Generalised Game Show Problem

We generalise a result in Professor David Mackay’s book on inference. Bayes theorem also plays a crucial role in decision-making:


Let us consider a worked example which demonstrates how unintuitive results following from this 260-year-old theorem can be.

In the Game Show example, we can simplify Bayes’ Theorem by using a form that expands out all individual doors, and then cancelling off the unconditional probability of each door in numerator and denominator as they are each P(any door)=1n, where there are n doors. We consider the two distinct representative cases:

P(prize behind my chosen doorm)=P(mprize behind my chosen door)P

P(prize behind another doorm)=P(mprize behind another door)P

where here


and where m means that any m doors were removed at the usual intermediate stage after I chose a door. When we start with n doors, and one has been chosen, and that door happens to have the prize behind it, then the Game Show Host is free to remove m doors from the set of n1 doors and so there are (n1m) available ways for the host to do so.

The probability P(anym1) is therefore 1/(n1m), since we equivocate between all the host’s options. For the other case, there are only n2 doors for the host to choose m from, so the number of ways is (n2m), and the probability P(anymnot 1) is therefore 1/(n2m).

after some algebra I find that

P(prize behind my chosen doorm)=n1mn1m+(n1)2


P(prize behind another doorm)=n1n1m+(n1)2

Thus, our probability factor is given by: P(prize behind another doorm)P(prize behind my chosen doorm)=n1n1m

This expression gives us back, from the original game, our game strategy factor of 2 times better if we shift door when n=3 and m=1. The factor rises to n1 better for shifting to another door for any n2 and m=n2 doors (all but one other than the one the player chose), and for the shift strategy a probability factor that tends to unity from above when n and m=1, i.e. only one door is removed.

An informal survey by Student-b has shown that quite often, intuition among a random sample of people asked, is lacking. Some will believe it is better to stick, and some say in the standard three-door game that it is slightly better to shift. The mathematics show that for positive n and m: it is always better to shift!