners.com 's april fools day prank :)

1 player simultaneous...

"SOLVE FOR X!!"

[gstelmack]

FOF 2006 will incorporate this change with a new technology never before found in a video game. DirectX 11.0, which is scheduled for release in 2005, will include a new "touch and feel" texture-pad interface on the mouse. So, when you're using the Microsoft IntelliTouchyFeelyMouse, you'll be able to sense when your receiver has been unfairly abused, and then you can instruct him to throw that flag.

Somehow I missed this last year, but there used to be a force-feedback mouse on the market (I've written code for it). In other words, except for the flag throwing bit, the rest of this is entirely plausible.


I've got to see if I can come up with a good angle for the Solecismic job application. Jim's just begging for us to make his day, and I think it's time we gave something back.

[Franklinnoble]

I've submitted my application. The rest of you might as well not bother - that internship is as good as mine.

[VPI97]

I'm too old to intern :(

[3ric]

Hope Jim publishes some of the best applications tomorrow - there are bound to be some good ones sent his way...

[Ksyrup]

You wouldn't want to commit a public solecism, would you?


LOL!

[Draft Dodger]

funny stuff from the applicants.

[Franklinnoble]

Second place?!? Bah.

My application was actually a lot longer... shoulda kept it for posterity.

[VPI97]

http://www.solecismic.com/

New Information about Our Current Project

Yes, Solecismic Software is involved in something exciting and new. Please visit our puzzle page for a set of challenging mind twisters that may lead to a surprising discovery. Hurry, time is running out.

[Tigercat]

Hmm what an interesting day for breaking info! Let me guess, the puzzle will hint at front office two man skeleton?

[sabotai]

And I guess this just happens to magically appear on April Fool's Day?

[sabotai]

For the lazy, here are the puzzles.

Yes, brand new information is available about Solecismic Software's current and pressing project. We'd love to share it with everyone, but it seems too simple just to put it out there. Almost anticlimactic.
So, we're going to make you work for it. We've placed this information in a file at the top level of our web site. The URL is a standard php file, but the prefix is a code. Something like http://www.solecismic.com/1234567.php, where the answers to the questions below will give you the 1234567 part.
If you solve all these problems correctly, you will be able to construct the URL we've hidden this information on. You have until April 2 to find this URL, when the information will disappear altogether.

Part 1

I'm into birthdays. Nothing makes me happier than to see two people randomly find out they have the same birthday. That's just way cool.
So, I'm holding a party, but I'm on a limited budget. It will cost me $7 in food and drink for each guest.
How much money, minimum, will it cost me to put together a party at which I'm at least 98% certain that two guests at the party will have the same birthday?

Part 2

I've gone crazy and decided to bet $30,000 on a best-of-seven baseball series between Boston and Philadelphia. That is, I win $30,000 if Boston wins the series and I lose $30,000 if Philadelphia wins the series.
There's one catch. I can only bet on each game individually and I must bet on each game. It's an antiquated betting web site and can't handle anything more complex. You can assume the odds of each game are set at 1:1, and the house does not take a cut.
So, game one is coming up, and Curt Schilling is facing Cory Lidle. How much money do I put down on Boston?

Part 3

I like playing bridge, but the mathematics is sometimes a lot tougher than it looks.
Anyway, I bid too aggressively, and when my partner's hand was laid out, I found he only had four hearts to go with my four hearts. The only way I can make this bid is if our opponents hearts are split 2/3 or 3/2.
Rounding down to the nearest whole number, what is my percentage chance of making the bid, assuming I make the bid if I get that split?

Part 4

Here's an easier one. Texas Hold 'em has received a lot of attention in the media lately, mainly because celebrities with brains the size of almonds seem to do all right competing against each other.
So, pretend you're Alexis Bledel, and you have an eight and a six in the pocket. Because you're adorable and cute, you somehow survive the first round of bidding, and you're encouraged to see a seven, a five and a king in the flop. Neato Cool!
There are six other players in the game, three of whom have folded already, including Martin Sheen and Brian Austin Green.
Suddenly, a math geek appears by your left shoulder and whispers something in your ear. Strangely, you're not disgusted, because while trying to peer down the front of your blouse, he tells you to bet 10,000 times the probability that you'll get your straight off the turn and the river, rounding down, of course. How much did he tell you to bet?

Part 5

If you want to receive a PhD in math, all together you're going to go through about 21 grades' worth of school material.
Well, math geeks tend be quite intelligent people overall, especially in ways that impress teachers.
Let's say we have a truly superior math geek student who can skip grades at will. But he can only skip one grade per year. Or not, if he chooses. So he can go one grade at a time and take 21 years to obtain his PhD. Or he can skip as often as possible, and need only 11 years. His choice.
Those are only two of literally thousands of ways he could choose to go. So, how many different combinations of skipping and not skipping are possible?

Putting it All Together

Just take the answers to the five questions, one through five, and create that URL. Remove all dollar signs, percentage signs and commas. As a hint, you should have 18 numbers in all. Good Luck.










Now solve them slackers!

[TazFTW]

And I guess this just happens to magically appear on April Fool's Day?

D'oh!

I had suggested in the past that Jim announce his next game via puzzle but this occuring on April 1st is not a good omen.

[Raiders Army]

And I guess this just happens to magically appear on April Fool's Day?
Good call.

