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  • #31
    Originally posted by Evan View Post
    Or you could just go 24,859.82 miles south.
    ????

    --- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
    --- Defend what you say with arguments, not by imposing your credentials ---

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    • #32
      Originally posted by TeeVee View Post
      1 in 147 as there is an equal chance that 146 other pax took his seat?
      It's easy but not that easy.

      The reason why that doesn't work is that passengers don't pick their seats at random except the first passenger or if they find their seats occupied.

      For example, if the first passenger picks any seat other than that of the second passenger, then the second passenger has a 0% chance to pick the last passenger's seat.

      --- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
      --- Defend what you say with arguments, not by imposing your credentials ---

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      • #33
        1 in 145. 1st pax took someone else's seat, so the assigned pax takes the 1st pax seat. 145 correct seats left?

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        • #34
          Originally posted by Gabriel View Post
          ????
          circumnavigate the globe to the south

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          • #35
            Originally posted by TeeVee View Post
            1 in 145. 1st pax took someone else's seat, so the assigned pax takes the 1st pax seat. 145 correct seats left?
            No. I don't understand your explanation very well, but for the sake of the argument, say that the first passenger picks the seat that of the 11th passenger. Then passengers #2 to #9 will find their assigned seats available and occupy them. Passenger 11 will find his seat occupied, and a total of 10 seats occupied and 137 empty seats to pick from. One of those empty seats will be the one original assigned to the first passenger, but he doesn't need to pick it. So "the assigned pax takes the 1st pax seat" is not necessarily true.

            --- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
            --- Defend what you say with arguments, not by imposing your credentials ---

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            • #36
              Hint: The number 147 is irrelevant. The answer will be the same regardless of the number of passengers and seats (as long as the flight is fully sold out). And the you don't need the number 147 (or whatever the total capacity is) for the explanation either.

              If you think a bit, this is a stronger hint than it might seem at first sight.

              --- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
              --- Defend what you say with arguments, not by imposing your credentials ---

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              • #37
                Originally posted by Gabriel View Post

                147 passengers are in a line to board a plane with 147 seats.
                Each passenger has a boarding pass with an assigned seat.
                The first passenger in the line is a first-time flier and he didn't notice that he has a seat assigned. So he gets in the plane and picks any seat at random.
                The rest of the passengers will take their own seat if available, or randomly pick any seat among the available ones if their own seat is occupied.

                What are the chances that the last passenger to board the plane sits in his assigned seat?
                To begin with, we have 1 in 147. Those are the odds that the first passenger will pick his correct assigned seat by sheer luck. If he does, the last passenger to board will also get his assigned seat. Then we have again 1 in 147. Those are the odds that the first passenger to board will pick the assigned seat of the passenger who ultimately picks the first passenger's originally assigned seat, in which case everyone else including the last passenger will get their assigned seat. (or is that 1 in 147 squared?) Then we have 0 in 147. Those are the odds that the last passenger will get his assigned seat if the first passenger had already taken it.

                There are two chances that he will sit in his assigned seat.

                The odds of me actually figuring this out are staggering.

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                • #38
                  Chances are "evens". He either does or does not sit in his assigned seat.
                  If it 'ain't broken........ Don't try to mend it !

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                  • #39
                    Originally posted by brianw999 View Post
                    Chances are "evens". He either does or does not sit in his assigned seat.
                    Right, the same than the chances to win the lottery: You either win it or you don't. (Not!)

                    --- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
                    --- Defend what you say with arguments, not by imposing your credentials ---

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                    • #40
                      Originally posted by Evan View Post
                      To begin with, we have 1 in 147. Those are the odds that the first passenger will pick his correct assigned seat by sheer luck. If he does, the last passenger to board will also get his assigned seat. Then we have again 1 in 147. Those are the odds that the first passenger to board will pick the assigned seat of the passenger who ultimately picks the first passenger's originally assigned seat, in which case everyone else including the last passenger will get their assigned seat. (or is that 1 in 147 squared?) Then we have 0 in 147. Those are the odds that the last passenger will get his assigned seat if the first passenger had already taken it.

                      There are two chances that he will sit in his assigned seat.

