-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA512 On 10/27/2012 04:49 AM, Deepa Mohan wrote: > I can TELL myself all this: > > On Sat, Oct 27, 2012 at 2:15 PM, Landon Hurley <[email protected]> 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 >
Probably isn't intuitive, in fact it's inherently counter intuitive. Unless you bother to understand the basic frequency mathematics, there isn't really any point in bothering at all in the long run. In the end, intuition is what we feel is simplest, and that's Occam's razor. That's fine, except that it's a rule of thumb not an absolute. It doesn't prove anything. Your argument is that flipping that coin will have a probability of .5 resulting in heads. What's the probability of coming up with the result of HHHHHHHHHHTHHTH? .5**15, definitely not .5, right? And that's because the probability of .5 for coins is a continuous string of states, not just independent. Very few things are independent in probability, and they are structured on a discrete set of foundations: 1. That the event in question happened, or else is included in the outcomes. This is where you have an intuitive problem, because you say that 1 will either happen or not happen. However, what about 2,3,4,5,6? They are all equally NOT 1, but they have their own discrete existences outside of ~1, right? They all have to exist as potential outcomes, and the probability that one of them occurs has to be 1, given that you rolled in the first place. Your interpretation has an inherent contradiction, since you can't have 6 equal probabilities of .5! Something cannot be 300% likely to happen. Obviously, if the event of rolling is even in question, then there are 7 potential outcomes, NULL,1,2,3,4,5,6. Number of desired outcomes divided by number of potential outcomes is the primitive of probability. To be honest I don't see a simpler way to break it down, maybe someone else on the list has a better method. However, I am interested in how you ameliorate the paradox between the probability of rolling a one is .5, i.e. half of the N times you role a die, you will come up with a 1, and the empirical fact that a you role a large N, you come up with a number that approximates 16.66 out of a 100 occurrences of 1. - -- Violence is the last refuge of the incompetent. -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.11 (GNU/Linux) Comment: Using GnuPG with Mozilla - http://www.enigmail.net/ iQIcBAEBCgAGBQJQi62eAAoJEDeph/0fVJWsUWsQAIueaVPqLw4BVcSSh4oX7k2M hzYZHjx7KNAmQJ9JWu1tj/+1VOgEywERFmPMMpEgTyel4pOSR6hIzUeHLBmC7uqy 1CRgbIlEcB9a1W5jGUV5qDGxB005f/lRjBgWqJEQkYVrxdhzae5+pxJghJDDbn/Z ypZFqJytvmu1OlRchfNDb21672sVQOlTaDuG5/Nyq8mCQKZgFeBIJEhnBQA49Wfa 0AxtCTbeSxOoDGWkElpdKyzjbOmwmTGYAGedUMfYQg+aI7MrphjgBbL9QZKDELZr AXS3+isQRpgHINA+zKnsTPstnaTHXtwGDbm4OMKopLqEZ9HfFVG1fnxCX0NKYvSN BFQXOAWSLoDzl9RY+o0kyK1nb6EJ6/qKp/sy7dA8uJAfzphUsfEkT3lSi4xuyGVU fLnOpDwoELg9uOTqMAbi9aaWeGvSpCOw8OdmA1xFjYfMqyM4yQz4IcLKer8kY3+M eRbBZ5pkMweBAO5ReFfIT/bsoJenZ29qzk+l7nRUaAdhAiVkF/dkjWcA8nZniHAB mQAkOWEiR5IXCwKW+Lwd7fNUjwItqvyO3afxumyvqBWMPUufZpZ/VCUchJ1E9g4h cZnh4tztt/FO+g+ATdKYdl9vEg1RZP7piCMqX9LupPPs19MjVwoTJRidc6cGty6e KU99+tv7d3bU9fLfCxxv =QuaT -----END PGP SIGNATURE-----
