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The scenario goes like this: a cell phone user has 1/3 of a chance to send a text message, and a 2/3 of a chance to receive a text message.

Question: what is the expected number of text messages received by the cell phone user before they send a text?

Is the answer 1/p = 1/(2/3) = 3/2 = 1.5 messages? (the expected value for a geometric random variable).

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2 Answers 2

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This indeed concerns geometric distribution but in your guess you make 2 mistakes.

Here "sending a message" must be interpreted as "success" so that $p=\frac13$.

We do not deal with the expected number of trials needed to arrive at a success but with the number of failures that preceed the first success. In that case the expectation is not $\frac1p$ but equals:$$\frac1p-1=\frac{1-p}p$$ So the correct answer is:$$\frac{1-\frac13}{\frac13}=2$$ A more direct way is solving the equation:$$\mu=\frac23(1+\mu)+\frac13\cdot0$$Here $\mu$ denotes the expectation.


Informal confirmation: in a situation like this it will not be surprising to meet sequences like:$$\cdots RRSRRSRRSRRSRRS\cdots$$ ($R$ for receive and $S$ for send)

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    $\begingroup$ Please notice that our policy for [self-study] questions is to give hints, not answers. $\endgroup$
    – Tim
    Oct 3, 2022 at 8:00
  • $\begingroup$ @Tim What part of my answer should be left out? Is it sufficient to delete the calculation leading to answer $2$ (starting with: "so the correct answer is...")? $\endgroup$
    – drhab
    Oct 3, 2022 at 8:07
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    $\begingroup$ I'm leaving it up to you, people have different styles of answering such questions. I just wanted to make you aware that we treat such questions slightly differently than others. $\endgroup$
    – Tim
    Oct 3, 2022 at 8:08
  • $\begingroup$ $$\frac{1-p}p$$ as in...odds? $\endgroup$
    – BCLC
    Oct 3, 2022 at 11:23
  • $\begingroup$ drhab full post: Mean of geometric distribution is odds? $\endgroup$
    – BCLC
    Oct 3, 2022 at 14:06
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You correctly noticed that the distribution is geometric, i.e. the distribution for "the number of failures before observing a success". The distribution takes two forms (described in Wikipedia), where the form that would be more useful for you is where $k$ stands for the number of "failures", so the probability mass function is $(1-p)^k p$ and where the expected value is $\tfrac{1-p}{p}$.

The tricky part is finding out what the "success" happening with probability $p$ is and the "failure" happening with probability $1-p$ mean in the problem.

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  • $\begingroup$ @drhab see the second paragraph. $\endgroup$
    – Tim
    Oct 3, 2022 at 8:03
  • $\begingroup$ that's interesting...what's the intuition behind why expected value of geometric distribution is $$\frac{1-p}p$$ as in...odds? $\endgroup$
    – BCLC
    Oct 3, 2022 at 11:25
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    $\begingroup$ @BCLC no, see en.wikipedia.org/wiki/Geometric_distribution#Properties , but if you want to ask a question, use questions not comments for it. $\endgroup$
    – Tim
    Oct 3, 2022 at 11:32
  • $\begingroup$ ok thanks. thought it would be a short story. $\endgroup$
    – BCLC
    Oct 3, 2022 at 13:06
  • $\begingroup$ Tim, full post: Mean of geometric distribution is odds? $\endgroup$
    – BCLC
    Oct 3, 2022 at 13:15

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