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  • #46
    Ok, I guess if the amount of pax doesn't matter, then you are looking for a percentage...

    so I see two possible success scenarios out of 148 basic scenarios.

    1.3513513513513513...etcetera% ??

    Com'on Gabriel, you are dealing with a bunch of dunderheads and one stick and rudder genius. What's the answer?!

    or do I have to get out my #2 pencil?

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    • #47
      Final hint:

      Think:
      - What event (let's call it A) will make you know for sure that the last passenger will take his own seat?, and what (let's call it B) will make you know for sure that he will not take his seat?

      That is the most important part. Once you figured this out, it's almost done.

      After that, think:
      - What are the occasions along the boarding process where A or B may happen?
      - What are the chances that one or the other happens when the occasions arises? (think comparatively: how the chances for A compare to those for B).

      Again, now that you know that the total number is irrelevant, trying with an easy number will give the right answer and help with the explanation.

      --- 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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      • #48
        Originally posted by guamainiac View Post
        Does the actual airline matter?...
        Yes, it has a large effect on the livery.
        Les règles de l'aviation de base découragent de longues périodes de dur tirer vers le haut.

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        • #49
          Originally posted by Gabriel View Post
          - What event (let's call it A) will make you know for sure that the last passenger will take his own seat?, and what (let's call it B) will make you know for sure that he will not take his seat?
          A: The first passenger has taken his own seat by chance.

          but also

          A2: If the first passenger sits in the wrong seat and the second passenger, who had that seat, sits in the first passenger's seat, then we also know for sure the last passenger will get his seat.

          B: The first passenger—or any passenger before the last one—has taken the last one's seat.

          A=1/147
          A2=1/146
          B=146/147
          (Ignoring of course the statistical probability that random seats will be chosen from front to back, so if the last passenger is seated in the last row his chances are much greater than if he is seated in the front rows)

          C=This better be good Gabriel cuz I'm fresh out of ideas.

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          • #50
            All I know is that if the first passenger sits in my favourite window seat then he's gonna get one chance to avoid a smack in the teeth. !!!
            If it 'ain't broken........ Don't try to mend it !

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            • #51
              Originally posted by Gabriel View Post
              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.
              herein lies the problem: assuming the 1st pax takes the wrong seat. it is possible that this causes a cascade effect and every other pax is forced to pick a seat they weren't assigned.

              i'm gonna go with the answer that this is not solvable without using some sort of permutation, which i've forgotten many years ago. of course, i'll probably be wrong and the answer will make me wince.

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              • #52
                Originally posted by TeeVee View Post
                herein lies the problem: assuming the 1st pax takes the wrong seat. it is possible that this causes a cascade effect and every other pax is forced to pick a seat they weren't assigned.
                It's worst than that, because it's also possible that the above doesn't happen. For example if the passenger picks his seat (just by luck), everybody will sit in their seat. If the first one picks the last passenger's seat, everybody but the first and last passenger will sit in their seats (who will have their seat "swapped"). If the first passenger picks the seat of pax #50 (for the sake of the argument), all passengers between #2 and #49 will sit in their seats. Pax 50 will find his seat occupied and we start all over again with the sames alternatives than for the first passenger.

                i'm gonna go with the answer that this is not solvable without using some sort of permutation, which i've forgotten many years ago. of course, i'll probably be wrong and the answer will make me wince.
                "Everybody picks a seat at random. What are the chances that the last passenger sits in his assigned seat?"

                That (which is not the case here) would be a permutation.
                The only way for the last passenger to sit in his seat would be that everybody before him MUST pick a seat that is not the last passenger's seat. What are the chances of that?

                To keep the hope alive, the first passenger must pick, out of 147 seats available, any of the 146 seats that are not the last passenger's seat. The chances of that is 146/157. Then, the next passenger would need to pick one of the 145 "not last passenger's" seat out of the 146 that remain available, That's a chance of 145/146. And so on. So the combined chance is:
                146/147 * 145/146 * 144/ 145 * .... * 3/4 *2/3 * 1/2 = (n-1)!/n!

                If you look closely you'll note that all the numbers in the numerator and in the denominator cancel each other, except the 1 in the numerator and the 147 in the denominator. So the result of that is 1/147, which is what TeeVee had said before.

                But in this case, the passengers don't take seats at random EXCEPT if they are the first passenger or one that their seats are occupied.

                If you want to solve it in a typical statistical way (using a probability tree, combinations, permutations, etc) it's much more difficult than the above because, as explained above, you need first to find all the possible combinations of which passengers will be picking their seats at random and which not, what will depend of the individual choices of the first passenger and other passenger may find his seat occupied.

