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People are waiting in line to board a 100-seat airplane. Steve is the
first person in the line. He gets on the plane but suddenly can’t
remember what his seat number is, so he picks a seat at random. After
that, each person who gets on the plane sits in their assigned seat if
it’s available, otherwise they will choose an open seat at random to sit
in.
The flight is full and you are last in line. What is the probability that you get to sit in your assigned seat?
It's 1/2. Shocked? Read how!
What was the puzzle?
The probability is really 1/2.
Let's assume that my name is Steve who was the first person who gets on the plane. And you are the person who is last in line. Also, let Seat No.1 was my assigned seat & Seat No.100 is your assigned seat.
Now if I choose seat no.1 itself, rest of all will get their assigned seat including you.
But what if I choose random seat say seat no.15? All the passengers from 2 to 14 will get their assigned seats.
Now, passenger no.15 can choose seat assigned to me i.e. seat no.1 or seat no.100 i.e. your seat. If he sits on my seat i.e. seat no.1 then rest of all including you will get the respective assigned seats. And if he chooses your seat i.e. seat no.100 then all passengers from 16-99 will get their respective assigned seats but you will be left with only option of seat left vacant by me i.e. seat no.1.
The passenger no.15 can choose any random seat as well. In that case, he leaves your 'result' to the passenger whose assigned seat he chooses. Suppose he chooses seat no.76. Again all passengers from 16 to 75 will be getting their respective seats. Now, passenger can choose seat no.1 or seat no.100 or any seat from 77-99.
Again, if he chooses my seat no.1 then rest of all i.e. 77 to 100 which includes you will be getting assigned seats. And if he selects your seat no.100 then again 77-99 will be getting their respective seats but my seat no.1 will be empty for you. However, he can leave your 'result' to any body from 77-99 whose seat he is going to occupy.
In short, your 'result' is in hand of the passenger whose assigned seat is already occupied. He can choose my seat to give you your assigned seat or he can select your seat leaving my seat for you as rest of all passenger will be getting their respective assigned seats. And he can leave your 'result' in hand of other passenger whose seat he randomly occupies.
Suppose, the passenger no.76 above selects seat no.99 in random. Now, all from 77-98 will have no problem in getting assigned seats but passenger 99 will have only 2 options either seat no.1 or seat no.99.
In the end, whenever the passenger ahead of you gets on the plane then either he will get his assigned seat if somebody occupied my seat hence you will get your assigned seat. And if my seat is vacant, then probability that he chooses my seat or your seat is 1/2.
So you will find either my seat or your seat empty when you get on the plane. That is it could be assigned seat or not assigned seat. The probability that the seat you have chosen is assigned must be 1/2.
Show that the numbers from 1 to 15 can’t be divided into a group A of 13
numbers and a group B of 2 numbers so that the sum of the numbers in A
equals the product of the numbers in B.
Here is the proof!
What was that fact?
For a moment, let's assume that such group of 2 numbers exists whose product is equal to sum of rest 13 numbers taken out of 15 numbers.
Let x and y be those numbers in group B. Now x and y can be any number from 1 to 15.
As per condition,
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 - x - y = xy
120 = xy + x + y
Adding 1 to both sides,
121 = xy + x + y + 1
121 = x( y + 1 ) + 1( y + 1 )
121 = ( x + 1 ) ( y + 1 )
Since x & y are the numbers in between 1 to 15, possible value of x & y satisfying the above equation is 10. But x & y are must be 2 different number. Hence, our assumption goes wrong here!
So, the numbers from 1 to 15 can’t be divided into a group A of 13 numbers
and a group B of 2 numbers so that the sum of the numbers in A equals
the product of the numbers in B.
