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Dec-10-09
 | | Domdaniel: <long strange trip> deadheads _ they are all deadheads_ the grateful dead song <truckin> from which this line comes is officially recognized as a <national treasure> in the usa _ contact your congresspersons and demand the same status for chessgames_ |
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Dec-10-09
 | | PinnedPiece: "Got my chips cashed in...."
Congratulations, guys! |
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| Dec-10-09 | | Jim Bartle: set up like a bowling pin |
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Dec-10-09
 | | keypusher: Happy birthday and many more! |
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| Dec-10-09 | | SPINK: zum geburstag viel glueck! |
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| Dec-10-09 | | WhiteRook48: happy birthday chessgames!! |
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Dec-10-09
 | | Once: Happy birthday CG! By a strange coincidence, you share your birthday with my son John, who is nine today. |
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Dec-10-09
| | Domdaniel: If you get 23 people together, the probability of a shared birthday is over 50%. Still, happy birthday, John Once. And CG too. |
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Dec-10-09
 | | jessicafischerqueen: Happy website birthday and thanks for this website, which is my favorite website on all of the internets. Thanks <D. Freeman> |
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Dec-10-09
 | | whiteshark: <Domdaniel: If you get 23 people together, the probability of a shared birthday is over 50%> You shurely can explain this enigma in layman's terms, right? :D |
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Dec-10-09
 | | WannaBe: <whiteshark> No fear, my friend, I am here to help! http://en.wikipedia.org/wiki/Birthd... |
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| Dec-10-09 | | Octal: What a long strange trip it's been!
I am probably one of the only chess players who have ever seen the Dead play this song live. And if not the only one, probably one of the only ones to see it while high. |
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| Dec-10-09 | | Jim Bartle: If I may one-up for a moment, I heard them play the song probably half a dozen times, all before it ever came out on record. Wasn't high, though. I don't remember the 23 people probability, but I do remember my fifth grade teacher saying there was low probablity that there would be no repeat birthdays in a class of 30. |
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Dec-10-09
 | | hedgeh0g: Happy birthday, Chessgames! |
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Dec-10-09
| | Domdaniel: <WannaBe> Thanks. Couldn't have explained the birthday paradox any better myself ... and I *play* the Birthday Attack. Wiki's approximation for the probability is 50.729%
If anyone wants more accuracy, the next two digits in this decimal are ... 23. Spooky, huh? |
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Dec-10-09
| | Domdaniel: If you get 23 chessplayers, what's the probability that two have been out of their skulls watching the Grateful Dead? High. Very high. |
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Dec-10-09
 | | Chessmensch: I know it's late in the day and few will see this, but here's the story on the birthday problem. http://en.wikipedia.org/wiki/Birthd... |
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Dec-10-09
 | | Jimfromprovidence: Saw the Dead many times. One memorial time was at Watkins Glen, NY in 1973, along with the Band, the Allman Brothers and 600,000 of my closest friends. Nobody was playing chess.
Happy Birthday, CG
"Truckin' got my chips cashed in. Keep truckin', like the do-dah man
Together, more or less in line, just keep truckin' on". |
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Dec-11-09
| | Once: <Domdaniel> Once John was grateful for your birthday wiahes, and has asked me to say "thanks" on his behalf. The birthday paradox has always been a favourite of mine. Most people get confused between the odds of everyone having the same birthday as me, with the odds of at least two people in the group having the same birthday as each other. This is how I explain it (asuming there are 23 people in the room): The odds of someone having the same birthday as person #1 are 22/365 - in other words, each person in the room has a 1 in 365 chance of matching person 1's birthday, multiplied by the 22 people in the room (other than person 1, natch). The odds of someone having the same birthday as person #2 are 21/365. The same odds as before, minus the chance of matching person #1, cos we've already tested that. Add those odds to the chance of matching person 1. So it's not too hard to see that the odds of finding any matches are 22+21+20+19 ..., all divided by 365. And all that lot looks like roughly half to me. Just don't get started me on the Monty Hall problem... |
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Dec-11-09
 | | SwitchingQuylthulg: <Once: So it's not too hard to see that the odds of finding any matches are 22+21+20+19 ..., all divided by 365. And all that lot looks like roughly half to me.> And the probability of finding matching birthdays in a group of 27 people is (27+26+25+24...+2+1)/365 ~ 1.036, or roughly 104%. |
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Dec-12-09
 | | kevin86: White's king rolls out of the pocket and is able to proect himself-black's is under the gun,with nowhere to go. |
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Mar-29-10
 | | Nightsurfer: Hola Daniel, FELICITATIONES belatedly! A great King's Walk, that King's Walk of yours - and, with colours reversed, a stunning replay of the rather cocky foray by White General deep into the Black camp 91 years ago and way across the Atlantic Ocean over there in rainy Scotland, back then a courageous attack that had been executed by Richard Teichmann versus Allies at Glasgow 1902, please check out the match here at Chessgames.com by surfing to the link www dot chessgames com slash perl slash chessgame ? gid=1250612 , an amazing case of DEJA-VU. My compliments, Daniel! |
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Apr-11-10
| | Nightsurfer: There is the missing direct link to >>R Teichmann vs Allies<<, check out and enjoy: Teichmann vs Allies, 1902 ! |
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Mar-18-13
 | | FSR: <Once: ... This is how I explain it (asuming there are 23 people in the room): The odds of someone having the same birthday as person #1 are 22/365 - in other words, each person in the room has a 1 in 365 chance of matching person 1's birthday, multiplied by the 22 people in the room (other than person 1, natch). The odds of someone having the same birthday as person #2 are 21/365. The same odds as before, minus the chance of matching person #1, cos we've already tested that. Add those odds to the chance of matching person 1. So it's not too hard to see that the odds of finding any matches are 22+21+20+19 ..., all divided by 365. And all that lot looks like roughly half to me.> As <SwitchingQuylthulg> so drolly indicated, your math doesn't work. The triangular number that is the sum of the integers 1 through 22 is (1+22)/2 x 22, i.e. 253. http://wiki.answers.com/Q/A_list_of... 253/365 is about .693, decidedly more than half. The correct way to figure it out (leaving aside the matter of Leap Day, and the fact that birthdays are not quite evenly distributed) is: when you have two random people in a room, the odds of them <not> having the same birthday are 364/365. If they don't, and a third person comes in, the odds of him not having the same birthday as either of the other two are 363/365. The odds of there not being a single matching birthday, with 23 people in the room, are the product of the first 22 such numbers, i.e. <364/365 x 363/365 x 362/365 x 361/365 x 360/365 x 359/365 x 358/365 x 357/365 x 356/365 x 355/365 x 354/365 x 353/365 x 352/365 x 351/365 x 350/365 x 349/365 x 348/365 x 347/365 x 346/365 x 345/365 x 344/365 x 343/365>. Paste that into Google, without the angle brackets (yes, Google also works as a calculator) and you'll find that it equals .49270276567. That is the odds of there <not> being a single match. The odds of there being at least one match are one minus .49270276567, i.e. 0.507297234. Leaving it at six decimal places, that's 50.7297%, as Wikipedia says. http://en.wikipedia.org/wiki/Birthd... |
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Mar-18-13
| | FSR: Cool finish. But I'm surprised that Laufer of all people would be so eager to surrender his dark-squared bishop (laufer in German) just to win a lousy pawn. |
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