On a mailing list that I belong to, someone attempted to explain it to me (as also to children who are facing the concept for the first time). But....
I can TELL myself all this:
On Sat, Oct 27, 2012 at 2:15 PM, L H <ljrh....@gmail.com> wrote:
> The probability that you will or will not roll a 1 is .5. The
> probability that the roll will result in a 1 is 1 in 6, given the event
> that you have rolled a six sided die. You can reduce most events to will
> or will not happen. That isn't the true representation of that scenario
> It can only be .5 if there are two equally probable events. How you are
> presenting them is not in terms of the probability of those outcomes,
> but in that given two discrete categories (1 or 2,3,4,5,6) the
> probability is .5 that you will land on 1.
> The inverse of rolling a one is important as well. The probability of
> rolling a 2 must also be .5, according to your interpretation. But since
> all possible probabilities must sum to 1, .5(6)=3 (or .5 added for each
> of the six possible events) is a contradiction, not to mention
> impossible given that probability is a continuum from 0 to 1.
> Given your terms:
> In terms of a lottery, there are n people who participate. My odds of
> winning must be .5, either I do win or I don't. However, what if there
> is only one person who participates in the lottery, and that person is
> me? Isn't my odds of not winning 0, since I would always win?
> However in a system where probability is always calculated by the
> desired events divided by the total number of possible outcomes, this
> contradiction is explained, and empirically valid with your perceived
> paradox with conventional frequency based probability. Given one desired
> outcome (rolling a 1) and two potential outcomes (rolling or not rolling
> a 1) the expression becomes 1 in 2. Charles' example though derives from
> rolling a 1 as opposed to rolling a 1,2,3,4,5,6. With the same equation
> structure of desired out of total outcomes, you get 1/6. Here is where
> your interpretation breaks down though: what is the probability of
> rolling a red? Your view would have either red, or not rolling a red,
> thus 50-50. However, you can never roll a red. So the probability is
> accurately reflected as 1/0.
But it ultimately makes as much sense to my intuition as this: