March 19th, 2018 at 6:49:33 PM
permalink

Quote:This is one of my favorite problems but I ask it in the form of a 100-seat airplane.

LOL.

Someone translated it into a question in Chinese and I translated it back. Skipped the airplane part.

Last edited by: andysif on Mar 19, 2018

March 19th, 2018 at 6:54:26 PM
permalink

Doc,

I was thinking along the same line.

It's hard to do it starting from 100, but if you do it starting from 3, and when you are up to 4 you use the result from 3 and build it up that way, it's more manageable.

I was thinking along the same line.

It's hard to do it starting from 100, but if you do it starting from 3, and when you are up to 4 you use the result from 3 and build it up that way, it's more manageable.

March 19th, 2018 at 7:10:51 PM
permalink

This is how I handle it

Assume there is 2 person instead of 100, then the chance of the last person siting in his own seat is 0.5

if there is 3 person, then p = 1/3*100%(1st guy picked his own seat) + 1/3*0% (1st guy picked the last seat and 1/3*0.5(pick any other seat, which reverts to the situation when there is 2 person involved, ie p(2))

when there is 4 person, p = 1/4*100% + 1*4*0% + 2/4 * (p(3))

And it always turn out to be 0.5 from 2 person all the way to 100 person

if there is 3 person, then p = 1/3*100%(1st guy picked his own seat) + 1/3*0% (1st guy picked the last seat and 1/3*0.5(pick any other seat, which reverts to the situation when there is 2 person involved, ie p(2))

when there is 4 person, p = 1/4*100% + 1*4*0% + 2/4 * (p(3))

And it always turn out to be 0.5 from 2 person all the way to 100 person

January 12th, 2019 at 8:27:51 AM
permalink

I used to have this problem on my web site. It brings back happy memories.

A number of complicated or incorrect answers have been submitted but

gordonm888 has the best one. Here I state it in as few words as possible

to help with understanding.

A sits in the wrong seat, displacing B

B sits in the wrong seat displacing C

C sits in the wrong seat displacing D

This continues until a displaced person sits in A's seat,

displacing no one and everyone else sits in his own seat, or

a displaced person sits in the last person's seat,

displacing the last person.

Either event is equally likely to happen, so

The probability of the last person getting his own seat is 1/2.

A number of complicated or incorrect answers have been submitted but

gordonm888 has the best one. Here I state it in as few words as possible

to help with understanding.

A sits in the wrong seat, displacing B

B sits in the wrong seat displacing C

C sits in the wrong seat displacing D

This continues until a displaced person sits in A's seat,

displacing no one and everyone else sits in his own seat, or

a displaced person sits in the last person's seat,

displacing the last person.

Either event is equally likely to happen, so

The probability of the last person getting his own seat is 1/2.

January 12th, 2019 at 2:59:37 PM
permalink

Tend to agree with above idea

If the first person sits at seat 1 then everyone will sit at their correct place : P=1.

If the first person sits at seat 100 then there's no way :P=0.

Suppose the first person sits at place 99 then it's 50/50 whether when 99 gets to choose he sits in 1 or 100; P = 1/2

Sippose the first person sits at place 98 then when 98 gets to choose it's 1/3 he sits at 1, 1/3 he sits at 100 and 1/3 he sits at 99. Thus P = 0*1/3 + 1 *1/3 + 1/2*1/3 = 1/2.

Similarly all the probabilities are 1/2.

As has been said another way of looking at it is as each person arrives whose seat has been taken (or first time for person #1) the situarion is resolved at that moment if they pick seat 1 or seat 100, otherwise the situation is resolved later on with the same chances. Thus it is 50/50.

If the first person sits at seat 1 then everyone will sit at their correct place : P=1.

If the first person sits at seat 100 then there's no way :P=0.

Suppose the first person sits at place 99 then it's 50/50 whether when 99 gets to choose he sits in 1 or 100; P = 1/2

Sippose the first person sits at place 98 then when 98 gets to choose it's 1/3 he sits at 1, 1/3 he sits at 100 and 1/3 he sits at 99. Thus P = 0*1/3 + 1 *1/3 + 1/2*1/3 = 1/2.

Similarly all the probabilities are 1/2.

As has been said another way of looking at it is as each person arrives whose seat has been taken (or first time for person #1) the situarion is resolved at that moment if they pick seat 1 or seat 100, otherwise the situation is resolved later on with the same chances. Thus it is 50/50.