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Math Help - Golf hole-in-one as a random variable

  1. #1
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    Golf hole-in-one as a random variable

    Viewing the time till my next hole-in-1 in golf as a random variable, I wondered if the recently covered fundamentals of exponential random variables in class, or what was informally referred to as the "waiting time" distribution, could be employed.

    NOTE: The probability of an amatuer hitting the ball into the hole from the tee-off is 1/12,500 i.e. once every 12,500 shots.

    I have thus far gathered the following framework:

    We consider a Poisson process with rate l per unit time and the random variable W, which is the time one must wait to see the next count, given by

    P(W < t) = 1 – exp(–l*t)

    I crunched in maple the following:

    1-exp(-(1/12500)*(87*10)), where 87 is the number of shots I typically hit per round of 18 holes of golf, multiplied by 10 to arrive at the possibly of hitting one in winter break i.e. 10 rounds of golf.

    Would appreciate corrections and suggetions to further the train of thought.

    Best,
    wirefree
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  2. #2
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    Quote Originally Posted by wirefree View Post
    Viewing the time till my next hole-in-1 in golf as a random variable, I wondered if the recently covered fundamentals of exponential random variables in class, or what was informally referred to as the "waiting time" distribution, could be employed.

    NOTE: The probability of an amatuer hitting the ball into the hole from the tee-off is 1/12,500 i.e. once every 12,500 shots.

    I have thus far gathered the following framework:

    We consider a Poisson process with rate l per unit time and the random variable W, which is the time one must wait to see the next count, given by

    P(W < t) = 1 exp(l*t)

    I crunched in maple the following:

    1-exp(-(1/12500)*(87*10)), where 87 is the number of shots I typically hit per round of 18 holes of golf, multiplied by 10 to arrive at the possibly of hitting one in winter break i.e. 10 rounds of golf.

    Would appreciate corrections and suggetions to further the train of thought.

    Best,
    wirefree
    The distribution of the number of hole-in-ones in 870 shots is binomial B(870,1/12500). Which is reasonably well approximated by a Poisson with mean 870/12500.

    CB
    Last edited by CaptainBlack; December 12th 2008 at 10:48 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    The distribution of the number of hole-in-ones in 870 shots is binomial B(870,1/12500). Which is reasonably well approximated by a Poisson with mean 870/12500.

    CB
    Appreciate the prompt response, CB.

    The suggested solution employing binomial does not address my query. As mentiond in my post, I am attempting to ascertain the time till my next hole-in-1 and not the number of hole-in-ones. Hence, the reference to the "waiting time" distribution.

    Would appreciate assistance with regards to my query.

    Best,
    wirefree
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  4. #4
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    Quote Originally Posted by wirefree View Post
    Appreciate the prompt response, CB.

    The suggested solution employing binomial does not address my query. As mentiond in my post, I am attempting to ascertain the time till my next hole-in-1 and not the number of hole-in-ones. Hence, the reference to the "waiting time" distribution.

    Would appreciate assistance with regards to my query.

    Best,
    wirefree
    The interval between events with a Poisson distribution in time with mean of N/period is exponentialy distributed with mean waiting time 1/N periods.

    Now re-read the last sentence in my earlier post.

    CB
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    Appreciate your patience, CB. You have a beautiful mind.

    As suggested, I arrive at 0.06492057290 chance of Hole-in-1 in the next 870 shots on evaluating poisson by:

    evalf(exp(-870*1/12500)*(870*1/12500))

    The above differs from the result arrived at by evaluating:

    evalf(1-exp(-(1/12500)*(87*10)))

    ...which comes to 0.0672331480. This latter results from the following:

    Now we consider a Poisson process with rate l per unit time and the random variable W, which is the time one must wait to see the next count. Hence, P(W > t) = e^–lt. Furthermore, the cumulative distribution of W is F(t) = P(W < t) = 1 – e^–lt.
    Could you please attempt to explain the difference in methodologies. Being math-junior I grasp the understanding better as explained in the quote above.

    Best,
    wirefree
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  6. #6
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    Quote Originally Posted by wirefree View Post
    Appreciate your patience, CB. You have a beautiful mind.

    As suggested, I arrive at 0.06492057290 chance of Hole-in-1 in the next 870 shots on evaluating poisson by:

    evalf(exp(-870*1/12500)*(870*1/12500))

    The above differs from the result arrived at by evaluating:

    evalf(1-exp(-(1/12500)*(87*10)))

    ...which comes to 0.0672331480. This latter results from the following:



    Could you please attempt to explain the difference in methodologies. Being math-junior I grasp the understanding better as explained in the quote above.

    Best,
    wirefree
    The Poisson calculation is for exactly one hole-in-one in the next 870 shots, not the probability of one or more. The exponential distribution calculatiuon automaticaly includes the multi hole in one cases.

    CB
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    If the trials are independent with probability of success p, then the number of trials to the first success (hole-in-one) has a geometric distribution with mean 1/p.
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  8. #8
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    Appreciate the assistance, CB.

    As a further exercise in fine-tuning, the formula should be reflective of the game of golf where lesser the number of shots I hit, better the player I am. And better the player, better his chances of a hole-in-1.

    So, in the formula above, increasing 870 to 1000 although implies a better chance of a hole-in-1, it is lacking in the obvious places.

    Could you suggest a method to address this concern?

    Best regards,
    wirefree
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  9. #9
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    Forum,

    I seek assistance with gaining an understanding of assigning weightages with regards to my query above. The same pertains to the hole-in-1 problem of my previous post.

    Would greatly appreciate advise.

    Best regards,
    wirefree
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