Crazy Passenger - June 15

[Amit Goyal]

June15_Prob1(CrazyP).png

[Chinni18]

Hi Rahul I am getting 1/2. Because each passenger's probability of finding his seat occupied depends on whether the one who came before him also found his occupied.

[aditi5000]

Yes I am also getting 1/2..!

[anon_econ]

Plz explain guys!!

[aditi5000]

if the crazy  passenger sits in seat 1, then all other passengers will be in their own seats including the 10th person... and if the crazy person takes the 6th seat, passengers 2 -5 will be seated normally, and when passenger 6's turn comes, they he will become the  crazy passenger because he will want to occupy any seay at random. So its like the problem reverses and changes into a crazy person problem for 5 people (6th-10th person).. and so the probability will be the same of the last person getting their own seat in the problem for 10 people, 5 people and so on. So probability has to be 1/2...

[rahul]

I am also getting 1/2( probability of 10th person to sit at 10th seat =probability of 10th person did not sit at 10th seat)

But Aditi  if 6th sit at 1st seat then other people will sit at their assign seat,

[ritu]

i think there are two broad cases.....
case 1:crazy one sits in his marked seat ...pr=1/2
case 2:crazy doesnt sits on his marked seat
 but since we have to find pr of 10th one sitting on his own seat so crazy one has eight seats to choose from:pr=1/2*1/8
now m getting confused hw to go from here like if crazy one sits on 2nd person's seat then we have to add the case that whn second when makes his decision he can choose from seven seats only.......there can be many cases ....hw do we incorporate that....:(

[anon_econ]

Ok this is a famous problem. U can google it.

[Amit Goyal]

Yes it is 1/2.