### A Physicist's Take on the M-Door Monty Haul Problem

The M-door Monty Haul Problem is very simply stated. You are a contestant of a game show in which a collection of M closed doors is present on the set. Unobserved by you, one door has an expensive car behind it and the others doors have goats behind them. Neither the car nor the goats are switched between doors or replaced with a different prize at any time. To start, you pick a door and then host Monty Haul opens one of the doors (not the one you picked, of course) to reveal the goat behind it. You then alternate between choosing to switch or not the door you have picked and Monty Haul opening doors to reveal the goats behind them. Once you are down to only two doors, you have one final chance to switch doors and then Monty Haul reveals the prize behind your picked door.

It turns out that my previous proof of the optimal strategy for the M-door Monty Haul problem has a few errors in it. The most important was an assumption that switching doors on any round other than the final two-door round hurt one's chances of winnings. This must be false. To see this, let's suppose that at the start of the game you've picked door #1. Suppose also that, after making a series of switches or non-switches, you switch back to door #1. Can it really be said that your odds of winning the car are now lower than if you had never switched from door #1 at all?

In physical terms, this property is called the Markovian Postulate:

In the M-Door Monty Haul problem, the probability of winning the car by making no switches at all except for one final switch in the two-door round is (M-1)/M. So this must be the probability of winning the car by switching doors in the final round, regardless or whether or not you switch or not on any previous round. So the optimal strategy is simple: switch doors in the two-door round.

It turns out that my previous proof of the optimal strategy for the M-door Monty Haul problem has a few errors in it. The most important was an assumption that switching doors on any round other than the final two-door round hurt one's chances of winnings. This must be false. To see this, let's suppose that at the start of the game you've picked door #1. Suppose also that, after making a series of switches or non-switches, you switch back to door #1. Can it really be said that your odds of winning the car are now lower than if you had never switched from door #1 at all?

In physical terms, this property is called the Markovian Postulate:

If K is any observational state and T is a trial whose preparation stage invariably ends with the system in the state K, then T is statistically regular. (O. Penrose, "Foundations of Statistical Mechanics", p.34)In particular, suppose you go into the two-door round having picked a certain door and with a certain other door remaining (call them #1 and #2). You can think of this as one long trial T which invariably end with you having picked door #1 with door #2 remaining at the end of a preparation stage of a certain length. The Markovian Postulate says that your probability of winning the car at this point will only depend upon whether or not you switch in this final two-door round, regardless of the past history of switches or non-switches in the previous rounds.

In the M-Door Monty Haul problem, the probability of winning the car by making no switches at all except for one final switch in the two-door round is (M-1)/M. So this must be the probability of winning the car by switching doors in the final round, regardless or whether or not you switch or not on any previous round. So the optimal strategy is simple: switch doors in the two-door round.

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