# 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:

$P\left(A|D\right)=\frac{P\left(D|A\right)P\left(A\right)}{P\left(D|A\right)P\left(A\right)+P\left(D|\mathrm{¬}A\right)P\left(\mathrm{¬}A\right)}$

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:

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

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

where here

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

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 $\left(\genfrac{}{}{0}{}{n-1}{m}\right)$ available ways for the host to do so.

The probability $\text{P(any}\phantom{\rule{0.278em}{0ex}}\text{m}\mid 1\right)$ is therefore $1/\left(\genfrac{}{}{0}{}{n-1}{m}\right)$, 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 $\left(\genfrac{}{}{0}{}{n-2}{m}\right)$, and the probability $\text{P(any}\phantom{\rule{0.278em}{0ex}}\text{m}\mid \text{not 1}\right)$ is therefore $1/\left(\genfrac{}{}{0}{}{n-2}{m}\right)$.

after some algebra I find that

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

and

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

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

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!