a problem about probability

Discussion in 'Trading' started by 0008, Dec 30, 2002.

  1. I think that the reason that some people find this confusing is because they think of the chance of a coin being flipped H 9 times in a row and then multiply it by .5 to get the answer for ten times in a row. Don't forget though that once you state that the coin has already come up H 9 times in a row then the chance of that happening is 1. so all the math you need is (1) (.5) = .5
     
    #31     Dec 30, 2002
  2. Young,

    That is an elegant way of illustrating the answer.
     
    #32     Dec 30, 2002
  3. It's been explained to death but here's another angle:

    For the 10 heads, you have to predict 10 events in a row, exactly right.

    For the one head, you only have to predict one event.

    The further out in the future you go, the harder it gets.

    Thus the greater odds.

    Once the 9 events have happened, they don't count anymore.

    Try to predict the next 10 ticks in the ES. :p
     
    #33     Dec 30, 2002
  4. Statistician joke :)

    "I read that there is about one chance in one million that someone will board an airplane carrying a bomb, and I started carrying a bomb with me on every flight I take. The way I figure it, the odds against two people having a bomb on the same plane are 1 in a trillion."

    Q1: If I toss a coin. What it the chance that I get 10 consecutive heads (or tail)? Is it (1/2)^10=0.00097?

    A1: True

    Q2: Many people think, if you see 9 consecutive heads, than the chance seeing next one is head is extremely small.

    A2: False


    Why ? Q1 is DIFFERENT from Q2 if you don't ask the same question you won't get the same answer :)

    If A and B are independants, then P(B) = P(B/A) this latter is the conditional probability of B knowing A, if B is independant from A, its probability and conditional probability are the same.

    and then P(A and B) = P(A) * P(B)


    Q1 is related to P(A)*P(B)*P(C) ....

    Q2 is related to P(B) = P(B/A)




     
    #34     Dec 31, 2002
  5. Ever heard of double headed coins guys ?

    Jack
     
    #35     Dec 31, 2002
  6. Some coins have a double headed eagle on one side and a human head on the other. I guess you could call them triple headed coins.
     
    #36     Dec 31, 2002
  7. rs7

    rs7

    Indeed the probabilities are 50/50 for the tenth flip. But the "appearance that tails would almost be a guarantee"?

    This has been discussed before here in the thread "Superstition, Luck and Voodo" at great length. (Too great).

    I said then, and still maintain, that I for one, would bet on heads again. Why? Illogical but convincing belief in "runs". When you are hot, you're hot.

    Streaks end, but with something like this, there is no rational basis. It will always be 50/50 on each event (flip of the coin). But experience has made me believe in runs. I have been in poker games where a player gets so hot that I just know I need to fold a hand (against the hot player) that I would normally play. Just happens. No logic to it. Better to wait 'til his run runs itself out.

    I think most serious gamblers would concur. This does not make me (or them) right. It just is the way we perceive the situation. When a streak is happening, it seems wisest to go with it, not against it. How many times have we all heard the expression "The Trend Is Your Friend".

    Peace,
    :)rs7
     
    #37     Dec 31, 2002
  8. *****
    Good one AAAinthebeltway, on the uptrend. I like the law of averages [or odds] when you get in a downtrend below 50 or 50.5 period moving average plus discretion.:cool: Helps if one gets 5 or 50 plus things in your favor also.
    ==============================
    ''The trend is your friend'' RS7 quote
    Ps got a .50 caliber Thompson Center from Mr. Meyers around Christmas.[thunderhawk]
     
    #38     Dec 31, 2002
  9. kztd

    kztd Guest

    If I toss a coin. What it the chance that I get 10 consecutive heads (or tail)? Is it (1/2)^10=0.00097?

    -> Yes

    Many people think, if you see 9 consecutive heads, than the chance seeing next one is head is extremely small. And you should bet tail. But some people said, the chance is still 1/2 because they are independent.

    -> The chances of seeing 9 consecutive heads is (1/2)^9, so chance of seeing an additional head is (1/2)^10 / (1/2)^9 or, 1/2.

    Those who disagree (parlais vous francais?) are the same ones that make big $ at roulette.


    :eek:
     
    #39     Mar 21, 2003
  10. Nicely said.
     
    #40     Mar 21, 2003