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

$$P(A|D)=\frac{P(D|A)P(A)}{P(D|A)P(A)+P(D|\mathrm{\neg}A)P(\mathrm{\neg}A)}$$

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 $\text{P(any door)}=\frac{1}{n}$, where there are n doors. We consider the two distinct representative cases:

$$P(\text{prize behind my chosen door}\mid \text{m})=\frac{P(\text{m}\mid \text{prize behind my chosen door})}{P}$$

$$P(\text{prize behind another door}\mid \text{m})=\frac{P(\text{m}\mid \text{prize behind another door})}{P}$$

where here

$$P=1/(\genfrac{}{}{0ex}{}{n-1}{m})+(n-1)/(\genfrac{}{}{0ex}{}{n-2}{m})$$

and where $\mid m$ 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 $n-1$ doors and so there are $(\genfrac{}{}{0ex}{}{n-1}{m})$ available ways for the host to do so.

The probability $\text{P(any}{\textstyle \phantom{\rule{0.278em}{0ex}}}\text{m}\mid 1)$ is therefore $1/(\genfrac{}{}{0ex}{}{n-1}{m})$, since we equivocate between all the host’s options. For the other case, there are only $n-2$ doors for the host to choose m from, so the number of ways is $(\genfrac{}{}{0ex}{}{n-2}{m})$, and the probability $\text{P(any}{\textstyle \phantom{\rule{0.278em}{0ex}}}\text{m}\mid \text{not 1})$ is therefore $1/(\genfrac{}{}{0ex}{}{n-2}{m})$.

after some algebra I find that

$$P(\text{prize behind my chosen door}\mid \text{m})=\frac{n-1-m}{n-1-m+(n-1{)}^{2}}$$

and

$$P(\text{prize behind another door}\mid \text{m})=\frac{n-1}{n-1-m+(n-1{)}^{2}}$$

Thus, our probability factor is given by: $$\frac{\text{P(prize behind another door}\mid \text{m})}{\text{P(prize behind my chosen door}\mid \text{m})}=\frac{n-1}{n-1-m}$$

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 $n-1$ better for shifting to another door for any $n\ge 2$ and $m=n-2$ 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 $n\to \mathrm{\infty}$ 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*!