[VPI97]

Part 1

I'm into birthdays. Nothing makes me happier than to see two people randomly find out they have the same birthday. That's just way cool.
So, I'm holding a party, but I'm on a limited budget. It will cost me $7 in food and drink for each guest.
How much money, minimum, will it cost me to put together a party at which I'm at least 98% certain that two guests at the party will have the same birthday?

The odds that two people have the same birthday are: (365/365) * (1/365) :: (100% chance the first person is born on their birthday) * (odds that second person is born on that specific day)...or 0.27%

You can also arrive at that number by figuring out the probability that the two people don't share a birthday and subtracting that from 1... (365/365) * (364/365) = .9973 = 99.73% chance they don't (0.27% chance they will). Figuring out three poeple would be (365/365) * (364//365) * (363/365) = .9918 :: 1 - .9918 = 0.82% chance of two people having the same birthday out of the three.

Plugging that calculation into Excel says that you need 53 people to get a 98.12% chance that two out of the 53 share the same birthday. Multiple that by $7 in food and drink and the answer to #1 is 371.

[MIJB#19]

The odds that two people have the same birthday are: (365/365) * (1/365) :: (100% chance the first person is born on their birthday) * (odds that second person is born on that specific day)...or 0.27%

You can also arrive at that number by figuring out the probability that the two people don't share a birthday and subtracting that from 1... (365/365) * (364/365) = .9973 = 99.73% chance they don't (0.27% chance they will). Figuring out three poeple would be (365/365) * (364//365) * (363/365) = .9918 :: 1 - .9918 = 0.82% chance of two people having the same birthday out of the three.

Plugging that calculation into Excel says that you need 53 people to get a 98.12% chance that two out of the 53 share the same birthday. Multiple that by $7 in food and drink and the answer to #1 is 371.You haven't taken into account the odds that two people are both born on the 29th of February...

[Fouts]

Part 4

Here's an easier one. Texas Hold 'em has received a lot of attention in the media lately, mainly because celebrities with brains the size of almonds seem to do all right competing against each other.
So, pretend you're Alexis Bledel, and you have an eight and a six in the pocket. Because you're adorable and cute, you somehow survive the first round of bidding, and you're encouraged to see a seven, a five and a king in the flop. Neato Cool!
There are six other players in the game, three of whom have folded already, including Martin Sheen and Brian Austin Green.
Suddenly, a math geek appears by your left shoulder and whispers something in your ear. Strangely, you're not disgusted, because while trying to peer down the front of your blouse, he tells you to bet 10,000 times the probability that you'll get your straight off the turn and the river, rounding down, of course. How much did he tell you to bet?


I believe it is 34% to complete an open ended straight draw on the turn or river (I could be wrong). .34 x 10000 = 3400.

[MIJB#19]

I believe it is 34% to complete an open ended straight draw on the turn or river (I could be wrong). .34 x 10000 = 3400.
52 cards.
5 are known.
47 are unknown.
8 are unknown and are a desired four or nine.

I think the odds to get a good card are:
8/47 + 39/47 * 8/46 = .31452

Multiplied by 10,000 that would be 3145.

[MIJB#19]

Part 3

I like playing bridge, but the mathematics is sometimes a lot tougher than it looks.
Anyway, I bid too aggressively, and when my partner's hand was laid out, I found he only had four hearts to go with my four hearts. The only way I can make this bid is if our opponents hearts are split 2/3 or 3/2.
Rounding down to the nearest whole number, what is my percentage chance of making the bid, assuming I make the bid if I get that split?
I think it's 33 percent.

With six options (0/5, 1/4, 2/3, 3/2, 4/1, 5/0), two winning options, it gives 1/3rd as the answer.

Edit: Nah, that can't be it, that sounds too easy.

[Fouts]

52 cards.
5 are known.
47 are unknown.
8 are unknown and are a desired four or nine.

I think the odds to get a good card are:
8/47 + 39/47 * 8/46 = .31452

Multiplied by 10,000 that would be 3145.

Ok, if you say so. He did say to round down though.

[Flasch186]

slowly backing out of thread as I think I just heard one of the seams in my brain pop

[Chubby]

www.solecismic.com (http://www.solecismic.com)/baseball.php

[JeeberD]

Ping: Quiksand

[TredWel]

I can do Part 3, being a bridge player myself. You'll face 3-2 distribution 68% of the time. That's one of the magic numbers bridge players learn.

[QuikSand]

You haven't taken into account the odds that two people are both born on the 29th of February...

I have always seen these puzzles solved with the convenient ignoring of leap years, for whatever reason. (Well, I guess the reasosn is simplification) Always struck me as a bit odd.

[QuikSand]

If you want to receive a PhD in math, all together you're going to go through about 21 grades' worth of school material.
Well, math geeks tend be quite intelligent people overall, especially in ways that impress teachers.
Let's say we have a truly superior math geek student who can skip grades at will. But he can only skip one grade per year. Or not, if he chooses. So he can go one grade at a time and take 21 years to obtain his PhD. Or he can skip as often as possible, and need only 11 years. His choice.
Those are only two of literally thousands of ways he could choose to go. So, how many different combinations of skipping and not skipping are possible?


I'm going to assume that both grade 1 and grade 21 are eligible to be skipped... though the wording sets up the possibility that either one or both are requred (i.e. how would this student prove his grade-skipping worth without attending year one?).

[Lonnie]

Well if you skipped year 1 and every other year after you would end up skipping year 21 also, which would make 10 years, not the 11 that he stated.

