A simple probability reasing to predict rain fall 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?

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.
