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Bertrand Paradox Method-1 and Method-2 may be equal

  1. Jul 13, 2015 #1
    I think that both methods (1 and 2) of Bertrand Paradox may be equivalent if the probability is represented by area of the circle enclosed by each method rather than by distances.
    Please see the attached file.

    In both methods, the probability is approximately 0.6.
     

    Attached Files:

  2. jcsd
  3. Jul 17, 2015 #2
    No, neither of your methods correspond to any of Bertrand's methods 1, 2 or 3 which are simply explained and calculated on the relevant Wikipedia page and you are therefore missing the whole point of the paradox, which is that the words "select at random" can be interpreted in more than one way which results in more than one answer for the probability.

    For instance in Bertrand's Method 1, first an arbitrary point is selected on the circumference (it does not matter how this first point is selected). Then another point is selected a distance ## d ## clockwise from the first point such that the probability of ## d ## falling in the interval ## (x, x+h) ## is ## \frac h{2\pi} ##. In your method you select the chord in a different manner such that this probability is different, so your calculation is not valid for Bertrand's Method 1.
     
  4. Jul 18, 2015 #3
    From this, it is concluded that the location of the second point on the arc between x and x+h is the random variable whose probability is to be measured.
    While according to the wikipedia this sentence from the second paragraph is quoted "What is the probability that the chord is longer than a side of the triangle?" That means the cord length is the random variable in question.

    That makes the whole change in the technique of calculating the probability. Again, if the area is adopted to represent the probability rather than the length or arc, then Method-1 is equivalent to Method-2.

    I do not think that I missed anything. The problem for me is well stated as quoted from wikipedia. I do not have to stick to any particular technique even if it was used by Bertrand himself. I am calculating probabilities of a well-defined random variable which is the chord length.
     
    Last edited: Jul 18, 2015
  5. Jul 18, 2015 #4
    You are missing the point. The techniques you have described can be called Method 4 and Method 5, and you have shown that the results for Method 4 and Method 5 are equal to each other. But the essence of Bertrand's Paradox is that Methods 1, 2 and 3 arrive at different results, and nothing you have written affects that.

    Yes, and the uncertainty over the distribution of that random variable is what causes the paradox.
     
  6. Jul 18, 2015 #5
    So let me continue missing the whole point by introducing another method called Method-6. In this method, the orientation is similar to method-2 in which a point is chosen on the radius and a cord is constructed to be perpendicular to the radius. To calculate the probability of chord longer than the side of the triangle, we follow the chord until it meets with the circle circumference and then calculate the length of the arc, which represent all chord longer than the triangle side, from the point on the circle where the diameter ends to the point on the circle where a parallel line to the diameter and to the side of the triangle that is bisecting radius also meets with the circumference. It is not surprising to see that the length of the first arc=1/2 the second one which means the probability of chords longer than the triangle length is 1/3 which is equivalent to the result of Method-1

    This means if we had to use the length of the arc rather than the length on a diameter in method-2, we would have ended to the exact same result of method-1, namely the probability amount of 1/3. This also means that in order to have a non-paradoxical results, it is important to have standard definition during the calculation. If we used arc lengths in method-1 we should have not changed to parts of the diameter in method-2.

    So missing the point by introducing more methods yielding the same probability would create no paradox, while sticking to the point would create a paradox. This means that sometimes it is important to missing the whole point in order to solve a paradox which is created by not missing any points. This itself may be a new "Missing Points Paradox".
     
    Last edited: Jul 18, 2015
  7. Jul 19, 2015 #6
    Yes! That is the essence of Bertrand's paradox.

    Like many paradoxes, Bertrand's paradox is not a puzzle that has a solution, it is a scenario that demonstrates that if we want consistent results we must define the question unambiguously. We say that this is the resolution (rather than solution) of the paradox.
     
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