[Dutch]

Worst April Fools Joke ever.

Here I am, packing to go on a trip and will have to wonder about this the entire time I'm on the road.

So tonight, when I get to the hotel, I'm going to be dying to open up FOFC and figure out what the link says....

And the over/under right now is that it will say, "Be sure to drink your Ovaltine."

FOF-Why do you torment me so???? Argh!!!

[bbor]

2/3

[QuikSand]

Well if you skipped year 1 and every other year after you would end up skipping year 21 also, which would make 10 years, not the 11 that he stated.

You're right... that does suggest that grade one can't be skipped. Thanks.

(Though I'm really not sure how to set this up, at least that sets me straight, I think)

[MJ4H]

Worst April Fools Joke ever.



not. even. close.

[vex]

I pick (C)

[Franklinnoble]

"A crummy commercial?!?! Son of a bitch!!!"

[albionmoonlight]

Re: Baseball betting

I think that the answer is $1,875 on Boston.

I think that trying to post my cyphering here would just confuse things, but I can at least outline my method. You start with the assumption that going into a game seven, your account must be at zero so that you can make a $30,000 bet for that game.

So then you ask yourself--for a series that is 3-2 after game 5, what do I need in my account such that a winning bet brings me to $30,000 and a losing bet brings me to $0? Then, once you have that number, you ask yourself the question about game 4, then about game 3, etc.

[QuikSand]

Re: Baseball betting

I think that the answer is $1,875 on Boston.

I think that trying to post my cyphering here would just confuse things, but I can at least outline my method. You start with the assumption that going into a game seven, your account must be at zero so that you can make a $30,000 bet for that game.

So then you ask yourself--for a series that is 3-2 after game 5, what do I need in my account such that a winning bet brings me to $30,000 and a losing bet brings me to $0? Then, once you have that number, you ask yourself the question about game 4, then about game 3, etc.

I buy that logic.

[QuikSand]

That means that question five has a five digit answer, right?

I thought about it a bit, and am not sure how to represent the concept of "adjacent" values into a combinatorial problem like this.

[MizzouRah]

http://home.teleport.com/~rasputin/ImageFiles/BeavisButthead.jpg

Uhhhhh... this is like hard and stuff.

Yeah, yeah...huhuhuhu...

[Katon]

Could you try doing it recursively? For instance:

Ways of passing 1 grade while skipping the last grade: 1
Ways of passing 1 grade while taking the last grade: 1

Then for 2 grades you can skip the last grade only if you took the one before it, while you can take the last grade whichever you did, so

Ways of passing 2 grades while skipping the last grade: 1
Ways of passing 2 grades while taking the last grade: 2

Ways of passing 3 grades while skipping the last grade: 2
Ways of passing 3 grades while taking the last grade: 3

Ways of passing 4 grades while skipping the last grade: 3
Ways of passing 4 grades while taking the last grade: 5

And this is pretty clearly going to stay as the Fibonacci numbers, so

Ways of passing n grades while skipping the last grade: the nth fibonacci number
Ways of passing n grades while taking the last grade: the n+1th fibonacci number

Ways of passing 21 grades while skipping the last grade: 10946
Ways of passing 21 grades while taking the last grade: 17711

So that would give a total of 28657 ways of passing 21 grades.

Can anyone spot any holes in my logic?

[AlexB]

Could you try doing it recursively? For instance:

Ways of passing 1 grade while skipping the last grade: 1
Ways of passing 1 grade while taking the last grade: 1

Then for 2 grades you can skip the last grade only if you took the one before it, while you can take the last grade whichever you did, so

Ways of passing 2 grades while skipping the last grade: 1
Ways of passing 2 grades while taking the last grade: 2

Ways of passing 3 grades while skipping the last grade: 2
Ways of passing 3 grades while taking the last grade: 3

Ways of passing 4 grades while skipping the last grade: 3
Ways of passing 4 grades while taking the last grade: 5

And this is pretty clearly going to stay as the Fibonacci numbers, so

Ways of passing n grades while skipping the last grade: the nth fibonacci number
Ways of passing n grades while taking the last grade: the n+1th fibonacci number

Ways of passing 21 grades while skipping the last grade: 10946
Ways of passing 21 grades while taking the last grade: 17711

So that would give a total of 28657 ways of passing 21 grades.

Can anyone spot any holes in my logic?

Miss Hoover, my worm went in my mouth and I then ate it, can I have another one?

[Lonnie]

I thought about it a bit, and am not sure how to represent the concept of "adjacent" values into a combinatorial problem like this.

Yeah that really makes it tough. I tried doing a few binary numbers by hand to see if I could find a pattern. Starting with 00,01,10,11 which leaves 3 valid choices. I then added a digit and eliminated double zeros to find valid combinations. I went through a six digit binary and got the pattern 3, 5, 8, 13, 21,.....

I guess since we are ignoring the first and 21st year, we can assume if you can expand that out to 19 digits, the last number would be the answer.

I have to go to a hockey game now. Maybe that will help, but I got stuck there and didn't want to do any more by hand.

[gstelmack]

For question 5, you have:

1 way he can skip no years
+
20 ways he can skip one year
+
pick any 2 from 19 years =
+
pick any 3 from 18 years =
...

does that logic make sense? It's a combinatorics problem then, correct?

