[gwnn]

The professor says to his students: "next week we will certainly have an exam. it can be on any weekday, but nobody will know in advance on which."

Now students start wondering...

"Friday is obviously impossible, because Thursday night it will become the only possibility, so all of us will know"

...

"Thursday is impossible, because Friday's impossible and thus on Wednesday we'd know"

et cetera

"We can't have a test! WTF"

[helene_t]

This dilemma has been studied extensively in the context of non-blinded randomized clinical trials with serial recruitment. The problem is this:

Suppose you can afford to recruit 100 test patients, and ideally 50 would be allocated to the treatment group and 50 to the control group, each having a 50/50 chance. Now if the first, say, 92 recruits accidentally are allocated as 50 in one group and 42 in the other, then you already know that all the 8 remaining will go to the 42 group. So you will be recruiting some under a 50/50 condition and others under a 100/0 condition, which causes some potential bias.

The theoretically optimal solution is to accept some compromise, i.e. the allocation will be somewhere between 50/50 and binomial, while the probabilities for the last recruit may be slightly off 50/50. (If feasible, one could also, for example, recruit the subjects as pairs and allocate each pair 1/1 randomly).

Your case is simpler because it is given that there will be exactly one test. There must be an, in some sense, optimal solution which assigns decreasing probabilities to the five days.

[MickyB]

Once you've realised the possibility that there might not be a test, all days become possible again.

[Fluffy]

you can also add that the test migth be at any hour unknown, them, over a continous range it won't work I think.

[BillHiggin]

Edit - changed mind

[bid_em_up]

helene_t, on Oct 25 2007, 07:14 AM, said:

This dilemma has been studied extensively in the context of non-blinded randomized clinical trials with serial recruitment. The problem is this:

Suppose you can afford to recruit 100 test patients, and ideally 50 would be allocated to the treatment group and 50 to the control group, each having a 50/50 chance. Now if the first, say, 92 recruits accidentally are allocated as 50 in one group and 42 in the other, then you already know that all the 8 remaining will go to the 42 group. So you will be recruiting some under a 50/50 condition and others under a 100/0 condition, which causes some potential bias.

The theoretically optimal solution is to accept some compromise, i.e. the allocation will be somewhere between 50/50 and binomial, while the probabilities for the last recruit may be slightly off 50/50. (If feasible, one could also, for example, recruit the subjects as pairs and allocate each pair 1/1 randomly).

Your case is simpler because it is given that there will be exactly one test. There must be an, in some sense, optimal solution which assigns decreasing probabilities to the five days.

Sometimes I wonder if you are serious, or just messing with us.

[inquiry]

bid_em_up, on Oct 25 2007, 09:27 AM, said:

helene_t, on Oct 25 2007, 07:14 AM, said:

This dilemma has been studied extensively in the context of non-blinded randomized clinical trials with serial recruitment. The problem is this:

Suppose you can afford to recruit 100 test patients, and ideally 50 would be allocated to the treatment group and 50 to the control group, each having a 50/50 chance. Now if the first, say, 92 recruits accidentally are allocated as 50 in one group and 42 in the other, then you already know that all the 8 remaining will go to the 42 group. So you will be recruiting some under a 50/50 condition and others under a 100/0 condition, which causes some potential bias.

The theoretically optimal solution is to accept some compromise, i.e. the allocation will be somewhere between 50/50 and binomial, while the probabilities for the last recruit may be slightly off 50/50. (If feasible, one could also, for example, recruit the subjects as pairs and allocate each pair 1/1 randomly).

Your case is simpler because it is given that there will be exactly one test. There must be an, in some sense, optimal solution which assigns decreasing probabilities to the five days.

Sometimes I wonder if you are serious, or just messing with us.

You do this as a double blinded assignment. You have 50 "chips" that are for the study, and 50 that are for the placebo. You let the participants (essentially) draw out a chip. The person allowing the "drawing" doesn't know the result, the chip is taken the pharmacy, where the medicine or the placebo is given. After 100 people have choosen, the pharmacy knows which is which, but not the clinician.

That isn't exactly the right way, but you never let the person analyzing the effectiveness of the treatment "assign" the people to the groups. If you do, there is a chance of bias in the assignment.

As for the exam issue, everyone has to show up on monday, prepared, because that might be the day. Etc, etc...so put the exam off to Friday to have a full class of prepared students every day of the week. What's the problem.

[helene_t]

The approximate solution is (probabilities for each day):
monday: 0.2582701
tuesday: 0.2259959
wednesday: 0.1934004
thursday: 0.1611666
friday: 0.1611670

csabaproblem.negsurprise = function (cond.p) { 
    p = c(cond.p,1)*c(1,cumprod(1-cond.p))
    return(sum(p*(c(cond.p,1))))
}
cond.p.opt=nlminb(rep(0.25,4),csabaproblem.negsurprise,lower=rep(0.00001,4),upper=rep(0.99999,4))$par
p.opt = c(cond.p.opt,1)*c(1,cumprod(1-cond.p.opt))

[matmat]

the test will be on tuesday? why? because bad things always happen on tuesdays...

