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Interesting puzzle involving prediction/preemption

  1. Oct 15, 2005 #1
    I came across this puzzle a while ago and recently recalled it but couldn't come up with any solid solution/explanation. What are your thoughts?

    I want to choose one day, in a five day window (we'll say monday to friday), on which an event will occur. (For simplicity of interpretation and wording we'll say the event will occur within the first hour of the day.) People know the event will occur within the five day window. I don't want anyone to know on the day befor that the event will occur the next day.

    It would clearly seem, then, that if ,on Thursday, the event hasn't occured people will know that the event will occur on Friday and therefor I shouldn't choose Friday. But if the people know I will not choose Friday, the last day they think I might choose is Thursday. Because of this, if on Wednesday the event hasn't occured, people will know that the event will occur on Thursday and therefor I shouldn't choose Thursday.

    It would seem this would continue until I decide that I simply can't choose any day on which the event will occur AND no one will know the day befor. What, if anything, is wrong?
  2. jcsd
  3. Oct 15, 2005 #2
    This puzzle sometimes goes under the name of the unexpected hanging. My take on it is that the event can occur on any day but Friday because of the following reasoning:

    There are two kinds of people. Those who are able to work out the implications of the logic, and those who are not. Those who are not will be unaware out of their own shortcoming and those who are will be unaware because logic told them it couldn't happen. The exception is Friday when everyone, even the dullest can figure out that the day for the event has come.
  4. Oct 15, 2005 #3
    Try looking at if from the perspective of the people you are trying to fool.

    If we assume they are logical, and they use the same logic as yourself, then which day do you think they predict that you will choose?

  5. Oct 15, 2005 #4
    I thought of another context for the problem that's a little more simple but no less interesting. Lets say I have 3 opaque jars lined up, under one of which I will hide an object. A person comes along who will check each jar starting with the one on the left progressing to the right. I don't ever want the person to be able to predict (meaning to be able to deduce logically) which jar it will be under befor he/she actually checks.

    Well obviously I don't want to choose the last jar, leaving me two jars to choose from, but the person knows this as well. Thus I shouldn't choose the second jar because the person will know after checking the first jar that the object is under the second. This leaves me with what would seem like only one choice, the first jar... but...

    After thinking about it what is clear to me is that there can be no one jar I "should" choose because by definition the person will know it will be that jar (he/she can use any logic I would have used in determining the jar I "should" pick)
  6. Oct 15, 2005 #5
    Ok, from the perspective of the guesser.

    If by the time I get to the last jar I haven't seen the object I will know it is under the last jar. If the chooser's goal is to prevent me from knowing befor hand which jar it is under (which indeed is his goal) he won't pick that jar.

    Seems like this will lead to the same conclusion (or inconclusion).
  7. Oct 15, 2005 #6
    Then follow this through to the logical conclusion.

    The guesser will perhaps not guess the last jar, for the reason you have given.

    Continuing with the logic, is there any reason why he/she should favour the first jar over the second, or vice versa?

  8. Oct 15, 2005 #7
    The explanation is you've established an axiom or a logical postulate:
    You cannot pick a random number manually.
    Therefore you cannot manually select an unknowable random event for all cases, within a finite set of possibilities.
    However, there is a solution, at least for an unexpected hanging:
    Your providing a morning hanging to a prisoner, ordered by the judge within the next five days. But the judge is also requiring “when” be unexpected by the prisoner. If you cannot show this to be true to the Judge, it will be cause for him to improve the local budget by downsizing you out of your job (or worse).
    Can you devise a method to satisfy the requirements of the judge, and save your job?
  9. Oct 17, 2005 #8


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    Sure. Convince the prisoner that the judge must be confused or lying.
  10. Oct 17, 2005 #9
    ?? How would that satisfy the Judge to allow you to keep your job, He might just hang you instead!
    No, I was thinking something more like: Show the prisoner and all others (most important the Hang ‘em High Judge leering out the second floor window at your work) that you’ve taken yourself out of the decision making processes. By showing that the hanging will take place only with a six rolled on a die dropped on a small table on the hangman platform while every thing is ready to go. If a six is not rolled the noose is taken off, the prisoner returned to jail and you try again the next day. Now since the prisoner must be hung on a weekday, if no six shows up by Friday – the prisoner will have to go free.

    Now, here’s where if he were to be let go after five days of no sixes, it could still put our hanging judge in a foul mood as well. So, by a private flip of a coin you decide between Wednesday or Thursday to “accidentally” lean against the lever that makes the trapdoor release go CLICK! Whereupon you clearly proclaim to the startled prisoner “oops, didn’t expect that!” while the die is still rolling!
    A Clint Eastwood accent and a spit of Ta’bacie would be a nice touch.

    Plus, if the judge has any doubt the hanging was unexpected by the prisoner, - just tell him he can interview the prisoner.
    Last edited: Oct 18, 2005
  11. Oct 18, 2005 #10


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    I feel dumb now. Hang the prisoner on Monday, and on Tuesday.
  12. Nov 16, 2005 #11
    its not a puzzle its a parradox-if the event was to present a gift..
    1 soloution is to hav a box for everyday of the week and open it ,
    another is to open it on the transition from 1 day to the next.....srry a bit vague im in a rush
    Last edited: Nov 16, 2005
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