Random Airplane Seats

I offer my explanation, since none of the answers below seems to explain the solution.

The probabilty is indeed 1/2. There are two things to realize:

1. The probabilty that Steve chooses his assigned seat is equal to the probability that he chooses your assigned seat.

2. In case that Steve would choose neither his own seat nor yours, then there are two alternatives: if somebody else would choose Steve’s seat at random, then you would get your assigned seat; otherwise you would be left with the Steve’s seat.

With that being said, we can go on to find out the probability. With every person choosing a seat at random (including Steve), there are there possible outcomes:

1. either he chooses your assigned seat, or
2. chooses the Steve’s seat, or
3. chooses someone else’s seat.

Notice, that the probabilty of choosing Steve’s seat is always equal to probabilty of taking your seat. That means that the probabilty of you getting your seat vs. not, is even. The case of a passanger choosing someone else’s seat doesn’t affect your final outcome in either way, it just passes that three possible alternatives to the next passenger.

Since the probability of someone taking your place is always equal to the probabilty of someone taking Steve’s place (and this also applies to the penultimate passenger with only two seats left), the probabilty of you getting your assigned seat is in the end 50%.

(1+2+4……………………………………………2^97 +1)/(2+4+8+16+32+64………………….2^98 +2)
it is  1/2 see clearly
suppose I named  100th person as 100 and so on
suppose steve sit on 99th person seat then there are 2 case 1 favorable
now if he sit on 98 then if 98 sit on 99 then similar case as before so 2 case 1 favorable if 98 sit on 100 1 case 0 favorable if 98 sit on 1 then 1 case 1 favourable so total 4 case 2 favourable
if he sit on 97 then if 97 sit on 98 then similar to previous 4case 2 favourable if 97 sit on 99 then similar to first case when steve sit on 99 2 case 1 favourable if 97 sit on 1, 1 case 1 favourable  and if he sits on 100 then 1 case 0 favourable so total 8 case 4 favourable

this thing goes on

please understand when I used number referring to person and when to seat
as it would have been too long to write
u know what to do if my answer is write

Answer is 1/2.

Let F(N) be the probability of last person getting the right seat if there are N seats left, and the last person’s seat is still unoccupied. Note that this is the initial condition, before first person has sat down.

If the first person sits in his right seat then last person gets their right seat. The probability that first person sits in his right seat is

1/N  –>(1)

If the first person sits in the last person’s seat then probability of last person sitting on their seat is zero.

1/N x 0 = 0 –>(3)

If the first person sits in one of the (N-2) remaining seats, then the probability of last person sitting on their right seat is dependent on which of the N-2 seats did the first person sit on. Did he sit on seat belonging to persons just behind him, or two person behind him, three person or N-2 persons behind. Each of which will produce a different result. Mathematically reducing it, it comes out to be

(N-2/N ) x (SUM of F(i) where i=2, N-1) x 1/(N-2)

1/N x (SUM of F(i) where i=2, N-1) –>(2)

Bringing them all together,

F(N) = (1) + (2) + (3)

F(N) = 1/N + 1/N x (SUM of F(i) where i=2, N-1)+ 1/N x 0

Reducing it we get,

F(N) = 1/N + 1/N x (SUM of F(i) where i=2, N-1) –> (4)

IF F(k)=1/2 where k=2, N, using induction we can prove that F(k+1) is also 1/2. –>(5) (induction proof given below)

We know F(2) = 1/2.  That is, if there were only two passengers, the probability that the last person will get the right seat is 1/2.

Using (5) F(3)=1/2, F(4)=1/2.. and F(100)=1/2.

Proof by induction of (5).

Assuming F(j)=1/2 where j=2, k , we will prove that F(k+1) is also 1/2. 

F(k+1) using (4) is

F(k+1) = 1/(k+1) + 1/(k+1) x (SUM Of F(j) where j=2, k)) –> (7)

We know,

(SUM Of F(j) where j=2, k)) = 1/2 x (k-1)  

as F(j) for j=2, k is 1/2

Replacing that in above equation (7) we get,

F(k+1) = 1/(k+1) x (1 + (k-1)/2)
                = 1/(k+1) x (k+1)/2
                = 1/2

Suppose there are only 2 people.
P(me getting my seat)= P(steve sitting on his seat)=0.5
if there are 3 people
P(me  getting my seat)=P(steve sitting on his seat)+P(steve sitting on wrong seat)*P(another person sitting on his seat)
=(1/3)+(2/3)*(1/2)=2/3
similarly if its 4 seats =3/4
P(n=100)=99/100

Kindly note these points as far as I understood the puzzle:
– Only Steve is choosing the seat randomly.
– Everyone else who is entering the plane is NOT choosing the seat randomly. As first preference, they are occupying their own seat.
– Only scenario in which a passenger is not likely to get his/her seat is if Steve has occupied that seat

Now let’s get probability calculation
Probability (out of 100 seats), that Steve occupies any particular seat number is 1/100 = 1%
Probability that Steve sits on his own seat by chance = 1% – But this probability is NOT important in this puzzle
Probability that Steve occupies my seat by chance = 1% – This is the only scenario in which I lose my seat, in every other scenario, I get to sit on my own seat
Probability that Steve does not occupy my seat = 99%

Thus probability that I get to sit on my own seat = 99%

My answer is also 1/2. By the time my turn comes, 99 seats are already occupied and only 1 is left. So the probability of having that 1 seat assigned or not is 1/2.

The probability of getting your seat is directly proportional to Steve remembers his seat no.
As we know that probability of Steve remembers and forgets seat no. is 1/2 i.e 50%-50%.

so the same with probability of getting your own seat.

so the answer is 1/2 i.e 50%.

Really? 1/2?  If that’s the answer, then all the things in the universe have a probability of 0.5. Whether we’ll reach Mars in 2016 is 0.5. Whether aliens will ever visit us is 0.5. Whether your wife (or girlfriend, depending how much responsible you are) will get pregnant or not is 0.5. Do you even math?

The probability is 1/2
You can either sit on your place or not

The probability is 1/100

Out of 100 seats  you may sit in any 1 of the seats, out of which  only 1 is your assigned seat.

No. of choices of sitting in right seat = 1
No. of choices of sitting in wrong seat = 99

So Probability of sitting in right seat = 1/100