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Probability of a coincidence - question

  1. Sep 17, 2008 #1

    I have been reading the book "200% of Nothing" by A K Dewdney, which explains various math abuses experienced in the real life. One of the examples explains the math behind a so called telepathy incidence; a coincidence in which, one gets the phone call from the very person whom she was thinking about just a minute ago. I am not able to get the final answer provided by the author. Can you please check my approach to the problem?

    The example goes something like this: Suppose, for example, that you know 200 people (family, friends and colleagues) well enough to find yourself thinking about each of them occasionally. Suppose that you think about 10 of those people per day and two of them call you each day on average. The calls could come at any time during those 16 waking hours. Now, when the phone rings, the probability of one of those 10 people, about whom you thought about during the day has called, is 10/200 or 0.05. Now, the probability that the call is from someone you thought of in the previous minute is 1/960 or 0.0001, where 960 is the number of minutes in 16 waking hours. Although the chances of having such a call on a particular day are very small, if we consider a period of 10 years, i.e. 3650 days, the probability over this period climbs to 0.52 (the final answer.)

    Now, this is how I approached the problem: I used a decision tree in my analysis. Let us say there are 10 minutes, spread evenly over the 960 minute period, during which I think of each of the 10 persons. Now the probability of receiving a call during any of those 10 minutes is 1/960. I further assume that I have a 2 line phone, which means I can receive two calls simultaneously, therefore, the probability of receiving a call during one of those 10 minutes is 1/960 + 1/960 = 2/960. Once I have received a call, within those 10 minutes, the probability that call being from one of those 10 persons of that day is 10/200 or 0.05. If the call is from one of those 10 people, then the probability that the call is from that very person whom I was thinking about earlier is 1/10 or 0.1, thus, the probability of receiving a call from the correct person is 2/960 x 10/200 x 1/10 or 0.00001. Since there are 10 such minutes during a day, the total probability per day is 0.0001. The probability of such an incident over a 10 year period is 0.0001 x 365 x 10 = 0.38 (my answer.)

    I am not sure how the author got his answer, 0.52. Can you please check where I am going wrong?


  2. jcsd
  3. Sep 17, 2008 #2


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    I don't know where the author's answer came from. I understand your process, and also find that the expected number of calls per 10 years is 0.38 based on those assumptions. (The probability of getting a call like that is somewhat lower, 31.7%, because some people will get several such calls.)

    The assumptions that you think of your friends for only an instant (leading to 1/960) and with equal probability both make the probability lower than I would have expected. If you think of each of the 10 friends for 1 minute each, that doubles the expected number of such calls/decade to 0.76; if you have some 'better friends' about whom you think more frequently and who are more likely to call, the expected number also rises.
  4. Sep 17, 2008 #3

    Thanks a lot for that.

    I don't understand what you mean by this quote.
  5. Sep 17, 2008 #4


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    If you roll three (normal 6-sided) dice, the average number of 1s is 0.5. If 100 people roll three dice each, the average expected number of 1s for the whole group is 100 * 0.5 = 50.

    But some people will get more than one of those 50, leaving more than half the people without any 1s.

    In particular, (5/6)^3 * 100 ≈ 58 people, on average, won't get any 1s, while 3*(1/6)^2(5/6)*100 ≈ 7 people will get two 1s and 3*(1/6)*(5/6)^2*10 ≈ 35 people will get exactly one.
  6. Sep 18, 2008 #5
    The population is 200.
    The persons thought of are 10 per day.
    The calls are 2 per day.
    When considering the thought of part, ask yourself, how many ways are there to form a unique set of 10 persons.
    When considering the calls, ask yourself, how many ways are there to form a unique set of 2 persons.
    Just figure the probability for a day, which will not change over
    time. The answer is much smaller than that given.
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