[bid_em_up]

helene_t, on Oct 25 2007, 10:19 AM, said:

The approximate solution is (probabilities for each day):
monday: 0.2582701
tuesday: 0.2259959
wednesday: 0.1934004
thursday: 0.1611666
friday: 0.1611670

csabaproblem.negsurprise = function (cond.p) { 
    p = c(cond.p,1)*c(1,cumprod(1-cond.p))
    return(sum(p*(c(cond.p,1))))
}
cond.p.opt=nlminb(rep(0.25,4),csabaproblem.negsurprise,lower=rep(0.00001,4),upper=rep(0.99999,4))$par
p.opt = c(cond.p.opt,1)*c(1,cumprod(1-cond.p.opt))

Ok, now I am certain you are just messing with us.

[gwnn]

So you mean the prof. said something silly and that it's nonsense to say "nobody will know" ?

[helene_t]

I should probably elaborate:

cond.p in my code are the probabilities of the test taking place on the individual days, from the viewpoint of the evening before. For example, cond.p[friday] must be 1.

The negsurprise is cond.p for the day when the test actually takes place. For example if cond.p[wednesday] is 0.4 and the test takes place on wednesday, the negsurprise is 0.4.

My script minimizes the expected value of the negsurprise, i.e. sum(p*cond.p).

[Echognome]

I've seen this paradox before. I'm sure it's discussed plenty on wiki.

I much prefer this simpler one. The statement "I am lying" can only be true if it's false.

[jdonn]

Echognome, on Oct 25 2007, 04:55 PM, said:

I've seen this paradox before. I'm sure it's discussed plenty on wiki.

I much prefer this simpler one. The statement "I am lying" can only be true if it's false.

Is that really a paradox? It is simply an untrue statement. "I am an elephant" is not a paradox.

[Echognome]

jdonn, on Oct 25 2007, 02:37 PM, said:

Echognome, on Oct 25 2007, 04:55 PM, said:

I've seen this paradox before.  I'm sure it's discussed plenty on wiki.

I much prefer this simpler one.  The statement "I am lying" can only be true if it's false.

Is that really a paradox? It is simply an untrue statement. "I am an elephant" is not a paradox.

It's not simply an untrue statement.

It can only be false if it's true.

The same is not true for your statement.

[jdonn]

Echognome, on Oct 25 2007, 06:02 PM, said:

It's not simply an untrue statement.

It can only be false if it's true.

The same is not true for your statement.

Oops duh. We have heard this one before, but that's what I get for answering at work.

[helene_t]

Strictly speaking, there will always be a chance that the students will know on the night before. What about the solution 0.249 - 0.249 - 0.249 - 0.249 - 0.004? The problem with that one is not that there's a 0.004 chance that the students will know that the test takes place on Friday. The problem is that there's a 0.253 chance that they will know with 0.984 confidence that it takes place on Thursday.

One has to forget about the dichotomous certain/uncertain distinction, and treat information as a quantitative thing.

[luke warm]

helene_t, on Oct 26 2007, 03:56 AM, said:

It's only a paradox because people look for a deterministic solution to an obvious mixed-strategy problem. Kinda like "which symbol is optimal in the stone-paper-scissors game".

Strictly speaking, there will always be a chance that the students will know on the night before. What about the solution 0.249 - 0.249 - 0.249 - 0.249 - 0.004? The problem with that one is not that there's a 0.004 chance that the students will know that the test takes place on Friday. The problem is that there's a 0.253 chance that they will know with 0.984 confidence that it takes place on Thursday.

One has to forget about the dichotomous certain/uncertain distinction, and treat information as a quantitative thing.

i love it when you talk like that

[helene_t]

Echognome, on Oct 25 2007, 11:55 PM, said:

I've seen this paradox before. I'm sure it's discussed plenty on wiki.

Not really.
http://en.wikipedia....hanging_paradox
says that the article is in need of expert attention.

[hrothgar]

helene_t, on Oct 25 2007, 06:19 PM, said:

The approximate solution is (probabilities for each day):
monday: 0.2582701
tuesday: 0.2259959
wednesday: 0.1934004
thursday: 0.1611666
friday: 0.1611670

csabaproblem.negsurprise = function (cond.p) { 
    p = c(cond.p,1)*c(1,cumprod(1-cond.p))
    return(sum(p*(c(cond.p,1))))
}
cond.p.opt=nlminb(rep(0.25,4),csabaproblem.negsurprise,lower=rep(0.00001,4),upper=rep(0.99999,4))$par
p.opt = c(cond.p.opt,1)*c(1,cumprod(1-cond.p.opt))

I want to make sure that I understand that nature of your "solution":

You are solving for the probability density function that minimizes the chance that a student can accurately guess which day the test will occur.