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Black, seito, black.

Welcome to EliteTrader, tradeingman. The following post may be of interest: http://www.elitetrader.com/vb/showthread.php?s=&postid=113954#post113954

Harry123/jay123, you should have consulted with me prior to starting this thread. You committed the classic "forgot to log out" gaffe, described in my following post under Alias Management II. Apart from the identical age of 33 already mentioned, the forensic giveaway is the frequent usage of...

This approaches 1/e (0.367879...) as the number of letters gets larger. 1/e = 1/2! - 1/3! + 1/4! - 1/5! + ................................... to infinity Since the letters are placed in the mailboxes at random, the probability that any single one is correctly matched is 1/n, where n is the...

My wife dresses to kill. She cooks the same way. -- Henny Youngman My wife and I were happy for twenty years. Then we met. -- Rodney Dangerfield A good wife always forgives her husband when she’s wrong. I was married by a judge. I should have asked for a jury. -- George Burns...

Worriers have ulcers.

This is the easy part - the answer to (c) is simply 1 minus the answer to (b). In fact, as the number of letters and mailboxes increases, the answer to (c) approaches 1/e (0.367879...). Another interesting fact - the average number of correct matches is 1. In other words, if the mailman...

In that case, we are finally in agreement :)

OK, you draw your card randomly. (1) Then what do you do? (2) How do you formulate (1) into a probability question?

I don't see the difference between the above and the original problem. Unless "one side of that card is red" means only one side is red, in which case the probability of it being the card with two red sides is zero.

Two elderly women were eating at a restaurant one morning. Ethel noticed something funny about Marge's ear and she said, "Marge, did you know you've got a suppository in your left ear?" Marge answered, "I have? A suppository?" She pulled it out and stared at it. Then she said, "Ethel, I'm glad...

So, as a trader, what do you do?

igsi, you dismissed my "intuitive" explanation without saying why, yet I distinctly peppered the post with logical exits whereby you could express disagreement. I therefore return your courtesy and will not "reply with comments" to your previous post. Rather, I append a numbered proof showing...

I'll try making it a little more intuitive for Bob, rs7 and igsi. An equivalent problem would be the following : 3 jars are placed into a hat. One jar contains 2 white balls, a second jar contains 1 white ball and 1 red ball, and the third jar contains 2 red balls. A ball is drawn out of one...

This is where I would resort to Bayes' Theorem : P(A|B) = (P(B|A)*P(A))/P(B) Let A = the event of drawing the card with 2 red sides Let B = the event that one side is red P(A|B) = (1*(1/3))/(1/2) = 2/3 or 66.67%

75% is a little high - one can make a nice living consistently hitting 60% as well (with a little size of course). This is assuming a fixed 1 point target, 1 point stop loss - DblArrow has a discretionary target with a 1 point maximum and on the slower pre-holiday days was using a 0.25 point...

Not even close - you're currently 8th. Good luck to you. http://www.elitetrader.com/vb/showthread.php?s=&threadid=11701 http://www.elitetrader.com/vb/showthread.php?s=&threadid=10903 http://www.elitetrader.com/vb/showthread.php?s=&threadid=10257...

"Spelling Bee Preparation" by Brandonf "The Art of Diplomacy" by daniel_m with a foreword by Cesko "Cryptography, A Compendium" by murray t turtle "Gone With the Wind" by 40yotrader

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More food for thought: In a bid to discover the trend, I drew seven lines end to end, From the resulting rectangle, I determined the angle, And confirmed the whole thing with a friend.