The probability that I will understand....

October 27, 2012

I canNOT understand probability theories. I think that everything is just 50-50…either it will happen or occur, or it will not.

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 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 though.

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:

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