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A friend told me the following about whether it will rain tomorrow (or not):

The probability that it will rain tomorrow is $1/2$ since it will either happen or not. But -even as a non mathematician- I do not find this argument a convincing one, because the same reasoning can be used to predict that I will be struck by a bolt of lighting in five minutes.

Can someone tell me what is mathematically wrong about this argument?

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Your friend is assuming that all probabilities are evenly distributed, which is simply observably not the case. The best way I can think of to demonstrate this is with a counterexample, as you did. – Ataraxia May 19 '13 at 22:52
As a joke my grandfather used to say the same thing about winning the lottery - either you win or you don't. As pointed out, these are not equally likely outcomes, particularly if you didn't buy a ticket! – Dale M May 19 '13 at 23:42

The two hypothesis -it will rain or it will not rain- aren't equiprobable. Separating the possibilities into two sets doesn't implies that the two set are equal, even if one of them must happen.

When you try to compute the probabilty of having an odd number when you roll a dice is computing by addition of the probabilities of having a 2, a 4 or a 6. You guess they are equals because there is no reason it would be otherwise, but if the dice is loaded, it will not be the case. It's not a theorem, it's an assumption.

Computing the probability it will rain or not depends of many factors, and we use statistics and meteorologics models to have a good estimation knowing a certain amount of informations. I would add that the probability depends of the amount of informations you have. If you know that there is no wind and no cloud 3000miles around, you can tell that the probability is very low that tomorrow will be rainy.

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If you toss a coin you have two possible outcomes that are as likely to happen so the probability of each is one half.

The difference for getting struck by lightening is that the two outcomes are not as likely to happen (in fact the probability you get struck is really close to $0$) so you can't separate the two outcomes in set of same probability.

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The argument is based on the premise that the chance of rain or no rain is equally likely.

There may be places in the world where this is true in general - i.e. of the 365 days in the year it rains on 182.5 of them - somewhere in the tropics perhaps?

However, even if this is the case, there will be seasonal variances to the rainfall pattern i.e. rain is more likely in the monsoon season than the dry season.

Alternatively in the desert of Antarctica it almost never rains. In the Sahara it rains a little more.

In general, the best predictor of rain tomorrow is if it is raining today. In Sydney, Australia you will be right with this about 75% of the time. This would translate to probabilities of (very roughly) $\frac{3}{4}$ if it is raining and $\frac{1}{4}$ if it isn't.

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