surprise examination
Richard raises the issue of the surprise examination paradox
your teacher tells you (I) she's going to give the class a surprise exam next week, and (ii) you won't be able to work out beforehand on which day it will be. Using this information, you work out that it can't be on Friday (the last day), or else you'd be able to know this as soon as class ended the day before, contrary to the second condition. With Friday excluded from consideration, Thursday is now the last possible day, so we can exclude it by the same reasoning. Similarly for Wednesday, Tuesday, and finally Monday. So you conclude that there cannot be any such exam. This chain of reasoning guarantees that when the teacher finally gives the exam (say, on Wednesday); you're all surprised, just like she said you'd be.
Richard proposes a solution he terms "epistemic confusion", where the statement "you cannot know" throws the listeners into confusion, thereby ensuring that they cannot be confident of their own reasoning or proofs.
This has some merit in a similar way to the examination of surprise I will discuss later but it doesn’t really seem to tie up the loose ends.
Here is an example of the problem working using the preconditions given.
Imagine this: Teacher watches you carefully to see if you show any sign of preparing for the test and if you do then he doesn’t put the test on that day.
Why doesn’t the kid "pretend to be preparing in order to fool the teacher?" well A - the teacher is a bit more experienced at these things - that's why they are the teacher and B- it isn’t worth any rational students time to worry so much about the "surprise" and so little about the test.
Which leaves just one way the strategy doesn’t work - if the students are in a constant state of expectation.
There is no requirement here for the students to be in any particular mental state or use any particular logic - as long as it doesn’t result in them being in a constant state of expectation.
And even better you can imagine a human being able to do it - while guaranteeing "confusion" is rather more demanding a task.
-----
But as well as this sort of argument there is always a more technical counter to the alternative hypothesis. I.e. why exclusionary logic doesn’t work.
I see the problem lying in the exclusionary principle which is based on contradictions and the use of only part of the available information.
1) You can only say P (=Friday) = zero if you define P (=mon-thurs) as 1 - this is a fundamental part of the problem. The exclusion of Friday is valid only in as far as the same logic does not also exclude other days.
2) On Thursday - you can't BOTH say P (=Friday) is 0 and also "if it is Friday the test is on Friday" because you have declared it CAN'T be Friday therefore there is no meaningful "if". Maths is a bit like 1*(test is on mon-thurs) + 0*(if it is Friday I know it is on Friday). But we can substitute ANYTHING into the zero probability set including things that do or don’t breach the rules of the game particularly since the current inhabitant of that set (which encourages us to define it as 0) clearly does breach the rules of the game anyway as per (1).
All the student has to do is to rationally reject the use of exclusionary logic based on the principle that it provides no useful information (i.e. it rejects all days).
-----
So that is why it contradicts itself so what happens as we apply it?
It is Friday - We can say the last day possible day is Friday - P=1.00 impossible (unless we get really confused)
It is Thursday P (=Friday) less than P (=Thursday) (otherwise your first step cannot be to exclude Friday) P (=Friday) is thus something less than .5. It is hard to define P (=Thursday)/P (=Friday) because your logic implies both are close to zero relative to any alternative.
This leaves two possibilities -
1) You can follow a policy of P (=tomorrow)/P (=today) = 0 but that requires each day to be infinitely more likely than the next AND it forces you to expect a surprise every day no matter how many millions of days there are.
Or you can say P (=tomorrow)/P (=today) is either indeterminate or some rational number in which case the exclusionary principle falls apart. You cannot reject Friday in favour of mon-thursday because you can’t prove mon-thursday is infinitly more likely or even finitely more likely. I suggest exclusion is only possible in a relitivistic sense not in an absolute one. (Relevant particularly if we see surprise relitivisticaly)
your teacher tells you (I) she's going to give the class a surprise exam next week, and (ii) you won't be able to work out beforehand on which day it will be. Using this information, you work out that it can't be on Friday (the last day), or else you'd be able to know this as soon as class ended the day before, contrary to the second condition. With Friday excluded from consideration, Thursday is now the last possible day, so we can exclude it by the same reasoning. Similarly for Wednesday, Tuesday, and finally Monday. So you conclude that there cannot be any such exam. This chain of reasoning guarantees that when the teacher finally gives the exam (say, on Wednesday); you're all surprised, just like she said you'd be.
Richard proposes a solution he terms "epistemic confusion", where the statement "you cannot know" throws the listeners into confusion, thereby ensuring that they cannot be confident of their own reasoning or proofs.
This has some merit in a similar way to the examination of surprise I will discuss later but it doesn’t really seem to tie up the loose ends.
Here is an example of the problem working using the preconditions given.
Imagine this: Teacher watches you carefully to see if you show any sign of preparing for the test and if you do then he doesn’t put the test on that day.
Why doesn’t the kid "pretend to be preparing in order to fool the teacher?" well A - the teacher is a bit more experienced at these things - that's why they are the teacher and B- it isn’t worth any rational students time to worry so much about the "surprise" and so little about the test.
Which leaves just one way the strategy doesn’t work - if the students are in a constant state of expectation.
There is no requirement here for the students to be in any particular mental state or use any particular logic - as long as it doesn’t result in them being in a constant state of expectation.
And even better you can imagine a human being able to do it - while guaranteeing "confusion" is rather more demanding a task.
-----
But as well as this sort of argument there is always a more technical counter to the alternative hypothesis. I.e. why exclusionary logic doesn’t work.
I see the problem lying in the exclusionary principle which is based on contradictions and the use of only part of the available information.
1) You can only say P (=Friday) = zero if you define P (=mon-thurs) as 1 - this is a fundamental part of the problem. The exclusion of Friday is valid only in as far as the same logic does not also exclude other days.
2) On Thursday - you can't BOTH say P (=Friday) is 0 and also "if it is Friday the test is on Friday" because you have declared it CAN'T be Friday therefore there is no meaningful "if". Maths is a bit like 1*(test is on mon-thurs) + 0*(if it is Friday I know it is on Friday). But we can substitute ANYTHING into the zero probability set including things that do or don’t breach the rules of the game particularly since the current inhabitant of that set (which encourages us to define it as 0) clearly does breach the rules of the game anyway as per (1).
All the student has to do is to rationally reject the use of exclusionary logic based on the principle that it provides no useful information (i.e. it rejects all days).
-----
So that is why it contradicts itself so what happens as we apply it?
It is Friday - We can say the last day possible day is Friday - P=1.00 impossible (unless we get really confused)
It is Thursday P (=Friday) less than P (=Thursday) (otherwise your first step cannot be to exclude Friday) P (=Friday) is thus something less than .5. It is hard to define P (=Thursday)/P (=Friday) because your logic implies both are close to zero relative to any alternative.
This leaves two possibilities -
1) You can follow a policy of P (=tomorrow)/P (=today) = 0 but that requires each day to be infinitely more likely than the next AND it forces you to expect a surprise every day no matter how many millions of days there are.
Or you can say P (=tomorrow)/P (=today) is either indeterminate or some rational number in which case the exclusionary principle falls apart. You cannot reject Friday in favour of mon-thursday because you can’t prove mon-thursday is infinitly more likely or even finitely more likely. I suggest exclusion is only possible in a relitivistic sense not in an absolute one. (Relevant particularly if we see surprise relitivisticaly)
posted by Genius at 4:19 PM
1 Comments:
I would also say that the act of excluding any day on the basis of any logical principle from being surprising in itself makes that day a target for surprise. That is, if you have a reason not to expect it, then by it being unexpected due to that reason, it should be expected.
Post a Comment
<< Home