[QuikSand]

(to Katon) If the implicit rules don't allow you to skip the first year, then it basically becomes a question of only 20 years, not 21. In that case, your logic leads to 17711.

[QuikSand]

I went through a six digit binary and got the pattern 3, 5, 8, 13, 21,.....

There are the Fibonaccis again...

[QuikSand]

pick any 2 from 19 years

But you can't just pick any two -- you can't skip both year 6 and 7, for instance.

[gstelmack]

So the guesses so far are:

Part 1: 371
Part 2: 1875
Part 3: 68
Part 4: 3145
Part 5: 28657

http://www.solecismic.com/371187568314528657.php cannot be found, so at least one is wrong. Right number of digits, though.

[gstelmack]

Based on Quik's correction for part 5:

Part 1: 371
Part 2: 1875
Part 3: 68
Part 4: 3145
Part 5: 17711

http://www.solecismic.com/371187568314517711.php (http://www.solecismic.com/371187568314528657.php) can't be found either.

[Katon]

(to Katon) If the implicit rules don't allow you to skip the first year, then it basically becomes a question of only 20 years, not 21. In that case, your logic leads to 17711.

There's also the question of whether you can skip the last year. If you can't skip either then it gives 10946. Still, that's more an issue with the phrasing of the question than with the basic logic of the solution.

edit: http://www.solecismic.com/371187568314510946.php can't be found either, so I could be wrong. Or maybe somebody else is.

[gottimd]

Face it, the answer will not show you anything, its a waste of time to figure this out.

[Deattribution]

Based on Quik's correction for part 5:

Part 1: 371
Part 2: 1875
Part 3: 68
Part 4: 3145
Part 5: 17711

http://www.solecismic.com/371187568314517711.php (http://www.solecismic.com/371187568314528657.php) can't be found either.

Ya know how ridiculous that url looks right? :)

And you guys know that you could potentially get the problem right and still never find the page because it doesn't exisit....

[MJ4H]

Face it, the answer will not show you anything, its a waste of time to figure this out.

Only if you dislike BOTH problem solving and Jim's fake announcement April Fools Day jokes. That describes me perfectly, of course.

[gottimd]

Only if you dislike BOTH problem solving and Jim's fake announcement April Fools Day jokes. That describes me perfectly, of course.

The answer will be a redirect to tubgirl.

[gstelmack]

Trying 10945 and 17710 for part 5 (sums of the first 20 and 21 Fibonnaci numbers) is page not found as well.

[gstelmack]

And you guys know that you could potentially get the problem right and still never find the page because it doesn't exisit....

Still some fun puzzles. And Jim did hint that he wanted to announce this way. But even if it is all a big joke, what's the harm in letting Jim have his fun and us try to solve some puzzles?

[Draft Dodger]

Still some fun puzzles. And Jim did hint that he wanted to announce this way. But even if it is all a big joke, what's the harm in letting Jim have his fun and us try to solve some puzzles?

especially considering WE are the people he made the puzzles for.

[jbmagic]

Its an April Fool Joke. :)

I can't believe your falling for this.

[EagleFan]

I get 196 for part 1, (28 people), though that actually yields over 98 percent but 27 people yields about 96 percent.

[3ric]

I assume Jim has the courtesy of creating an "April Fools" page to let you know you solved the riddle.

[Ben E Lou]

I assume Jim has the courtesy of creating an "April Fools" page to let you know you solved the riddle.Of course, the April Fool could be that this year it is NOT a joke, but a real announcement, given that he does an April Fools' joke every year. :p

...and I agree with Todd on the 53-person answer, yielding $371.

[Ben E Lou]

I'd like to see the mathematics on the 68% chance of a 3-2 split, by the way. I'm lost on that one.

[cthomer5000]

Its an April Fool Joke. :)

I can't believe your falling for this.

You can't be serious, can you?

[Barkeep49]

Ok so even if it is an April's Fools joke it still behooves us, I think, to solve this puzzle. Better safe then sorry and all that.

[GrantDawg]

Y'all aren't through yet? Us mathematically challenged folks are waiting.

[ISiddiqui]

Jim is laughing somewhere :D.

[kurtism]

Jim is laughing somewhere :D.

Wait, I know this one!

Michigan

[MizzouRah]

I believe it's real, but since we only have one day to complete all 5 puzzles, Jim knows the probability of all 5 right answers are slim.

[Maple Leafs]

Re: Baseball betting

I think that the answer is $1,875 on Boston.

I think that trying to post my cyphering here would just confuse things, but I can at least outline my method. You start with the assumption that going into a game seven, your account must be at zero so that you can make a $30,000 bet for that game.

So then you ask yourself--for a series that is 3-2 after game 5, what do I need in my account such that a winning bet brings me to $30,000 and a losing bet brings me to $0? Then, once you have that number, you ask yourself the question about game 4, then about game 3, etc.
What happens if Philadelphia is leading 3-2 instead of Boston?

Also, couldn't you get the same result by having $15K in your account before game seven and wagering that?

[Ben E Lou]

What happens if Philadelphia is leading 3-2 instead of Boston?

Also, couldn't you get the same result by having $15K in your account before game seven and wagering that?No, because you'd then losing nothing. The puzzle stipulates winning or losing 30K.

However, this just brought me to a problem with EF's solution: what if the series goes less than 7 games?

