Should the Faithful vote off Richard Grant?

Last night on the Traitors, Richard Grant said to his fellow contestants that he was a traitor.

If you were a traitor, then to announce you were a traitor is extremely risky; it is necessary for the success of the traitors for their identity to remain a secret.

There followed huge discussion about whether he was a traitor or not. Had he made a slip in his excitement? Had he wished desperately to be a traitor, so said he was just to feel it for a moment?

We can use math to help us!

Using Bayesian probability, we can estimate the probability that Richard is a traitor, and decide whether we — as the unknowing faithfuls — should vote him off or not.

Bayes’ theorem will tell us the probability that Richard is a Traitor, given that he said he was a traitor, which we can write as $P(T|S)$.

To calculate this, we need to know, or have estimates for, the following probabilities:

• $P(T)$ — the probability that Richard is a Traitor
• $P(S)$ — the probability that Richard says he is a Traitor, regardless of whether he is a traitor or not
• $P(S|T)$ — the probability that Richard says he is a Traitor, assuming that he is a Traitor

We can plug these values into Bayes’ theorem:

$$ P(T|S) = \frac{P(S|T)P(T)}{P(S)} $$

Let’s estimate some values.

$P(T)$ is already a bit complicated. In a usual series, there are 3 traitors, and 22 contestants. So the probability that Richard is a traitor would be $P(T) = \frac{3}{22}$.

However, this series started by only assigning two traitors, adding a third later. The third was added after Richard announced he was a traitor, but the contestants didn’t know how many traitors had been announced, having been told only that the set of traitors would be completed later.

I would suggest that 2 traitors is a sensible estimate.

There could be 1, 2, or even 3 traitors assigned. If we give equal weight to them, the expected number of traitors would be 2. There could be more, but I wouldn’t have expected them to assign 4 traitors and add more later, so I wouldn’t let that have much bearing.

So we have $P(T) = \frac{2}{22} \approx 0.09$.

Now we need to estimate the probability that Richard would say he is a traitor, regardless of whether he is a traitor or not. If we are being strictly evidence based, we could take the number of contestants that have played the game before, and see how many of them have said they were a traitor.

There have been 6 series of the UK version of the show, including the current series. I have not watched them, but I’ll assume no one else said immediately following the choosing of the traitors that they were a traitor. That would mean only one out of 132 contestants has said they were a traitor, so $P(S) = \frac{1}{132} \approx 0.008$.

However, in the celebrity version of the show the stakes are lower, and the personalities and relationships of the contestants are different. To get a good estimate, we would need to really understand game theory and the psychology of the contestants. I don’t have that knowledge.

Instead, I shall break down the problem.

As being a traitor and being a faithful are mutually exclusive, we can say that $P(S) = P(S|T)P(T) + P(S|F)P(F)$, where $F$ is the event that Richard is a faithful.

The formula for our estimate will now be

$$ P(T|S) = \frac{P(S|T)P(T)}{P(S|T)P(T) + P(S|F)P(F)} $$

So, now we need to estimate the probability that Richard would say he is a traitor, assuming that he is a traitor, and the probability that he would say he is a traitor, assuming that he is a faithful.

Let’s start with the first, $P(S|T)$. I think someone who is a traitor would be fairly unlikely to say there were a traitor, as they should know that they are going to be voted off if they are found out, and it would bring an unnecessary amount of heat. So I will assign a fairly low probability, say 0.1. It would probably be lower for the civilian version, but in a celebrity version with lower stakes, the celebrity might be willing to experiment and have some fun.

For $P(S|F)$, I would say this is a very low probability, say 0.01. I just wouldn’t expect a faithful to want to put their position in the game at risk, unless they felt that they were a target for the traitors, and were applying it as a strategy to stay in the game. I don’t think that would be the case this early in the game, so I will stick with the low probability.

In summary, I have the following estimates:

$$ \begin{aligned} P(T)&=0.09\newline P(F)&=1-P(T)\newline &=0.91\newline P(S|T)&=0.1\newline P(S|F)&=0.01\newline \end{aligned} $$

Putting them into our formula:

$$ \begin{aligned} P(T|S) &= \frac{0.1 \times 0.09}{0.1 \times 0.09 + 0.01 \times 0.91 }\newline &= 0.5 \end{aligned} $$

That is, given that Richard said he was a traitor, there is an approximately 50% chance that he is a traitor.

Good enough for me to vote out.

What if our estimates are wrong? What if we have overestimated the probability that a traitor would reveal themselves, and underestimated the probability that a faithful would want to have some fun?

Let’s use the following:

$$ \begin{aligned} P(T)&=0.09\newline P(F)&=0.91\newline P(S|T)&=0.01\newline P(S|F)&=0.05\newline \end{aligned} $$

That gives us

$$ \begin{aligned} P(T|S) &= \frac{ 0.01 \times 0.09}{0.01 \times 0.09 + 0.05 \times 0.91 }\newline &= 0.019 \end{aligned} $$

That’s LESS than the original chance that he was a traitor!

So, possibly a cunning ploy by Grant.

Are humans rational? Will Professor Hannah Fry whip out the whiteboard and do the maths? What prior probabilities should she use?