1. SubscriberProper Knob
    Cornovii
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    13 Oct '10 12:32
    It's been a long time since i did my probability module in college.

    If i have two independent events occuring that both have a probability of 1/10. The probability of both of them happenning is 1/100?

    Is that right?

    And if i have three independent events occuring, again each event at a probability of 1/10. The probability would be 1/1000?

    Have i remembered correctly?
  2. Joined
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    13 Oct '10 12:58
    That's right. As long as they're independent.
  3. weedhopper
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    13 Oct '10 23:06
    So, that means my chances of winning the "Pick 3"Lottery is 1 in 1000? {choose 3 numbers from 0-9 and match them all}
  4. Standard memberrandolph
    the walrus
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    14 Oct '10 02:37
    Originally posted by PinkFloyd
    So, that means my chances of winning the "Pick 3"Lottery is 1 in 1000? {choose 3 numbers from 0-9 and match them all}
    Yes.
  5. Melbourne, Australia
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    14 Oct '10 02:47
    Originally posted by PinkFloyd
    So, that means my chances of winning the "Pick 3"Lottery is 1 in 1000? {choose 3 numbers from 0-9 and match them all}
    Not sure on that one. Have you taken into account that after the first ball is picked there are less balls etc. Unless the lottery replaces the ball taken. (Haven't played this sort of lottery.) Don't ask me to calculate it though.
  6. Subscribercoquette
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    14 Oct '10 05:19
    Originally posted by mtthw
    That's right. As long as they're independent.
    Probably.
  7. Germany
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    14 Oct '10 07:17
    Originally posted by Taoman
    Not sure on that one. Have you taken into account that after the first ball is picked there are less balls etc. Unless the lottery replaces the ball taken. (Haven't played this sort of lottery.) Don't ask me to calculate it though.
    If the numbers can only be picked once (though it does not appear to be the case here), it's still quite easy to calculate; the chance would be 1/(10*9*8).
  8. SubscriberProper Knob
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    14 Oct '10 12:44
    Originally posted by mtthw
    That's right. As long as they're independent.
    So if i have 200 independent events each with a probability of 1/100million, that would be 100million x 10`200? (that should be 10 to the power of 200)
  9. Joined
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    14 Oct '10 13:19
    Originally posted by KazetNagorra
    If the numbers can only be picked once (though it does not appear to be the case here), it's still quite easy to calculate; the chance would be 1/(10*9*8).
    If the order mattered. If it didn't, multiply by 6 (the number of permutations of three different numbers).
  10. Joined
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    14 Oct '10 13:211 edit
    1/(100 million)^200...or 1/10^1600

    Also know as "damn near to impossible".
  11. Cape Town
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    14 Oct '10 16:06
    Originally posted by mtthw
    1/(100 million)^200...or 1/10^1600

    Also know as "damn near to impossible".
    Yet we shouldn't misinterpret that claim. We almost certainly do have over 200 lotery winners of loteries with worse than 1 in 100 million odds.
    What is 'damn near impossible' is for the same person to win all those lotteries (and hasn't happened).
  12. Joined
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    14 Oct '10 16:521 edit
    Originally posted by twhitehead
    Yet we shouldn't misinterpret that claim. We almost certainly do have over 200 lotery winners of loteries with worse than 1 in 100 million odds.
    What is 'damn near impossible' is for the same person to win all those lotteries (and hasn't happened).
    Or, equivalently, to predict in advance who the 200 winners will be.

    It's true, though, you've got to keep an eye on your sample space. Massively unlikely events happen all the time. As in the Richard Feynman quote in a lecture: "I was walking to class today and the funniest thing happened: I saw a car with the license plate ‘ARW 357’. Can you imagine? Of all the possible license plates, what are the chances of seeing that one?"
  13. Standard memberadam warlock
    Baby Gauss
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    14 Oct '10 18:13
    Originally posted by mtthw
    As in the Richard Feynman quote in a lecture: "I was walking to class today and the funniest thing happened: I saw a car with the license plate ‘ARW 357’. Can you imagine? Of all the possible license plates, what are the chances of seeing that one?"
    The point of that quote is that it doesn't make sense to calculate the probability of an event after it already happened.
    He says so himself when retelling that story.
  14. weedhopper
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    15 Oct '10 04:06
    The Black Swan.
  15. Joined
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    15 Oct '10 10:40
    Originally posted by adam warlock
    The point of that quote is that it doesn't make sense to calculate the probability of an event after it already happened.
    He says so himself when retelling that story.
    I know.

    The other part of the point is that you have to consider probabilities in relation to the sample space your interested in. Because if there are millions of possible things that can happen then million-to-one incidents are not unlikely.

    There's a very good essay on this by Ian Stewart and Jack Cohen. Unfortunately I can't find a reference to it. But I remember one part where one of them had a bet that "a coincidence will occur" while walking through an airport in Sweden. They then bumped into someone they knew and hadn't see for ages.
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