[MizzouRah]

The odds that two people have the same birthday are: (365/365) * (1/365) :: (100% chance the first person is born on their birthday) * (odds that second person is born on that specific day)...or 0.27%

You can also arrive at that number by figuring out the probability that the two people don't share a birthday and subtracting that from 1... (365/365) * (364/365) = .9973 = 99.73% chance they don't (0.27% chance they will). Figuring out three poeple would be (365/365) * (364//365) * (363/365) = .9918 :: 1 - .9918 = 0.82% chance of two people having the same birthday out of the three.

Plugging that calculation into Excel says that you need 53 people to get a 98.12% chance that two out of the 53 share the same birthday. Multiple that by $7 in food and drink and the answer to #1 is 371.

Nice work!!

[TredWel]

I'd like to see the mathematics on the 68% chance of a 3-2 split, by the way. I'm lost on that one.
I confess that I never worked the math out myself. It's just one of those things that is pounded into your brain when you work on declarer play.

Google found me the following link about the math behind "Eight ever, Nine never" which should explain how we get 68%.

hxxp://www.durangobill.com/BrSplitHowTo.html

[cthomer5000]

No, because you'd then losing nothing. The puzzle stipulates winning or losing 30K.

However, this just brought me to a problem with EF's solution: what if the series goes less than 7 games?

I'm with SkyDog here. I'd at least like to hear some further explanation from albion before I'm willing to go with his answer.

[EagleFan]

With 28 people you get:
27/365 + 26/365 + 25/365 ..... 1/365 = 378/365 or a 103% chance

with 27 people you get 351/365 or a 96% chance.

[VPI97]

However, this just brought me to a problem with EF's solution: what if the series goes less than 7 games?
FWIW, I came up with this table...when the series record is the first value, bet the second.

(0-0) - $9375
(1-0) - $9375
(2-0) - $7500
(3-0) - $3750
(0-1) - $9375
(1-1) - $11250
(2-1) - $11250
(3-1) - $7500
(0-2) - $7500
(1-2) - $11250
(2-2) - $15000
(3-2) - $15000
(0-3) - $3750
(1-3) - $7500
(2-3) - $15000
(3-3) - $30000

No matter how the series goes, you'll end up either $30000 richer or $30000 poorer when the thing is concluded. So my answer for #2 is 9375.

[DanGarion]

We could just brute force the answer with a script that tries every page with the # of digits he gave us. I'm sure we'd finish by the end of the year.

Quick someone write a program!

[MrBigglesworth]

The odds that two people have the same birthday are: (365/365) * (1/365) :: (100% chance the first person is born on their birthday) * (odds that second person is born on that specific day)...or 0.27%

You can also arrive at that number by figuring out the probability that the two people don't share a birthday and subtracting that from 1... (365/365) * (364/365) = .9973 = 99.73% chance they don't (0.27% chance they will). Figuring out three poeple would be (365/365) * (364//365) * (363/365) = .9918 :: 1 - .9918 = 0.82% chance of two people having the same birthday out of the three.

Plugging that calculation into Excel says that you need 53 people to get a 98.12% chance that two out of the 53 share the same birthday. Multiple that by $7 in food and drink and the answer to #1 is 371.
Are you sure about this one? This is a play off of the classic birthday paradox, and plugging the numbers into this formula:
http://upload.wikimedia.org/math/0/a/c/0ac7f1cd70e2573cd65a36269c207cc8.png
I get 53.4 people, so you'd have to go with 54 people, not 53. It woudl make it $378.

[Katon]

The way I think albion did it:

If the series goes 7, then you want to be even after six so that you can bet $30,000.

This means that if the Sox get 3-2 up, you want to be at $15,000 so that you can bet $15,000 on the Sox; if you do that, then you'll either be even heading into game 7 or be $30,000 up with the series over. Conversely, if the Sox go 3-2 down then you want to be at -$15,000.

This means that if the Sox go 3-1 up you want to have $22,500, so a $7,500 bet puts you either at $30,000 or $15,000; on the other hand, if the series goes to 2-2 then you want to be at $0 so a $15,000 bet will put you at the right value. If the Phillies go 3-1 up, you want to be at -$22,500.

Then you keep tracing it back and, if you're me, wind up not with $1,875, but with $9,375.

Also, has anyone checked what happens to #1 if you use 366 days to account for leap years?

[TredWel]

I agree with $9,375 for part 2, just worked it out myself.

[Katon]

With 28 people you get:
27/365 + 26/365 + 25/365 ..... 1/365 = 378/365 or a 103% chance

with 27 people you get 351/365 or a 96% chance.

But the 28 people answer is plainly wrong - a 103% chance makes no sense in the context of the problem, and it's perfectly possible to have 28 people with different birthdays so it has to be less than 100%.

The factorial way of calculating (365/365*364/365*...) really is the proper way of figuring the odds.

[Barkeep49]

So the bridge and the poker problem seem to be set answers. I had hoped that the birthday one was as well, but that doesn't seem as true anymore, though there seem to only be 2 possiblities. That gives us 10 of the 18 numbers. Wouldn't that make brute forcing it somewhat more realistic?

[TredWel]

We're getting close

http://www.solecismic.com/371937568314517711.php

[VPI97]

Edit - Tred wins. :)


http://www.solecismic.com/371937568314517711.php



No Information about our Current Project


I've been watching the discussion, and you're close, but you're making an assumption about one of the problems that is not quite correct. Keep at it, you'll get there eventually.