                For example, look what happens with just 3 passengers.
                Let's call the passengers 1, 2, 3 and their assigned seats A, B, C respectively. Something like "1B (1/3)" means "there is a 1 in 3 chance that passenger 1 sits in seat B".

                The probability tree looks like this:

                1A (1/3) (pax 1 randomly picks seat A)
                2B (1/1) (given the above, there is a 100% chance that passenger will pick seat B because it's his assigned seat and it is available)
                3C (1/1) - TOTAL of this branch: 1/3*1*1=1/3
                1B (1/3)
                2A (1/2) (given the above, pax 2 find his seat B occupied and will randomly pick between A and C)
                3C (1/1) - TOTAL of this branch: 1/3*1/2*1=1/6
                2C (1/2)
                3A (1/1) - TOTAL of this branch: 1/3*1/2*1=1/6
                1C (1/3)
                2B (1/1)
                3A (1/1) - TOTAL of this branch: 1/3*1*1=1/3

                The bold and underlined branches are where "the last passenger sits in his assigned seat". There are two branches (possible cause-effect sequences) with this outcome, one with a chance of 1/3 and the other with a chance of 1/6. So the chances that either one of these branches happen is 1/3 + 1/6 = (2+3)/6 = 3/6 = 1/2 = 50%.

                That's the answer to the question: The chances that last passenger sits in his assigned seat is 50%. (or "even", as Brian said before but for the wrong reason).

                Now, this answer is correct for 3 passengers, and only because I said it as a hint ("147 is irrelevant, the result is independent of the number of pax") you know that it will be true for any number, including 147. But how can we prove it?

                Imagine doing the above cause-effect tree with 147 passengers to start with and all the possible ways that they can combine. Let alone doing it with a generic number "N".

                However, a little of common sense and imagination let's one walk around all that statistical nightmare and find a very very easy layman-level solution. What is what I'll post later if nobody does it first. Not that I'm playing mysterious here, but I have things to do now and this post already took a lot of time.

                --- 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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                • #53
                  The answer is 1 in 147....

                  But here's the logic...If the first dude sits in the wrong seat, then however the heck it works out the last guy will be SOL....
                  In the unlikely event that passengers 2 through 146 all get to their assigned seats, the last passenger is SOL because the first dude took his seat.

                  Any other permutation of this event eventually cascades into the last guy being SOL....(more likely it quickly cascades into all sorts of folks picking random seats- but the odds of the last guys seat being open will always hit zero right before he gets there (if not sooner)...always (and I don't often say always...I guess a meteor could hit the plane first...so almost always).

                  (it doesn't matter if some folks in the middle do get to thier seats, the luck always runs out for the last guy sooner OR later...all the bizillion permutations in the middle 'cancel out to zero' for the last dude on)

                  I think......
                  Les règles de l'aviation de base découragent de longues périodes de dur tirer vers le haut.

                  Comment


                  • #54
                    Originally posted by 3WE View Post
                    The answer is 1 in 147....
                    No. I've already revealed the right answer is that the chances are 1 in 2, 1:2, 1/2, 50%, 0.5, "fifty-fifty", even.

                    The missing part is why.
                    But here's the logic...If the first dude sits in the wrong seat, then however the heck it works out the last guy will be SOL....
                    In the unlikely event that passengers 2 through 146 all get to their assigned seats, the last passenger is SOL because the first dude took his seat.
                    First passenger randomly picks the seat assigned to the 4th passenger.
                    Passenger 2 finds his assigned seat available and sits there.
                    Passenger 3 finds his assigned seat available and sits there.
                    Passenger 4 finds his assigned seat occupied by passenger 1 so, among all the seats that remain available, he randomly picks one that, by sheer luck, happens to be seat originally assigned to passenger 1.
                    Passengers 5 to 147 (the last one) all find their assigned seats available so they occupy their seats.
                    This is one out of many sequences where the first passenger picks other than his own seat and the last passenger is not SOL.

                    Remember, passengers will only pick a random seat if they find they assigned seat occupied (except the first one who will pick one at random no matter what). If not, they'll sit in their assigned seats.
                    If the first passenger randomly picks the seat assigned to pax #146, all pax from 2 to 145 will find their assigned seats available and occupy them. Pax 146 will find his seat occupied and 2 empty seats to pick from: The ones originally assigned to #1 and to #147. So again he has 50% chances to make a SOL out of the last passenger.

                    --- 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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                    • #55
                      Ha!

                      I was close... On the reasoning...
                      Les règles de l'aviation de base découragent de longues périodes de dur tirer vers le haut.

