There are 100 people in line to board an airplane. Each person is holding a ticket corresponding to both his place in line and his seat (i.e. Person 1 is supposed to sit in Seat 1, Person 2 is supposed to sit in Seat 2, etc.). All of the people in line are normal, except for Person 1, who is an idiot. When Person 1 boards the plane, he will ignore his seat number and sit in a random seat. Following Person 1, each person will sit in his own seat, unless his seat is occupied. In this case, the Person who's seat is occupied becomes the new idiot and will sit in a random seat. For example, if Person 1 sat in Seat 2, Person 2 will become the idiot and sit in a random seat. What is the probability that Person 100 will end up in Seat 100?
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Are we assuming a uniform distribution over the seats for the idiot? If so, this is pretty simple. 50%.
Not hard to prove rigorously, but just look at the base cases of 2, 3 and 4 people to convince yourself.
yea 50%....50% chance the guy sits in his seat or not...but based on the fact that he's an idiot, i highly doubt he'll walk all the way to the back of the plane without falling down or tripping
Sorry, can someone walk me through it? I feel like I'm over-thinking it.
If there are only two passengers, it's the chance that person 2 ends up in seat 2 is 50%, because this is the chance person 1 will randomly choose seat 1.
If there are three passengers, the chance the person 1 chooses seat 1 is 1/3, and the chance that person 1 chooses seat 2 is 1/3. If person 1 chooses seat 2, the chance that person 2 chooses seat 1 is 50%. So the chance neither person 1 nor 2 end up in seat 3 is 1/3 + 1/3*1/2) = 0.5.
If there are four passengers, the chance that person 1 chooses seat 1 is 1/4. The chance he chooses seat 2 is 1/4. If he chooses seat 2, person 2's chances of choosing seat 1 are 1/3 and choosing seat 3 are 1/3. If he chooses seat 3, person 3s chances of choosing seat 1 are 1/2. If person 1 chooses seat 3, person 2 will choose seat 2, and person 3 has a 1/2 chance of choosing seat 1. Add up all these probabilities and you get 50%.
There is a similar analogy (which gets more and more cumbersome as the number of passengers goes up) for every number of passengers.
Not sure why I still cant wrap my head around this. Looks like I'm the idiot on this plane.
Let's call P(x,y) the probability that person x ends up in seat y and P( (x,z) | (x-1,y) ) to be the conditional probability that person x will end up in seat z given person x-1 ends up in seat y. i.e. P( (2,1) | (1,2) ) is the probability that person 2 ends up in seat 1 given person 1 picks seat 2.
So the three person case is:
P(3,3) = P(1,1) + P( (2,1) | (1,2) ) = 1/3 + 1/21/3 = 1/3(1 + 1/2) = 1/2
The four person case is:
P(4,4) = P(1,1) + P( (2,1) | (1,2) ) + P( (3,1) | (1,3) ) + P( (3,1) | (2,3) ) = 1/4 + 1/41/3 + 1/41/2 + 1/41/31/2 = 1/2
And so on...
Degree & Major: math
Haha, figures. I wish I would have studied math to become a critical thinker like you. Kudos, bro!
MBP how did you get so good at math
The probability is 100%!! Because person 1 is such an idiot that he boarded a train instead of an airplane. This means when person 2 boards the plane is empty and all passengers will be seated correctly.
Hahahhahaha omg. Great attention to detail.
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