[Katon]

The bridge, poker, and betting problems seem set. The birthday one's down to two cases - 365 or 366 - and I think the grades one's down to three. At this point it should be possible just to try all the possible combinations.

[Barkeep49]

Katon: Are we set on 378 and not 371 as the right answer to the birthday?

[albionmoonlight]

I'm with SkyDog here. I'd at least like to hear some further explanation from albion before I'm willing to go with his answer.

Never trust a lawyer to do math. But here's what I did, as best as I can explain it.

Going into game 7, you need 0 in your account. That means that going into game 6 (where Boston will be up 3-2)*, you need to bet a certain amount such that if your bet pays off, you win $30,000 and if your bet loses, you need 0 in your account (to set up game 7).

GOING INTO GAME 6:

So, going into game 6 (where X is what you have in your account and Y is what you bet):

X+Y = 30,000
X-Y = 0

Solving for X gives us 15,000. So, going into Game 6, we need 15,000 in our account.

GOING INTO GAME 5:

Going into Game 5, the series will either be 3-1 or 2-2.

If the series is 3-1, then

X+Y = 30,000
X-Y = 15,000

Solving for X gives us 22,500.

If the series is 2-2, then

X+Y =15,000
X-Y = -15,000

Solving for X gives us 0.

So, going into game 5, you will need 22,500 in your account if the series is at 3-1 and 0 in your account if the series is tied.

GOING INTO GAME FOUR:

Going into Game 4, the Series will either be 3-0 or 2-1.

If the series is 3-0, then

X+Y = 30,000 (you win the bet)
X-Y = 22,500 (you go to 3-1)

Solving for X gives us 26,250

If the series is 2-1, then

X+Y = 22,500 (you go to 3-1)
X-Y = 0 (you go to 2-2)

Solving for X gives us 11,250.

GOING INTO GAME THREE:

Going into game 3, the series will either be 1-1 or 2-0.

If the series is 2-0 then

X+Y = 26,250 (you go to 3-0)
X-Y = 11,250 (you go to 2-1)

Solving for X gives us 3,750

If the Series is 1-1 then

X+Y = 11,250
X-Y = -11,250

Solving for X gives us Zero

GOING INTO GAME 2:

Going into Game 2, the series will be 1-0

X+Y = 3,750
X-Y = 0

Solving for X = 1,875.

So you need to bet 1,875 in order to have 1,875 going into Game 2.

Feel free to pick apart while I try to format it a little better.


*You can just make all of the bets negative and have Boston losing. It works out the same.

[Barkeep49]

DOLA -- And it's nice to know whatever the message is that Jim does at least want us to find the answer.

[twothree]

http://www.solecismic.com/371937567314517711.php

[EagleFan]

But the 28 people answer is plainly wrong - a 103% chance makes no sense in the context of the problem, and it's perfectly possible to have 28 people with different birthdays so it has to be less than 100%.

The factorial way of calculating (365/365*364/365*...) really is the proper way of figuring the odds.


That's a Statistics 101 problem. The very first question our professor asked in the class.

[Barkeep49]

Very good twothree. About what I expected.

[Katon]

Going into game 3, the series will either be 1-1 or 2-0.

If the series is 2-0 then

X+Y = 26,250 (you go to 3-0)
X-Y = 11,250 (you go to 2-1)

Solving for X gives us 3,750


Er, no it doesn't. It gives you 18,750. Since Y>0, having X < X-Y is silly.

Apart from the calculation error, this is exactly what I did, only explained more clearly.

[cthomer5000]

http://www.solecismic.com/371937567314517711.php

He's an evil man.

[MrBigglesworth]

Katon: Are we set on 378 and not 371 as the right answer to the birthday?
I'm not sure I'm right, the formula I used was an approximation, I'm checking again...

[DanGarion]

BOOOOOOOO!

[VPI97]

http://www.solecismic.com/371937567314517711.php
http://www.fof-ihof.com/phpBB2/images/smiles/icon_party.gif

[Katon]

That's a Statistics 101 problem. The very first question our professor asked in the class.

Then how does 103% make any sense? The chance always has to be less than 100% for any number of people less than 365, because it's always possible for everyone to have different birthdays.

[Ben E Lou]

So, it was 67 and not 68 on the bridge problem.

[Barkeep49]

For those who care:

Birthday problem $371

Boston betting: $9375

Bridge 67%

Poker: 3145

Skipping grades: 17711

[albionmoonlight]

Dola: See my mistake now. I just mis-solved for X at one point. I now agree with $9375

[SirFozzie]

EEEVIL!

[MrBigglesworth]

Also, has anyone checked what happens to #1 if you use 366 days to account for leap years?
That makes is pretty complicated. You can't just make it 366, because that leap year day is 1/4 as likely to be a birthday as all the other days. It would be 1/1461 compared to 4/1461 for all the other days.

[Ben E Lou]

EEEVIL!Not really. I half-wondered if it would be house info, considering the way it was worded, but there were some nice mathematics involved there that made it worth it. Good stuff.

[Ben E Lou]

That makes is pretty complicated. You can't just make it 366, because that leap year day is 1/4 as likely to be a birthday as all the other days. It would be 1/1461 compared to 4/1461 for all the other days.More complicated than that, I think. 1900, 2100, 2200, 2300, 2500, 2600, 2700 aren't leap years, no?

[jbmagic]

Maybe all those questions are found on a web site.

Anybody try google the answers?

