-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA512 On 10/27/2012 03:16 AM, Deepa Mohan wrote: > On Sat, Oct 27, 2012 at 8:54 AM, SS <[email protected]> wrote: >> On Sat, 2012-10-27 at 04:46 +0200, Charles Haynes wrote: >>> Do you have a six sided die? What do you think the chances are >>> that you will roll a 1? > > 50-50. Either I will roll a 1, or I will not. And since I turned > 98 on FB, thank you for calling me (intellectually, at least) a > child! > > By the way, it was Founder's Day on this mailing list on October > 23. Happy Foundering! > > Deepa. >
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. landon - -- 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/ 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 =5clz -----END PGP SIGNATURE-----
