0

While solving Monty Hall Problem,

In possible positions/choice table,

There is a case where player initially chooses car (winning) position and host has to reveal either goat 1 position or goat 2 position. Here, do we need to consider both these possibilities as separate choice and calculate the probability (that leads to 1/2 probability before and after switch). Or, do we need to consider both as same choice (that leads to 2/3 probability after switch).enter image description here

Murali
  • 1
  • Not sure what you are asking. Of course it matters which door Monty opens. – lulu Sep 04 '23 at 11:10
  • Should add: it's often left out of the usual statement, but it is important to assume that, in the scenario you describe, Monty chooses between the two worthless doors with equal probability. If Monty "prefers" to open, say, the door with the lower number, that changes the computation. – lulu Sep 04 '23 at 11:11
  • It does not affect the probability of winning the car with the "always switch" strategy: if you initially choose the door with a car (probability $\frac13$), then switching after Monty opens another door and shows a goat always loses; if you initially choose a door with a goat (probability $\frac23$), then switching after Monty opens another door and shows a goat always wins. – Henry Sep 04 '23 at 11:18
  • 1
    Have you read all the other questions about Monty Hall on this website? There have been many, quite possible that your question has already been answered. – Gerry Myerson Sep 04 '23 at 12:58

0 Answers0