[MizzouRah]

I believe it's real, but since we only have one day to complete all 5 puzzles, Jim knows the probability of all 5 right answers are slim.

http://nelson.oit.unc.edu/~alanh/images/dumbass-293x184.jpg

[Solecismic]

So, it was 67 and not 68 on the bridge problem.

Yes, the question asked people to round down, and the probability is .6783. So that's why I added the other link, because 68 is essentially correct.

For those wondering about the math, because intuitively the answer is 62.5%, you can't look at the distribution of the hearts as a separate problem. That probability is affected by the other 21 cards remaining.

You have to set it up looking at all [13/26] hands. I expected this one to hang people up more than any, didn't think we'd have a real bridge player here.

[EagleFan]

Then how does 103% make any sense? The chance always has to be less than 100% for any number of people less than 365, because it's always possible for everyone to have different birthdays.


It's statistical probability. Take sets of 28 random people and bet on each one having at least one set of duplicate birthdays and you'll have a 103% return.

[Solecismic]

That makes is pretty complicated. You can't just make it 366, because that leap year day is 1/4 as likely to be a birthday as all the other days. It would be 1/1461 compared to 4/1461 for all the other days.

Yes, that is how you solve it. But it ends up not changing the final result. The difference is only about a tenth of a percentage point.

[Senator]

Look at the big brains on FOFC.

[twothree]

Part 5

If you want to receive a PhD in math, all together you're going to go through about 21 grades' worth of school material.
Well, math geeks tend be quite intelligent people overall, especially in ways that impress teachers.
Let's say we have a truly superior math geek student who can skip grades at will. But he can only skip one grade per year. Or not, if he chooses. So he can go one grade at a time and take 21 years to obtain his PhD. Or he can skip as often as possible, and need only 11 years. His choice.
Those are only two of literally thousands of ways he could choose to go. So, how many different combinations of skipping and not skipping are possible?

Another way to view this problem is...

How many distinct ways can you empty a basket of 21 apples, if you can only take one apple or two apples from the basket at a time.

[Katon]

It's statistical probability. Take sets of 28 random people and bet on each one having at least one set of duplicate birthdays and you'll have a 103% return.

But that's not the question. The question is what is the possibility that at least two of them have the same birthday. Your system gives extra weight to the cases where four or five or all 28 of them have the same birthday, because that makes sense for the problems it's designed to solve, but in the context of this question all 28 of them sharing a birthday is exactly the same as just two sharing a birthday.

[Young Drachma]

Wow, this is crazy.

[MIJB#19]

Wow, this is crazy.
Yes, indeed. One crazy signature you've got there. :D


Nice to see the progress of the project, Jim. :)

[gstelmack]

Still some fun puzzles. And Jim did hint that he wanted to announce this way. But even if it is all a big joke, what's the harm in letting Jim have his fun and us try to solve some puzzles?

Knowing what the final answer was and what we got out of it, I still stand by this. 'twas fun Jim.

[Franklinnoble]

http://www.solecismic.com/371937567314517711.php

Yeah, I'm sure of it now. You're going straight to hell for that.


:D

[M GO BLUE!!!]

The answer will be a redirect to tubgirl.
This brings to mind a question that is very perplexing: Would it be possible for Jim to be put in the box?

[Flasch186]

if you look closely at the dirt you can make out a nascar oval...hmmmmm

[EagleFan]

Hmmm "The house that Jim built" .... "The house that Ruth built" It's obvious, he's doing a baseball sim.

;)

[Noop]

Look at the big brains on FOFC.

Plup Fiction.

[Ragone]

So I'm Confused... Jim is making a Virtual reality version of the sims?

[Hurst2112]

Glad I missed this thread in its popularity.

http://www.bewarethecheese.com/nerds.jpg

[Anthony]

i rest assured knowing that i'm not nerdy enough to have even attempted to answer those questions. too much work for a saturday.

[Solecismic]

i rest assured knowing that i'm not nerdy enough to have even attempted to answer those questions. too much work for a saturday.

The rest of us rest assured as well. We'd be disappointed if you weren't too cool to hang with us.

Now, where are Lindsay Weir and the rest of the Mathletes? We have a competition coming up and need to study differential equations.

[sabotai]

Now, where are Lindsay Weir and the rest of the Mathletes? We have a competition coming up and need to study differential equations.

I heard she went to a Grateful Dead concert.

[Solecismic]

I heard she went to a Grateful Dead concert.

She had better not be hanging out with that stoner who thinks he going to make a living as a drummer. He's only going to break her heart once he meets that girl from band camp.

[Anthony]

"Mathletes".

HAHAHAHAHAHA. llol (literally laughed out loud). sounds like a mega-nerd, or some type of bionic geek.

[kcchief19]

She had better not be hanging out with that stoner who thinks he going to make a living as a drummer. He's only going to break her heart once he meets that girl from band camp.
She's my Lady L.

[M GO BLUE!!!]

http://www.solecismic.com/essentials.php

[JonInMiddleGA]

Somebody had waaaaaaaaaay too much time on their hands ;)

[TazFTW]

"I am loving your Post Game Pleasure soap. All my lady friends can get enough of me after the game. It is truly a party in a package," - Michael V., Atlanta.

Greatness.

[Swaggs]

Haha!

Morning Turf
Every football player gets a little misty-eyed on that first crisp fall morning when there's just enough chill in the air to excite the senses. It's time for football. We've formulated a special soap that captures that feeling and leaves you energized, excited and ready for a game.

