How to calculate probability of rain today given today percentage Given this: If the probability of rain in a given day is 0.65, what are the odds that it will rain today ?
Would I setup as P(A) = it rains 65%
But with the missing today forecast, what and how would I setup P() for the conditional?
Is not this answer just, since in a given day the  probability is 65%, then the odds for rain today is 65%?
P(rain tomorrow|rain today)=P(rain tomorrow | rain today)P(rain today) = (0.65 * 0.65) / 0.65 = 0.65. For this 65%, I will assume that we can only make at best an educated estimate or educated (or ignorant) guess base upon this one dimension data P(A); as missing event data may constitute an different estimate or confident for the event of rain today.
 A: Conditional probability is handled fairly well via Bayes' rule.
$$ P(\mbox{rain tomorrow} \vert \mbox{rain today)} = \dfrac{P(\mbox{rain tomorrow} \cap \mbox{rain today})}{P(\mbox{rain today})} $$
The numerator is the probability it rains today and tomorrow. If days are independent, then the probability it rains tomorrow is the same as the probability it rains any day.  If you believe that raining today somehow increases the chances of it raining tomorrow, you need to specify that relationship.  Your problem needs more detail on the assumptions about raining.  We can not provide those to you.
A: Such probabilities are always dependent on the background for the statistical model (the information). 

There isn't such a real thing like the probability of rain as if God is rolling dice which will turn up rain with 65% probability.
Or actually you could see it a bit like that as if the machinery behind the weather is a random number generator, that shows 65% probability rain (and in reality it is deterministic and either 100% sure rain or 100% sure no rain, but with some level of abstraction, a level of knowledge that is not all-knowing, you can speak of random behaviour for nature). But that is often not what is meant with a weather forecast. 

The 65% is part the random behaviour of nature, but foremost it is a large part an indication of the uncertainty of the model. 
For the given knowledge/measurements/observations there is 65% probability of rain.
This '65%' means, 'we think it might rain but we are not sure'.

So to give a prediction for the weather today based on the prediction of weather on another day is very limited information. It is statistically not sensible to make predictions based on a single number. If you would be placed on a strange planet and they asked you 100 days ago the probability of zizozazu was 65% what is it today? Then you could not give a reasonable answer. You could guess 65% as a try, but you do not really know how zizozazu behaves on that planet. Maybe it only occurs once every 1000 days and the peak of probability 100 days ago should indicate a low probability today. Maybe zizozazu occurs typically with 10% probability and the 65% 100 days ago should be weighted with the mean value of typical weather (regression to the mean).
So whenever you make a prediction of an event and express it as a probability, then this probability is often an expression of uncertainty in your statistical model and data (which do not allow you to make predictions with certainty). 
E.g. you normally observe a certain frequency and this is what you base your computation on. It is not like a real frequency for a system deviced such that you can estimate it accurately based on theory and with very little variation away from that theory (like a fair roulette game, or a roll with a fair dice).
The expressions of probability are a conversion of the data into numbers about frequency. With the absence of data (or knowledge derived from data), you can not make a reasonable expression of probability.
