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  1. #1
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    Anybody want a game of darts?

    Below is a pictureof a standard dartboard. You throw six darts at the board, completely at random, except that you are accurate enough to get them all actually on the board. You do not score any bullseyes, and doubles and trebles do not count ( so that '7','double 7' and 'treble 7' all count the same) What is the chance that two of them will stick in the same segment?
    Jerry

  2. #2
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    Re: Anybody want a game of darts?

    It has been many, many years since I studied probability, but in my current alcohol enhanced state I will step up to the oche and say <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">56.395%</span hide>

  3. #3
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    Re: Anybody want a game of darts?

    Tony

    It appears you maybe going in the same direction as me <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16>

    Ok

    I have done this slightly differently.

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">The position of the first dart can be in any of 20 sectors. The probability of the second dart will not strike the same sector is 19/20. The probability of the third dart will not strike either of the first two is 18/20. For the fourth dart, the probability is 17/20 and so on. The chance of that some pair of six darts will stickin the same sector is 1 minus the probability that none of them do so. So it is:

    1 - 19 x 18 x 17 x 16 x 15/20^5= 1 - 0.436 = 0.564 =56.4%</span hide>
    Jerry

  4. #4
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    Re: Anybody want a game of darts?

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">The probability that all three will end up in different segments is (19/20)*(18/20) = 0.855 or 85.5%:
    - It doesn't matter where the first one hits.
    - There are 19 segments out of 20 left for the second dart to hit, so a probability of 19/20 that it will end up in a different segment than the first one.
    - After that, there are 18 segments out of 20 left for the third dart to hit, so a probability of 18/20 that it will end up in a different segment than the first two.

    So the probability that at least two will stick in the same segment is 100% - 85.5% = 14.5%.
    The probability that all three will stick in the same segment is (1/20)*(1/20) = 0.0025 or 0.25%.
    The probability that exactly two darts will stick in the same segment is 14.5% - 0.25% = 14.25%</span hide>

    Added: I misread the original question and somehow assumed 3 instead of 6 darts. <img src=/S/stupidme.gif border=0 alt=stupidme width=30 height=30>
    The method I described can easily be extended to 6 darts.

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    Re: Anybody want a game of darts?

    Jerry

    I may regret sticking my neck out here but it seems that your answer includes situations where 2, 3, 4, 5 and 6 darts share the same segment. I interpret the wording of the puzzle to mean two, and only two, darts share the same segment - which would seem to make the calculation somewhat more complex... (explaining why I have not posted an answer)

    Ian

  6. #6
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    Re: Anybody want a game of darts?

    I interpreted it as "2 or more darts" rather than "exactly 2" (because the calculations were easier) <img src=/S/grin.gif border=0 alt=grin width=15 height=15>

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