Morning Turf has a 5% Goat's Milk base added to the glycerine, which provides that extra Vitamin D protection for your skin. We've added Cypress to enhance concentration, and Mandarine Orange for a burst of energy. This is truly a soap worthy of game day.

"I can't do without my Morning Turf. Nothing I've ever tried, has ever made me feel more prepared for football. And I need a little something extra smooth, because my supplements make my ----- shrink and acne form on the inside of my teeth," - Shawne M., San Diego.

[Young Drachma]

Niice.

[Ben E Lou]

2004-2007 merged into one thread. Still looking for earlier ones...

[JeeberD]

"Tobacco Withdrawal Paste"

Gold! :D

[Ksyrup]

LOL! Good job Jim.

These are also the pefect gift for the FOF player who enjoys the affinity/conflicts astronomy feature.

[M GO BLUE!!!]

2004-2007 merged into one thread. Still looking for earlier ones...

I wondered what happened to my post! :D

[Izulde]

:D Hilarious!

I literally LOLed.

[gottimd]

Give me one of everything, I don't care about the price. Should I email them or post my credit card # and expiration date here with the security code?
http://tim.cexx.org/images/jimmycards.png

[Ben E Lou]

http://www.solecismic.com/placement.php

[Ben E Lou]

Heh. Easily my favorite part:



1-5 points: A thank-you card signed by our cat and the unofficial Solecismic Software mascot: Penelope
6-10 points: An autographed 8-by-10 glossy of Gridiron Greta
11-25 points: A copy of The College Years, our college football simulation
26-50 points: A copy of Front Office Football 2007, our professional football simulation
51-100 points: Your name on a new piece of white lego for our home (we need to build a kitchen)
101-500 points: Free copies of all our football games for as long as we're in business
501 or more points: A copy of our new card game: Up and Down the River

[Mota]

I'd totally be in for that picture of Gridiron Greta. Booking my appointment for the tattoo right now, and then I'm going upstairs to make Solecismic #1 with my wife.

[Raiders Army]

Who's the ugly woman with the Solecismic logo on her forehead?

[Raiders Army]

dola, is the mailing list new? That's the first I've seen it.

[Young Drachma]

Love it. Excellent stuff.

[Ben E Lou]

dola, is the mailing list new? That's the first I've seen it.
That has been there for a few weeks.

[gottimd]

and then I'm going upstairs to make Solecismic #1 with my wife.

This is just a pure set up for the inevitable..."Wait in line" joke.

[Raiders Army]

That has been there for a few weeks.

Ah. Glad to know it's not part of the April Fool's Day happenings.

[Passacaglia]

I went straight to the first post, and read something about Front Office Football 2006. I was like, "FOF 2006? Wtf is he talking about?" until I realized this was a bump. You fooled me again, Gindin!

[Ben E Lou]

nuthin?

[Logan]

Probably busy with TCY2.

[Chubby]

Probably busy with TCY2.

/rimshot

[Ben E Lou]

<blockquote class="twitter-tweet" lang="en"><p>Brand new feature for <a href="https://twitter.com/hashtag/FrontOfficeFootball?src=hash">#FrontOfficeFootball</a>... Ctrl-Alt-8 (numpad). Enjoy...</p>&mdash; Solecismic Software (@Solecismic) <a href="https://twitter.com/Solecismic/status/583167325777616896">April 1, 2015</a></blockquote>
<script async src="//platform.twitter.com/widgets.js" charset="utf-8"></script>

[chinaski]

<blockquote class="twitter-tweet" lang="en"><p>Brand new feature for <a href="https://twitter.com/hashtag/FrontOfficeFootball?src=hash">#FrontOfficeFootball</a>... Ctrl-Alt-8 (numpad). Enjoy...</p>&mdash; Solecismic Software (@Solecismic) <a href="https://twitter.com/Solecismic/status/583167325777616896">April 1, 2015</a></blockquote>
<script async src="//platform.twitter.com/widgets.js" charset="utf-8"></script>

woohoo!

[korme]

Can someone spoil this, please?

[cartman]

Can someone spoil this, please?

You got it.

Can someone spoil this, please?

[Barkeep49]

You got it.



Well played.

[Pyser]

but then he really released 7.1c on april fools day? confusing

[MizzouRah]

<iframe title="Embedded Tweet" style="border-width: 1px; border-style: solid; border-color: rgb(238, 238, 238) rgb(221, 221, 221) rgb(187, 187, 187); -moz-border-top-colors: none; -moz-border-right-colors: none; -moz-border-bottom-colors: none; -moz-border-left-colors: none; border-image: none; max-width: 99%; min-width: 220px; margin: 10px 0px; padding: 0px; display: block; position: static; visibility: visible; border-radius: 5px; box-shadow: 0px 1px 3px rgba(0, 0, 0, 0.15); width: 500px;" allowfullscreen="" class="twitter-tweet twitter-tweet-rendered" allowtransparency="true" scrolling="no" id="twitter-widget-0" height="188" frameborder="0"></iframe>
<script async="" src="//platform.twitter.com/widgets.js" charset="utf-8"></script>

Thumbs up!

<iframe style="display: none;" allowtransparency="true" scrolling="no" id="rufous-sandbox" frameborder="0"></iframe>

[Tasan]

Oh man that was funny. Really added a feature everyone had been begging for.