c**********e
发帖数: 2007 | 1 There are 100 seats on a plane, and 100 ticketed passengers.
The first 2 passengers are blind. They board the plane first,
taking random 2 seats. Other passengers boards one by one: if
he finds his seat taken, he will take a random seat. If his
seat is not taken, he takes his own seat.
Question, what is chance of the last passenger gets his own
seat? | z*******e
发帖数: 122 | 2 The probability is 1/2. Actually the the 100th person can only sit on the
first seat or the 100th seat. Prove it by contradiction. If say the 100th
person sit in the 70th seat, it means that when the 100th person get into
the plan, the 70th seat is empty, this means that when the 70th person goes
into the plan, the 70th seat is empty, so the 70th person will seat in the
70th seat.
As a result, the 100th person can only seat in either the 1st or the 100th
seat. so the probability of seating in th | c**********e
发帖数: 2007 | 3 There are 2 blind men. The answer should be 1/3.
Your idea is good. The 100th passenger can only seat in 1st, 2nd or 100th
seat. But why the chances are equal?
goes
【在 z*******e 的大作中提到】
: The probability is 1/2. Actually the the 100th person can only sit on the
: first seat or the 100th seat. Prove it by contradiction. If say the 100th
: person sit in the 70th seat, it means that when the 100th person get into
: the plan, the 70th seat is empty, this means that when the 70th person goes
: into the plan, the 70th seat is empty, so the 70th person will seat in the
: 70th seat.
: As a result, the 100th person can only seat in either the 1st or the 100th
: seat. so the probability of seating in th
| G******y
发帖数: 1831 | 4 There are two blinds, so you mean the probability is 1/3? Your logic is
right except one important thing: you need to show that the chance the two
or three open seat available have the same probability to be taken.
It can be shown this way: for everyone who is going to take the seats, these
three seats will be taken only when people randomly take seats. Thus it is
equally likely for each one to be taken (or not taken). The last person may
only sit on one of the three seats, with equal probabilit
【在 z*******e 的大作中提到】
: The probability is 1/2. Actually the the 100th person can only sit on the
: first seat or the 100th seat. Prove it by contradiction. If say the 100th
: person sit in the 70th seat, it means that when the 100th person get into
: the plan, the 70th seat is empty, this means that when the 70th person goes
: into the plan, the 70th seat is empty, so the 70th person will seat in the
: 70th seat.
: As a result, the 100th person can only seat in either the 1st or the 100th
: seat. so the probability of seating in th
| x******e
发帖数: 1428 | 5 nice!!! can also be extended to multiple blinds cases.
goes
【在 z*******e 的大作中提到】
: The probability is 1/2. Actually the the 100th person can only sit on the
: first seat or the 100th seat. Prove it by contradiction. If say the 100th
: person sit in the 70th seat, it means that when the 100th person get into
: the plan, the 70th seat is empty, this means that when the 70th person goes
: into the plan, the 70th seat is empty, so the 70th person will seat in the
: 70th seat.
: As a result, the 100th person can only seat in either the 1st or the 100th
: seat. so the probability of seating in th
| c**********e
发帖数: 2007 | 6 Guys, thank you for the good discussion and solution.
Good luck for your job search! |
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