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?

## 24 comments:

The probability is .5 u can either sit on your place or not

100% because I kept my ticket!

1/100

He should check his seat number which is on his ticket and take his seat!

1/100

if steve guess the right seat out of 100 then the last seat will be urs

1/50

1/2

1/2Great site!!!Carry on the Excellent work..

That depends, do I really want my assigned seat? If it is a good one I will use my ticket to keep it!

So which is the right answer? I was thinking 1/2 as well. Between the last seat, and the seat the guy is in.

is it 0?? the flight is full..

I think zero is correct

You could work it our for 100 passengers but it's easier if you do it for smaller numbers and see if there is a general case that can be applied to greater numbers.

For 2 passengers it is obvious that you have a 1/2 chance of getting your seat.

For 3, 1/3 of the time Steve will pick his own seat and you'll get yours; 1/3 of the time he'll pick your seat (and you don't); and 1/3 of the time he'll take passenger #2's seat. Now the problem is exactly the same as for 2 passengers, with passenger #2 now playing the role of Steve. We already know the chance of you getting your seat in this case is 1/2.

So the possible outcomes are:

Steve picks his own seat / you get yours - 1/3

Steve picks #2's seat / you get yours - 1/3 * 1/2 = 1/6

Steve picks #2's seat / you don't get yours - 1/3 * 1/2 = 1/6

Steve picks your seat / you don't get yours - 1/3

Total chance of you getting your seat: 1/3 + 1/6 = 1/2

Next, you do the same thing with 4 passengers. Again, Steve can either pick his own seat or yours with equal probability. If he picks one of the others, the problem is now equivalent to the 3-passenger problem with it's 1/2 probability of you getting your seat.

Do the same for 5 passengers and it reduces to the same as for 4 and so on.

So, no matter how many passengers there are, as long as the plane is full and there is only one "Steve", your chances of getting your own seat are 1/2.

The answer of 1/100 (presumably obtained by (99/100 * 98/99 *...*2/3 * 1/2) would only be true if *all* of the passengers chose their seat at random. In fact, it would be zero for me because I would not get on a plane full of crazy passengers.

well if steve initially chooses his correct seat at random, then everyone following would get there correct seat. including yourself. so it cant be 0

I believe it is 49% = 50% minus the 1/100 chance that the first guy picks the right seat.

Secret Squirrel had an error in his proof where he stated 1/3+1/6 = 1/2.

Ok it is 50/50.

What is wrong with 1/3+1/6=1/2

Wow. I was way off my game on this riddle. My humble apologies....

100%... you are the pilot

where is the like button? the pilot answer is extremely funny!

the first guy has a 1 in 100 chance to choose the right seat if he does then everyone else will sit in their assigned seat because they will all be available so for the last person its 1/100

n!= the # of possibile outcomes

(n-1)!= # of outcomes that will give you your seat

# of favorable outcomes/ # possible outcomes = P(getting you seat)

P(getting your seat) = n-1!/n!

(n-1)!/n! = 1/n

P(getting your seat = 1/n... 1/100

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