
crazy guy on a plane
Let us consider the various possibilities one by one 
1. If the crazy guy sits on seat 1 [ which he will not], then all is well [ all other people will sit on their respective seat numbers]. So the 100th guy will definitely find his seat vacant and he will then occupy it.
2. If the crazy guy seats on seat 100, then the 100th guy can not occupy his seat [ although he will find exactly one other seat ].
3. Now consider other 98 events [ crazy guy seating on any seat number between 2 and 99 [ both inclusive]. Let us start from 99.
1. If he seats on 99, then all the people between [298] will occupy their respective positions. The 99 guy will come and will see his seat occupied. At this point, he has 2 choices 1 and 100. He can choose to sit on any one of them with equal probability of 1/2. So the 100th guy getting 100th seat is also 1/2.
2. If he seats on 98, then people between [297] will occupy their seats. 98th guy comes and he has 3 choices 1, 99 and 100 with equal probability of 1/3. Suppose he chooses 1, then 99th guy will seat on his seat and 100th guy on his. So we have 1/3 possibility in this case.
Suppose the 98th guy seats on 99 [with probability 1/3], then 99 has 2 choices 1 and 100 which as we have already seen got equal probability of 1/2. So the 100th guy getting his seat in this case is 1/3 + 1/3*(1/2) = 1/2
Here also we got the probability of 1/2 again.
If we continue like this backwards for 97, 96, 95... we would get probability of 1/2 each time. 
3. For 97, it will be like
P(1) = 1/4
P(98) = 1/4 so 100th guy getting his seat is (1/4) * (1/2) = (1/8)
P(99) = 1/4 so 100th guy getting his seat is (1/4) * (1/2) = (1/8).
So total = (1/4) + (1/8) + (1/8) = 1/2.
Probability of crazy guy 1 choosing any number between 2100 is 1/99. So total probability is
1/99 * [ 1/2 + 1/2 + ... + 98 times + 0 ] = (1/99) * (98//2) = 0.494949....
P.S. Since this is my first posting, let me know whether my analysis is correct or wrong !
shailesh nirgudkar
Friday, December 10, 2004
The answer is one chance in two. You're basically right.
Stephen Jones
Friday, December 10, 2004
Everything was fine until this statement.
>Probability of crazy guy 1 choosing any number between >2100 is 1/99. So total probability is
>1/99 * [ 1/2 + 1/2 + ... + 98 times + 0 ] = (1/99) * (98//2) >= 0.494949....
Why only consider seats 2100? Seat #1 is very important. So the conclusion is like this.
Probability of crazy guy choosing a particular seat between 1100 is 1/100.
Crazy guy chooses seat #1, probability that 100th guy get own seat is 1.
Seats between 299, probability that 100th guy get own seat is 1/2.
Seat #100, probability that 100th guy get own seat is 0.
Sum up all probabilities, 1/100 * (1 + 1/2 * 98) = 1/2
JHY
Friday, December 10, 2004
As the first guy was going to ignore their seat number, I assumed that he will also igonre his own seat number. If he sits on his own number i.e. 1, can we say that he ignored his own number but still choose '1' randomly ??
shailesh nirgudkar
Monday, December 13, 2004
Yes. His random action just happens to line up with his actual assignment.
Imagine you're a school teacher and you've carefully assigned seats to each child by placing a piece of cardboard on each desk in your room. The children enter and randomly seat themselves. One may appear to be in the correct position to you, but to the child the position is completely random. Ask them to randomly shuffle and you may find another coincidental alignment, but it doesn't signify special knowledge on the part of the children.
Lou
Monday, December 13, 2004
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