riddle of the day

A classic probability problem that defies intuition: the answer is the same for any number of passengers, and arriving at it requires looking at the problem from the right angle.

The last passenger

Strategist
Chance and uncertainty

Riddle statement

A plane has \(n\) seats and \(n\) passengers, each with their assigned seat.

Passenger 1 loses his card and sits randomly in any seat. Starting with passenger 2, each one acts like this:

  • if his seat is free, he sits in it;
  • if it is occupied, he chooses at random from the free seats.

What is the probability that the last passenger ends up in his own seat?

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Solution

Answer:

$ \mathbb{P}(\t\t\text{último en su asiento})=\tfrac{1}{2} \quad (n\ge2). $

Explanation:

Mientras the proceso sigue abierto, only importan two asientos: the 1 and the n.

Cada vez that a passenger find ocupado its sitio and must elegir to the azar, can ocurrir three cosas:

  • toma the asiento 1, and from that momento the last acabará in the suyo;

  • toma the asiento $n$, and the last already no podrá sentarse in the suyo;

  • toma cualquier otro, and the problem it traslada to the dueño of that asiento without cambiar its estructura.

El proceso termina in the momento in that aparece by first vez one of esos two asientos críticos. As ambos it encuentran siempre in situation simétrica, each one is igualmente probable.

De ahí that the probability of that the last passenger ocupe its propio asiento sea exactly $\tfrac{1}{2}$, independently of $n$.