The probability that I will understand…

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:

> iQIcBAEBCgAGBQJQi58mAAoJEDeph/0fVJWsN84P/0GaRnK0CH8SR7IQsWoQcsx8
> 4t9kaO59i7l/vgw9PVc4AU7Vkoixj0Q3W/jiw7IYIgejCVWdzvPWJynUhd/U+gDF
> 3nMK+iTNOOUjZjEZRv8e6Oki/io2AHfRZRjP/ugNOkOspdo+H78r1fmXOs5yXlqh
> i++NP2DAXJ3k+BTc7043PrLvdIOtlrryGNPXG4qs8tvkvtC3v/Wjqq0k0d34RE3T
> O5AnymocsoBDm9pYAOuxRveXcphMb1zA0zE3zBQmHW9JDfvNqad5o+/QbLjjpwQw
> NWjLkR7PIOA/CUv9XTyVQwLw9LjHr1m+y68ZNMdsgQR4VF1CK0F4qL7QPKLJEIJs
> 4GPGlPqg1XUK54PkQqQDSQYSj/rosQqBVUVUxlQPALxIHpfxLWqtG3T67j2Rk8Oo
> qObTNoBksJTYu81Ii1VX3Pvt4bjogZgUcH6HJBPc0aIxOnHK4LGQ6p55xuSKLMSy
> vfcf6NyGiyoQcziUHMGVdMjdKoy9PPGvRLsov3ezNxvePw/cunMwVnJacBZgf6+/
> mU9FLmTpeH18phe+QorheMwA/M9nnS13C/48fGoRFqKt19x+fVOVrtUiubgkAlS7
> sx8J9WL8FbT0D5HmrRB6DRavYxQIO4PCsGODBqag22GYBp6vR6GU8/dJe4hz95Yi
> PbIz9P4Wml4NillJLvwk

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