# Random Airplane Seats

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?

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%.

Yes, if this happens 500 times, you’ll find your seat 250 times.

Let’s try it once again.

Any passenger taking your seat = you’ll end up with wrong seat.

Any passenger taking Steve’s seat = you’ll end up with your seat.

These two options are always equally likely to happen, when a passenger chooses his seat at random (i.e. his seat is already taken).

Any passenger taking someone else’s seat = one more passenger choosing seat at random, doesn’t affect your chances.

If it is still not clear to you, let’s suppose penultimate passenger chooses his seat at random. It means there are two seats left and it only has to be yours and Steve’s. There can’t be someone else’s seat free instead, otherwise that person would took his seat when he enters the plane. So the penultimate passenger can take either your seat or Steve’s, chances are 50/50.

Now suppose the third to last passenger chooses his seat at random. There must be three seats left and these have to be yours, Steve’s and the seat of penultimate passenger. Thus, the probabilities are 1/3 to choose Steve’s seat, 1/3 to choose your seat and 1/3 to choose penultimate passenger’s seat, in which case the penultimate passenger will also have to choose seat at random and we know the probabilities for him are 50/50 your seat vs. Steve’s seat.

Now suppose fourth to last passenger chooses seat at random. Now there are your seat, Steve’s seat and seats of a penultimate and third to last passeneger. Chances: 25% of taking your seat, 25% of taking Steve’s seat and 50% of taking one of the other two seats, in which case a penultimate or third to last passenger will also have to choose seat at random and we already know that in both cases the probabilities end up being 50/50 of a passenger taking your seat vs. Steve’s.

Now suppose fifth to last passenger chooses his seat at random, and so on… up to the first passenger who is in fact Steve, and you can clearly see that regardless of which passengers choose seats at random, the probabilities of you getting your seat or Steve’s seat will always end up being 50:50.

Along with the above mentioned 3 possible ways to choose a seat, there is one more outcome in my view.

4. Steve chooses his own seat in random.

Which got 1/100th probability.

So, the answer must be 1/2 + 1/100. i.e, 51/100 if I am not wrong.

(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**

Actually it should be (1+2+4……………………………………………2^97 +1)/(2+4+8+16+32+64………………….2^98 +2) and this indeed equals 1/2 😉

not + 2 at end since if steve sit on his seat then then there is only 1 case and 1 favorable

not + 2 at end as it is if steve sit on his seat then it would be 1 case 1 favorable

okay but steve can as well sit on your place in which case there is 1 case and 0 favourable

I accept my mistake

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%.

I beg to differ from the above answer.

Its stated that Steve forgot his seat and chose a random seat. So, the probability of him taking his seat is 1% and not 50% because out of 100 only one is correct. Now, if Steve sits on a right seat certainly I will get my correct seat, probability of this happening is 1%.

For all other cases (where some one before me sits on Steve’s seat to balance things) probability reduces further.

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

### Your Answer

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