1. Joined
    02 Apr '07
    Moves
    2911
    20 Jun '07 02:01
    Originally posted by FabianFnas
    Actually you did, but you didn't know how or when.
    When using the characters ; and ) together, the editor change it into 😉 automatically.
    Try it youself and you'll see.

    You can easily get rid of the face by finding the ; and ) and delete the two characters, or putting an extra space in between.
    Originally posted by FabianFnas
    Actually you did, but you didn't know how or when.
    When using the characters ; and ) together, the editor change it into 😉 automatically.
    Try it youself and you'll see.
    You can easily get rid of the face by finding the ; and ) and delete the two characters, or putting an extra space in between.


    But if the editor did it automatically despite my ignorance of the existence of the automatic process, and did so contrary to my intent, I think it can be argued that, in a case where the posted version of my text is supposed to express my thoughts on the matter (not all of them, but those which I saw fit to incorporate into that post at that time), *I* did not put the smiley face in. What I put in, with a reasonable expectation that it would not be altered, was a pair of symbols whose use by me was syntactically correct and meaningful (as they appeared in the composition process), but which was changed *by the editor* during the conversion of my composed text into posted text, into something cryptic and ultimately meaningless.

    However, I do thank you for solving the mystery!
  2. Joined
    15 Feb '07
    Moves
    667
    20 Jun '07 13:312 edits
    Here's one I recall which is similar in vein to the Monty Hall problem.

    I run a little betting game invloving 3 simple identically shaped cards and a hat. One of the cards is all white, one is all black, and one has white on one side and black on the other.

    I place all 3 in the hat and mix them up. I will then draw a random card and place it so that both of us only see the top side, setting aside the hat without looking.

    Let us suppose then that you see a white face. I then say to you, "This card is one of two cards, White/White or White/Black. I will bet you even money that the face on the opposite side of this card is white."

    Is this a good bet?

    (Just as a hint, if it came up black instead, I would swap black for white in what I tell you. I also do not know what card it is, nor do I know what's in the hat.)
  3. Joined
    31 May '07
    Moves
    696
    20 Jun '07 14:58
    No, the side has 66% probability of being on a white/white card as there are three possibilities:
    side 1 on white/white
    side 2 on white/white
    side 1 on black/white
  4. Joined
    07 Sep '05
    Moves
    35068
    22 Jun '07 15:00
    I'm not sure this really qualifies for the thread, but...

    How many (randomly chosen) people do you need before you have at least a 50% chance that two of them have the same birthday?

    Try guessing before working it out, see how close you get. If you don't know how to work it out, try guessing anyway.

    [Assume birthdays are evenly distributed throughout the year for simplicity, and ignore leap years]
  5. Standard memberagryson
    AGW Hitman
    http://xkcd.com/386/
    Joined
    23 Feb '07
    Moves
    7113
    22 Jun '07 21:55
    oh, this one turns out very small, like 20 people or something doesn't it?
    Anything qualifies for the thread, it's hard to define what I'm looking for, but a good example is the aeroplane taking off a few posts back. A real world example that doesn't require too much math that you can answer off the top of your head, but will probably be wrong. The birthday question fits that perfectly as long as you say to just give a ball park figure (which you did) I don't like maths though... urgh.
  6. Joined
    31 May '07
    Moves
    696
    24 Jun '07 14:00
    The birthday question is a good one.
    If you ask 20 people, the chances of you having the same birthday as them is 5.4%
    If you want a 50% chance of having the same birthday as someone, you need to ask 183 people.
    However, if everyone is asking everyone else we have the use of triangle numbers, and the probability rockets up.
    With 2 people, you have the probability of 0.27%
    With 5 people, you have the probability of 2.7%
    With 23 people, you get to 50%
    Really quite interesting.
  7. Standard memberagryson
    AGW Hitman
    http://xkcd.com/386/
    Joined
    23 Feb '07
    Moves
    7113
    24 Jun '07 14:12
    Yeah, my teacher in school once bet €5 that two of us would share a birthday in the class, there were about 50 in the class and his money was safe. He was gentlemanly enough to decline taking the €5 off each of us!
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