# Does “50/50 chance of.. . ” convey information?

I distinctly remember the professor in the undergrad introductory systems & control course saying that "when weather forecasters say there's a 50% chance of precipitation, they are conveying no information".

At the time (freshman) I didn't know Shannon or Kolmogorov but it struck me as a strange comment, after all, if true, why bother reporting the numbers in the first place?

More recently, in Itzhak Gilboa's Theory of Decision under Uncertainty I found the comment, p.25:

First, if we have a random variable X on [0, 1], and we know nothing about it, we cannot assume that it has a uniform distribution on [0, 1] and pretend that we made a natural choice.

So even though the uniform distribution is the maximum entropy distribution (eg among binomial distributions, {rain,no rain}), doesn't saying 50/50 chance convey more information than not knowing the distribution at all?

I'm 50% sure some will vote to close this question, but note I'm not asking for a quantification of how much more (if any) information, but rather just a binary yes/no, which should be within the realm of mathematics.

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If somebody tells me there's a 50% chance of rain, I take an umbrella with me. Without that information, I assume the usual distribution (based on experience) in which there is a much smaller chance of rain, and I leave the umbrella at home. – Brett Frankel Feb 8 '13 at 3:44
@BrettFrankel, ok but maybe you don't live in Seattle? – alancalvitti Feb 8 '13 at 4:02
I find it as a valid question. What principle of indifference says is that if no information is known, then assign uniform probabilities to the (disjoint) events. So in this angle what your professor said looks correct. On the other hand @BrettFrankel's point also seems valid. Isn't there a contradiction here? – Kumara Feb 8 '13 at 4:12
@Kumara The difference is that, in the absence of the weatherman's report, I make other assumptions about the distribution. – Brett Frankel Feb 8 '13 at 4:17
@alancalvitti If I lived in Seattle, I'd be less inclined to bring an umbrella if there were only a 50% chance of rain. – Brett Frankel Feb 8 '13 at 4:18

It does convey information- you must take conditional probability into account. Suppose that I have some information $I$, and that in general the probability of a certain point having rain is $P(R)$. What the weatherman is giving is $P(R|I)$, the probability of rain given the information. It could be that $P(R)=0.1$, while the $P(I)=0.1$, $P(I|R)=0.5$- then by Bayes rule $P(R|I)=\frac{P(I|R)P(R)}{P(I)}=\frac{0.5*0.1}{0.1}=0.5$, which is rather a lot more information than simply knowing that $P(R)=0.1$.