George Rebane

From some reader comments it seems that there continues to be some confusion about how the 3-door Monty Hall relates to Bayes theorem, and how that relationship is different from some more usual applications of Bayes.  We have covered Bayesian probability extensively in these pages over the years (for example here).

The fundamental takeaway in the use of Bayesian probability is that 1) all of our knowledge, compiled of countless propositions, is and can be formulated in terms of their probability of being true; and 2) knowledge (i.e. any given proposition) is most correctly updated with new information/data by combining its prior truth probability with the truth probability of the newly obtained info/data according to the Bayes formula (see above link).  The updated (posterior) probability of the proposition may be an increased or decreased value depending on the nature of what new stuff was learned.  All critters know how to do that instinctively with an evolved Bayes-like algo that has promoted their species’ survival.

A more usual application of Bayes can be illustrated with a wet driveway which you discover when you awaken and look out the window.  Did it rain or did your neighbor again soak your driveway with his wayward lawn sprinkler.  You know that rain was predicted to start overnight with a 20% chance, and you also recall that your neighbor sets his sprinkler to do a predawn watering two or three times a week.  You decide to set that probability at 2.5/7 = 0.36.  Should you still carry an umbrella today?

Now for Monty Hall.  I explain the solution to my students by upping the number of doors to a thousand behind one of which is a new care, each of the others hiding a smelly goat.  The contestant nominates a door to open.  It’s obvious that the probability of having a car behind it is 1/1000, and the probability that it’s behind one of the other unopened doors is 999/1000, almost certainty.  Now for the duration of the problem it is important to realize that no matter what happens before the prized door is opened, THESE TWO PROBABILITIES DON’T CHANGE.

Monty knows the door with the car, and he won’t open it until the contestant makes the final pick of the door he wants opened.  So in the first version of our game Monte opens 998 doors with nothing but goats behind them.  Only two doors remain closed – one initially nominated by the contestant, and the other belonging to the set NOT nominated by the contestant.  Knowing that the probabilities have not changed, the contestant knows that he will almost certainly (999/1000) win the car by switching to the door Monty has left closed for it’s the only other door that could possibly hide the car.  The remaining unopened door will always have the complimentary probability (N-1)/N no matter the value of N representing the original number of closed doors.

In the classic Monty Hall problem we had N = 3.  So, of course, it always paid the contestant to switch since his probability of winning doubled from 1/3 to 2/3.  In fact, no matter the value of N, it always pays for the contestant to switch after Monte has opened M doors where now M can range from one to N – 2.  This can be seen from our N = 1000 door problem even when Monte opens only one door.  Switching then will increase the contestant’s win probability from 0.00100000 to 0.00100200.  We see this by writing the formula for winning after switching to a door selected from the N – 1 – M remaining closed doors – i.e. the new evidence is opening M of N-1 prizeless doors.  In that case the win probability of switching is easily shown to be (1/(N-1-M))*(N-1)/N.  In our case of N = 1000 and M = 1 calculates to (1/998)*(999/1000) = 0.00100200.  And when we use Monte Hall’s N = 3, M = 1 we have (1/1)*(2/3) = 2/3, where switching doubles the probability of winning.