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The New Coffee Room

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  3. Puzzle time- Numbers on foreheads

Puzzle time- Numbers on foreheads

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  • HoraceH Offline
    HoraceH Offline
    Horace
    wrote on last edited by Horace
    #7

    In the case of a billion prisoners, you would guess the mode, similar to jon's intuition. If from your perspective there were two equally common numbers, you could choose the lowest, or second lowest, etc, based on an agreed upon assignment beforehand. Such as, you could be given a number between zero and (number of prisoners) which would, when modded by the number of equal modes, tell you which mode to choose. That should be guaranteed to work in most cases. More than the 75% chance that random guessing will work. Essentially you are betting on the existence of collisions.

    Education is extremely important.

    1 Reply Last reply
    • jon-nycJ jon-nyc

      My intuition is that choosing the number closest to the mean of those you see is the best solution but I can't back it up. Maybe tomorrow's hint will help.

      KlausK Offline
      KlausK Offline
      Klaus
      wrote on last edited by
      #8

      @jon-nyc said in Puzzle time- Numbers on foreheads:

      My intuition is that choosing the number closest to the mean of those you see is the best solution but I can't back it up. Maybe tomorrow's hint will help.

      I don't quite get the logic behind that.

      Why should the median of the other numbers influence the distribution of your number?

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      • jon-nycJ Online
        jon-nycJ Online
        jon-nyc
        wrote on last edited by jon-nyc
        #9

        The idea is that the median is only slightly affected by your number, less so if your number is close to it.

        Imagine if there were 1,000 people doing that. Clearly the differences in median that each would calculate would be a tiny rounding error, thus they would all converge on the same number, including at least one person who had that number.

        Only non-witches get due process.

        • Cotton Mather, Salem Massachusetts, 1692
        KlausK 1 Reply Last reply
        • jon-nycJ Online
          jon-nycJ Online
          jon-nyc
          wrote on last edited by
          #10

          You can also improve over the random probability by doing something with modes like Horace suggested. Even with just 10.

          I still haven't been bothered to work out probabilities. I'm hoping the hints reveal a more clever intuitive approach.

          Only non-witches get due process.

          • Cotton Mather, Salem Massachusetts, 1692
          1 Reply Last reply
          • HoraceH Offline
            HoraceH Offline
            Horace
            wrote on last edited by
            #11

            A lot would hinge on whether the median was a single-occupancy number or multi-occupancy. In the case of a single-occupancy median, that strategy is a guaranteed loss. It's a guaranteed win for multi-occupancy median when it's still the median after removing one of them. It begins to be similar to the mode idea from that perspective.

            Education is extremely important.

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            • KlausK Offline
              KlausK Offline
              Klaus
              wrote on last edited by Klaus
              #12

              I think I got a solution that works with probability 1. Good old pigeonhole principle.

              click to show

              The prisoners number themselves 0 to 9.

              Let s the the sum of all numbers on all heads and let d be its rightmost digit.

              Now consider prisoner number p. He computes the sum t of all numbers he sees on the other's heads.
              He guesses that d = p. Hence he guesses (p - t) modulo 10 as his number.

              This will yield a correct guess for prisoner d.

              Proof. Consider prisoner d.

              His guess will be (d-t) modulo 10, which is the same as s - t, which is the number on his forehead.

              1 Reply Last reply
              • jon-nycJ Online
                jon-nycJ Online
                jon-nyc
                wrote on last edited by
                #13

                That's clever.

                Only non-witches get due process.

                • Cotton Mather, Salem Massachusetts, 1692
                1 Reply Last reply
                • jon-nycJ jon-nyc

                  The idea is that the median is only slightly affected by your number, less so if your number is close to it.

                  Imagine if there were 1,000 people doing that. Clearly the differences in median that each would calculate would be a tiny rounding error, thus they would all converge on the same number, including at least one person who had that number.

                  KlausK Offline
                  KlausK Offline
                  Klaus
                  wrote on last edited by
                  #14

                  @jon-nyc said in Puzzle time- Numbers on foreheads:

                  The idea is that the median is only slightly affected by your number, less so if your number is close to it.

                  Imagine if there were 1,000 people doing that. Clearly the differences in median that each would calculate would be a tiny rounding error, thus they would all converge on the same number, including at least one person who had that number.

                  I don't think so.

                  Let's consider a uniform distribution of the numbers. Then the median of the 1000 would very likely be 4 or 5. But that number is not more likely to show up than any other number.

                  1 Reply Last reply
                  • jon-nycJ Online
                    jon-nycJ Online
                    jon-nyc
                    wrote on last edited by
                    #15

                    You choose the visible number closest to the median.

                    Only non-witches get due process.

                    • Cotton Mather, Salem Massachusetts, 1692
                    1 Reply Last reply
                    • jon-nycJ Online
                      jon-nycJ Online
                      jon-nyc
                      wrote on last edited by jon-nyc
                      #16

                      I think your solution isn't right. They each (potentially) have a different value of t.

                      Only non-witches get due process.

                      • Cotton Mather, Salem Massachusetts, 1692
                      KlausK 1 Reply Last reply
                      • jon-nycJ jon-nyc

                        I think your solution isn't right. They each (potentially) have a different value of t.

                        KlausK Offline
                        KlausK Offline
                        Klaus
                        wrote on last edited by
                        #17

                        @jon-nyc said in Puzzle time- Numbers on foreheads:

                        I think your solution isn't right. They each (potentially) have a different value of t.

                        I made a small correction and elaboration on why this is correct.

                        1 Reply Last reply
                        • jon-nycJ jon-nyc

                          Each of 10 prisoners will have a digit between 0 and 9 painted on his forehead (they could be all 2's and 3's, for example).  At the appointed time, each prisoner will be exposed to all the others, then taken aside and asked to guess his own digit.

                          All the prisoners will be freed, provided at least one prisoner guesses correctly.

                          The prisoners have an opportunity to conspire beforehand; how can they maximize their probability of success?

                          AxtremusA Offline
                          AxtremusA Offline
                          Axtremus
                          wrote on last edited by
                          #18

                          @jon-nyc said in Puzzle time- Numbers on foreheads:

                          The prisoners have an opportunity to conspire beforehand; how can they maximize their probability of success?

                          If prisoners get to choose their own number, then in the “conspire” phase all prisoners agree to one number (e.g., every prisoner chooses to paint himself with the number 0), then every one knows that his own number is 0.

                          1 Reply Last reply
                          • taiwan_girlT Offline
                            taiwan_girlT Offline
                            taiwan_girl
                            wrote on last edited by
                            #19

                            Not sure if I understand, but to me, there is no way to make a guess any more accurate. If it is a true random number on the forhead, then there is a 10% chance of guessing correctly.

                            It doesnt matter if every other number is "8", your number could still be anything from 0-9. etc

                            What am i missing?

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