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I will preface this with the actual question taken from lecture:

An unusual pseudo-random generator outputs integers between 1 and 4, inclusive, in such a way that each value, v, occurs with a probability that is inversely proportional to v. For example, 4 is half as likely to occur as 2, and 1 is three times as likely to occur as 3. Determine the probability P(v) for all possible values of v. Explain in detail, and give the final answer(s) explicitly, in decimal notation.

This seems like a very simple probability question; however, I just want to make sure I am looking at this the correct way. It says that each value occurs with a probability that is inversely proportional to v. I understand that the sample space here is S={1,2,3,4}, but wouldn't this not be possible due to the fact that the probability of 1 would be 1 (which is the inverse of 1). Thus, P(v) would not sum up to be equal to 1, which is a basic property of probability on events.

Just double checking that I am looking at this question the correct way, and whether or not it is a trick question. Thanks!

EDIT: I understand that this question lies more as a statistical problem; however, it is from a CS class, so I thought it would still be appropriate to post here.

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The key here is the part of the problem that says "each value, $v$ occurs with a probability that is inversely proportional to $v$". You can do this by assigning $$ P(1)=c, P(2)=\frac{c}{2}, P(3)=\frac{c}{3}, P(4)=\frac{c}{4} $$ Then, since these values are the only ones possible, we'll have $$ c\cdot 1+\frac{c}{2}+\frac{c}{3}+\frac{c}{4} = 1 $$ so $$ 1 =c(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}) = c\cdot\frac{25}{12} $$ and so $c=12/25$, giving us $P(1)=12/25, P(2)=6/25, P(3)=4/25, P(4)=3/25$. It's easy to see that these satisfy the requirements of the problem.

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  • $\begingroup$ Thank you for the help! I misunderstood what the purpose of v was in the problem. Makes perfect sense now. $\endgroup$ – E. Otero Jan 28 '16 at 1:19

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