Comments on: Determinism makes one more likely to be cheater, pumpkin eater http://www.oregoncommentator.com/2008/04/18/determinism-makes-one-more-likely-to-be-cheater-pumpkin-eater/ Free Minds, Free Markets, Free Booze Sun, 20 Jul 2008 07:07:04 +0000 http://wordpress.org/?v=2.3.3 By: Betz http://www.oregoncommentator.com/2008/04/18/determinism-makes-one-more-likely-to-be-cheater-pumpkin-eater/#comment-98783 Betz Fri, 18 Apr 2008 19:36:50 +0000 http://www.oregoncommentator.com/2008/04/18/determinism-makes-one-more-likely-to-be-cheater-pumpkin-eater/#comment-98783 I feel the need to respond to this, because I'm a math major, and also because I just got out of a stats quiz this morning. Lets consider two events, the event that the person did in fact cheat (call this event A), and the second event that the person does not believe in free will (Event B). Lets assume that in the experiment, half of the population was in the control group, and the other half in the experimental group. Therefore, the liklihood that the person believes in free will or does not believe in free will (ie, P(B) and P(-B) ) is the same... 50%, or 0.50. Lets also assume that most students at the university are "fairly" honest, and that they will cheat only 15% of the time. What is the likelihood that a student will have cheated (ie, P(A) )? To calculate this, the formula goes: P(A) = P(B)P(A | B) + P(-B)P(A | -B) The 'P(A | B' notation means the probability of A given that event B has occured. We know P(B) and P(-B) = 0.5. P(A | B) means that the person has a 45% more liklihood to cheat than not to... Since a students liklihood to cheat is only 0.15 anyways, we can get this by adding 45% of 0.15, or multiply 0.15 by 1.45. P(A | -B) is the liklihood that the student will cheat if they do not believe in free will, which is just our initial assumed value of 0.15. Therefore, P(A) = (0.5)((.15 * 1.45)) + (0.5)(0.15) = .18375 = 18.4% that a given will cheat, which is not too different from our assumption of 15% to begin with. Therefore, the statistic given is not too great a change. If we assume a much higher liklihood to cheat, say along the lines of 50% (So, a student will cheat half of the time), we get: P(A) = (0.5)((0.5) * 1.45)) + (0.5)(0.5) = .6125 = 61.3%, compared to our assumption of 50%. This is a little bit different (10%+ is noteworthy), but I don't know if it is jaw-dropping. If the liklihood were smaller, the difference between the new findings and the assumption would be much closer. Therefore, If ever caught or accused of cheating in any other dept. other than the math dept., you probably stand a pretty good chance of convincing the accuser that Determinism made you do it. If that fails, call the accuser "insensitive" to your beliefs and incite an accusation of bigotry, and they should back down. ;) I feel the need to respond to this, because I’m a math major, and also because I just got out of a stats quiz this morning.

Lets consider two events, the event that the person did in fact cheat (call this event A), and the second event that the person does not believe in free will (Event B). Lets assume that in the experiment, half of the population was in the control group, and the other half in the experimental group. Therefore, the liklihood that the person believes in free will or does not believe in free will (ie, P(B) and P(-B) ) is the same… 50%, or 0.50. Lets also assume that most students at the university are “fairly” honest, and that they will cheat only 15% of the time.

What is the likelihood that a student will have cheated (ie, P(A) )? To calculate this, the formula goes:
P(A) = P(B)P(A | B) + P(-B)P(A | -B)
The ‘P(A | B’ notation means the probability of A given that event B has occured.

We know P(B) and P(-B) = 0.5. P(A | B) means that the person has a 45% more liklihood to cheat than not to… Since a students liklihood to cheat is only 0.15 anyways, we can get this by adding 45% of 0.15, or multiply 0.15 by 1.45. P(A | -B) is the liklihood that the student will cheat if they do not believe in free will, which is just our initial assumed value of 0.15. Therefore,
P(A) = (0.5)((.15 * 1.45)) + (0.5)(0.15) = .18375 = 18.4% that a given will cheat, which is not too different from our assumption of 15% to begin with. Therefore, the statistic given is not too great a change.

If we assume a much higher liklihood to cheat, say along the lines of 50% (So, a student will cheat half of the time), we get:
P(A) = (0.5)((0.5) * 1.45)) + (0.5)(0.5) = .6125 = 61.3%, compared to our assumption of 50%. This is a little bit different (10%+ is noteworthy), but I don’t know if it is jaw-dropping. If the liklihood were smaller, the difference between the new findings and the assumption would be much closer.

Therefore, If ever caught or accused of cheating in any other dept. other than the math dept., you probably stand a pretty good chance of convincing the accuser that Determinism made you do it. If that fails, call the accuser “insensitive” to your beliefs and incite an accusation of bigotry, and they should back down. ;)

]]> By: Vincent http://www.oregoncommentator.com/2008/04/18/determinism-makes-one-more-likely-to-be-cheater-pumpkin-eater/#comment-98776 Vincent Fri, 18 Apr 2008 17:19:04 +0000 http://www.oregoncommentator.com/2008/04/18/determinism-makes-one-more-likely-to-be-cheater-pumpkin-eater/#comment-98776 An interesting post that touches on some issues of determinism (or "luck") vs. free will can be found <a href="http://illtelligent.com/stereo/2008/04/17/just-my-luck/" rel="nofollow">here</a>. An interesting post that touches on some issues of determinism (or “luck”) vs. free will can be found here.

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