                      The odds of me actually figuring this out are staggering.
                      I am sorry. Are you giving an answer? If so, I don't get it. Two chances out of how many? I was expecting an answer of the type "1 in 147", "fifty-fifty", "13%", etc.

                      Your reasoning, on the other hand, is definitively on the good path.

                      --- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
                      --- Defend what you say with arguments, not by imposing your credentials ---

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                      • #41
                        Ok, an additional hint:

                        The "traditional" statistical way would be to make a probability tree where all the possible decisions are considered (in each decision a branch of the tree is divided in as many branches as options for that decision).

                        But a probability tree with this example is unmanageable:
                        The first pax has 147 options, so you have 147 branches. For 146 of these branches, the second pax can only decide to pick his assigned seat, but for the remaining branch, the second passenger can pick from 146 options that are 146 new branches, and so on.

                        But...

                        As I've said before that the answer is the same with any number of passengers=seats. I didn't say it, but in fact that'c correct for any number equal to or greater than 2, because if there is only one pax and one seat, the pax will seat in the only seat and the whole idea of assigning seats becomes meaningless. That means that you'll find the right answer trying with any number of passengers=seats.

                        So why not to try with easy numbers like 2, 3 and 4 pax and see what happens?
                        If my hint is true, the answers will be the same, and while this doesn't give you the explanation of why this answer will be valid for any number (including 147), seeing how it works can help.

                        If needed, I have one more hint to give (before running out of them).

                        --- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
                        --- Defend what you say with arguments, not by imposing your credentials ---

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                        • #42
                          Originally posted by Gabriel View Post
                          I am sorry. Are you giving an answer? If so, I don't get it. Two chances out of how many? I was expecting an answer of the type "1 in 147", "fifty-fifty", "13%", etc.

                          Your reasoning, on the other hand, is definitively on the good path.
                          I see only two possible scenarios where the last person gets his assigned seat: either the first person chooses his correct assigned seat (1 in 147) or the first person chooses the seat of the person who later chooses that seat (1 in 147). On the other hand I see only one scenario in which the last person will not get his assigned seat: the first person does not choose his correct assigned seat AND the first person does not choose the seat of the person who will later choose that seat (but the odds of this happening are 145 in 147).

                          So two chances out of three he gets his assigned seat, but with the odds stacked heavily against him.

                          I'm sorry if this is all over your head Gabe

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                          • #43
                            Originally posted by Evan View Post
                            I see only two possible scenarios where the last person gets his assigned seat: either the first person chooses his correct assigned seat (1 in 147) or the first person chooses the seat of the person who later chooses that seat (1 in 147). On the other hand I see only one scenario in which the last person will not get his assigned seat: the first person does not choose his correct assigned seat AND the first person does not choose the seat of the person who will later choose that seat (but the odds of this happening are 145 in 147).

                            So two chances out of three he gets his assigned seat, but with the odds stacked heavily against him.

                            I'm sorry if this is all over your head Gabe
                            And what if the first pax picks the seat of the person who later chooses the last passenger's seat?

                            Or what if he picks the seat of a pax that later picks the seat of the pax than later picks the seat of the pax that later....?

                            For example, first pax picks the seat of the 7th pax, so pax 2 to 6 take their assigned seats, then pax 7 finds his seat occupied so he picks the seat of pax 12, so pax 8 to 11 take their assigned seats, then pax 11 finds his seat occupied and picks the seat of pax.... You get the idea.

                            --- Judge what is said by the merits of what is said, not by the credentials of who said it. ---
                            --- Defend what you say with arguments, not by imposing your credentials ---

                            Comment


                            • #44
                              Originally posted by Gabriel View Post
                              And what if the first pax picks the seat of the person who later chooses the last passenger's seat?

                              Or what if he picks the seat of a pax that later picks the seat of the pax than later picks the seat of the pax that later....?
                              Sure, but what are the odds of that happening...

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                              • #45
                                Does the actual airline matter?

                                When this is over, please let us know which one. I'll never book with them if the cabin crew allows such a cluster fudge happen on every flight.
                                Live, from a grassy knoll somewhere near you.

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