                      Comment


                      • #56
                        Wait... Should it not be 0.5 + 1/147???
                        Les règles de l'aviation de base découragent de longues périodes de dur tirer vers le haut.

                        Comment


                        • #57
                          Solution:

                          I'll use my hints as guidance:

                          a- What event (let's call it A) will make you know for sure that the last passenger will take his own seat?, and what (let's call it B) will make you know for sure that he will not take his seat?

                          That is the most important part. Once you figured this out, it's almost done.

                          After that, think:
                          b- What are the occasions along the boarding process where A or B may happen?
                          c- What are the chances that one or the other happens when the occasions arises? (think comparatively: how the chances for A compare to those for B).
                          a-
                          B: The moment that a pax sits in the seat assigned to the last passenger I know that the last passenger will not take his assigned seat.
                          A: The moment that a pax sits in the seat originally assigned to the first passenger, I know that the last passenger will take his assigned seat (because all the passengers after him will find their seats available).

                          The answer to the problem is that there is a 50% chance that the last pax will occupy his assigned seat, because each time there is an opportunity for A or B to happen, A and B have equal chances.

                          (And I didn't use the number 147 or any complex statistics, did I?)

                          If not convinced yet, then we must move to b- and c-.

                          b- The only occasions where A or B can happen are when the first passenger boards, or when another passenger whose seat is already occupied boards.

                          c- In each of these occasions, the chances that the concerned pax occupies the first pax's seat or the last pax's seat are equal between them. That doesn't mean that they are fifty-fifty, but they are equal.

                          For example, when the first passenger boards, he can randomly pick his own seat or, with equal chance, the last passenger's seat. If he does one of them, the situation is finished. But there is a good chance that, instead, he will pick a seat assigned to another pax (neither the last pax nor himself).
                          In this case, the situation is not finished. The following paxs will sit in their assigned seats until we reach to the one whose seat is occupied by the first pax. Now this pax can again randomly pick the first pax's seat or, with equal chance, the last pax's seat. If he does one of them, the situation is finished. Of course, again, he can pick another empty seat that is neither the first pax's or the last pax's. And so we go.

                          Eventually, someone will HAVE to pick either the first pax's seat or the last pax's eat. The last time this can happen is to the second-to-last pax. If he boards the plane and nobody picked the first or last pax's seat yet, then he will find his seat occupied and only these two seats available. So he will have to pick one of them and, again, with equal chance among them.

                          Since there were no winners, the thread is open to the first one to post a quiz.

                          --- 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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                          • #58
                            Originally posted by Gabriel View Post
                            For example, when the first passenger boards, he can randomly pick his own seat or, with equal chance, the last passenger's seat.
                            Can I say 1/147 + 0.5 - 1/147????

                            Conceptually...

                            By the way- this still isn't AVIATION...You could just as well ask this about a Greyhound Bus or train, or sports stadium....
                            Les règles de l'aviation de base découragent de longues périodes de dur tirer vers le haut.

                            Comment


                            • #59
                              Originally posted by 3WE View Post
                              Can I say 1/147 + 0.5 - 1/147????

                              Conceptually...
                              I understand the above mathematically, but not conceptually.

                              By the way- this still isn't AVIATION...You could just as well ask this about a Greyhound Bus or train, or sports stadium....
                              No kidding!?
                              Originally posted by Gabe in the opening post
                              The idea is to present brain teasers. The rule is that they must be related to aviation somehow, even if modified from its original version to replace a train for a plane, for example.

                              --- 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


                              • #60
                                Pax 1 (1 / 147)
                                Pax 2 (1 / 147) * (145 / 146)
                                Pax 3 (1 / 147) * (145 / 146) * (144 / 145)
                                Pax 4 (1 / 147) * (145 / 146) * (144 / 145) * (143 / 144)
                                Pax 5 (1 / 147) * (145 / 146) * (144 / 145) * (143 / 144) * (142 / 143)

                                If you repeat this 147 times and add up the figures you end up with 0,50.


                                Bit more explanation.


                                The odds that passengers one takes your seat are 1 in 147.

                                The odds that passenger two does NOT take your seat are 100% if passenger one has the correct seat. However, if passenger one has the wrong seat then passenger two will pick one seat from the remaining 146 seats. That means that the odds of him NOT taking your seat are 145 to 146 IF passenger one has the wrong seat.

                                So the odds of passenger two NOT taking your seat are (1/147) * (145/146)

                                Repeat this for the amount of passengers and your done. It does not matter how many passengers there are, the end result will always add up to 50%.
                                Please visit my website! http://www.schipholspotter.com/

                                Don't make me use uppercase...

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