Birthdays on mafiascum

[Dragon Phoenix]

April 8, 1957.

[Quailman]

Happy Birthday, Kerplunk!

[Cadmium]

Ooh, another birthday boy .

Van harte gefeliciteerd met je verjaardag, Kerplunk!!

[Dragon Phoenix]

Lang zal hij leven, lang zal hij leven....

[Kerplunk]

Lang zal hij leven, lang zal hij leven....

Dank iedereen!

[translate]Thanks everyone![/translate]

Got 6nimmt!/Take6!/Take5! on my birthday. Great fun.

[MeMe]

oooh! Love that game! And ignore anyone who says it's all luck...they only say that because they're

losers

.

[Untrod Tripod]

Same as mathcam, September 29th

[Someone]

November 15, monday


Is there any reason that the names aren't in any particular order?

Just because I'm bored: the birthdays in chronological order are

Wacky:Janruary 30th
MeMe:March 13th
Dourgrim:March 28th
Dragon Pheonix: April 8th
shelper:April 13th
JereIC:April 14th
Cuban Smoker: April 30th
PolarBoy:May 11th
Stewie:June 14th
d8p:June 23rd
Corsato:June 23rd
PBuG:July 13th
Sugar:July 19th
rite:July 27th
gslamm:August 4th
Macros:August 8th
bloojay:August 15th
Fletcher:August 23rd
mlaker:August 30th
Quagmire:September 8th
Talitha:September 11th
mathcam:September 29th
untrod Tripod:September 29th
Thoth:October 10th
Kerplunk: November 2nd
Quailman:November 7th
Someone:November 15th
vIQles: 5th December
God: December 8th
Cadmium:December 16th
shadyforce:December 22nd
genku:December 23rd

[MeMe]

Excellent - now if we could just get mlaker to cut and paste that into the first post...but where's Kerplunk (Nov. 2)?

[edit]Oh - I see, you just took the current front post without adding in the extra submissions since the last update...I see now some others are missing too[/edit]

[Dourgrim]

All I gotta say is: us March people rock.

[Cadmium]

::wonders why noone's parents here felt horny in May::

[Someone]

Updated

[Kerplunk]

Maybe it's an idea to put the one which birthday is next on top and then order chronologically the rest under it. When that day is passed, say "Happy Birthday <...>!" and put him/her down on the list and the next birthday-boy/girl is on top of the list.
Or would this be too much trouble? I could do it if mlaker or Someone doesn't want to.

[Someone]

Well I can't edit other people's posts...

[Kerplunk]

Well I can't edit other people's posts...

True. We/I could make a new topic if mlaker doesn't show up. Where is mlaker anyway?

And Happy Birthday to Quailman!

[Quailman]

Thanks! It's great to be alive!

[Dourgrim]

Thanks! It's great to be alive!

Well I would certainly

hope

so!

Anyway, happy birthday! :confetti and so forth:

[mathcam]

There's a very cool question in probability theory that asks "How many people do you have to get together before you'd expect to have two people with the same birthday?" You might naivley guess the answer to be 365 (one for each day of the year), but it's surprisingly smaller than that. If there's only 23 people getting together, therel's already a 50 percent chance of two or more people sharing a birthday. Neat, huh? And you only need like 41 people to have a 90 percent chance.

We have a mere 32 people and have two repeated birthdays. (Me/Untrod and d8P/Corsato). Just a neat little tidbit.

Cam

[Quailman]

I saw that demonstrated in a class once. The teacher selected a section of 25 students and instructed them to say their birthdays, one at a time, and if any of the remaining students in that group shared that birthday, they should speak up at once. He only got to one student, and someone near the back of the group said "stop - that's my birthday, too". He continued anyway, and there was one more match.

[Dragon Phoenix]

We had twins in the class..

And once again, happy birthday Senior.

[Thoth]

There are lots of interesting probability examples like that. I always like the one about a game show: There are 3 doors, behind one is a car. You have to pick one. After you picked one tha game show host opens one of the other 2 that is empty and then asks you whether you want to switch your choice. Switching gives you double the odds to get the car, but in practice only about 20% of the guests went with the odds. The rest stuck with their original choice.

[gslamm]

In the Monty Hall problem switching increases your chances from 1 in 3 to 1 in 2. This seems counter intuitive. Keep in mind Monty knows where the uber-prize is and will not reveal it.

*Start with 100 doors
*pick one
*Monty reveal one of the doors that does not contain the uber-prize
*switch or stay? SWITCH!
*Let Monty reveal another door
*Repeat till you switch to the last door.

Clearly you've increased your odds from 1 in 100 to 1 in 2

[mathcam]

Monty Hall is definitely another cool problem. Here's a simple also very cool related one.

A guy comes up to you with two envelopes with money, one has, say, X dollars, and the other has 2X dollars. He gives you one of the envelopes, you open it, count the money, and then he offers you a chance to switch envelopes if you want. Do you take it? (Say, for example, the envelope you opened had 50 bucks in it).

Cam

[PolarBoy]

Well there are basically two possibilities. Either you lose $25 or gain $50(the other envelope is either going to have $25 or $100 in it). I suppose the best guess would be to switch, as you can't lose everything, but you stand to gain 100% if you're lucky.

[Norinel]

In the Monty Hall problem switching increases your chances from 1 in 3 to 1 in 2.

I thought it went 1/3 to 2/3. A better 100-door analogy is that you pick a door, Monty opens (And will always open) 98 of the other doors that don't contain prizes, and you choose then to switch or not.