And since nobody has brought this up yet, my favourite example of the counter intuitive probability puzzle: http://en.wikipedia.org/wiki/Monty_Hall_problem
Venky, the Second. On Saturday, 27 October 2012 at 3:17 PM, Landon Hurley wrote: > -----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] > > (mailto:[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-----
