Making Two Tribes fairer
Two Tribes is a TV game show in which an initial seven contestants are whittled down to a single player, who then has the chance to win a cash prize. In each of the first three rounds, the contestants are split into two teams ("tribes'') on the basis of a fairly arbitrary criterion, such as going or not going on caravan holidays. The teams compete, and one member of the losing team is eliminated. A different criterion leads to new teams in the next round.
An odd number of contestants means unequal teams.
When six contestants remain, each tribe has three players, and, all things being equal, every player has the same chance of being eliminated. But with seven or five players, the team numbers are inevitably unequal: is the quiz format fair to all players in those rounds, or does it advantage players in one team over the other? If it is unfair, what changes might make it fairer?
It's not fair!
Under the current rules, one team will have one more member than the other. Each team in turn is asked a series of quiz trivia questions, a correct answer from any team member scores one point, but as soon as the team fails to answer a question correctly, or a one-minute time limit is reached, their turn ends. If the scores are equal, each team selects a "champion'' to go head-to-head to determine the winning team: one member of the losing team is then eliminated (in a fair manner).
With five or seven players, denote the larger team by $A$ and the smaller one by $B$. Assume all players are equally competent, and let $Pr(B)$ denote the chance that team $B$ wins; plainly $Pr(B)0.5$. How big a disadvantage is it to belong to team $B$? Look first at the case of five players, with team $A$ having three players and team $B$ just two. Then a given member of $B$ survives if either \noindent (i) team $B$ wins — the chance is $Pr(B)$, or \noindent (ii) team $B$ loses, but it is the other player who gets eliminated — the chance is $(1-Pr(B))/2$. \noindent Summing these, the chance a given member of $B$ survives is $(1+Pr(B))/2$. A given member of team $A$ survives if \noindent (i) team $A$ wins — the chance is $1-Pr(B)$, or \noindent (ii) team $A$ loses, but some other player gets eliminated — the chance is $2Pr(B)/3$, \noindent giving an overall survival chance of $1-Pr(B)/3$. The latter survival chance exceeds the former when $$1-Pr(B)/3\geq (1+Pr(B))/2,$$ i.e. when $Pr(B)\leq 3/5.$ A similar calculation for seven players shows that being in team $A$ is advantageous whenever $Pr(B)\leq 4/7$. Since $Pr(B)0.5$, members of the smaller team are significantly disadvantaged. How best might we overcome this?A little mathematical analysis provides some ideas. Read on if you want to see the calculations — if not, you can skip straight to the conclusion.
Can we make it fairer?
First simplify matters and ignore the time limit. For a given team, let $x$ be the chance of success at any question. We assume that this chance is constant and independent across questions. This team scores exactly $m\geq 0$ points if it gives the correct answers to the first $m$ questions asked, and then fails on the next question: the chance this happens is $x^m(1-x)$. This is a standard statistical format, known as the Geometric distribution; also, the chance of scoring at least $m$ is $x^m$, as this happens with success on the first $m$ questions.
We assume that every player has the same chance of giving a correct answer.
Under the suggested new rule, team $B$ wins if either \noindent (i) $A$ scores 0 or 1. The total chance of this is $$q^n + (1-q^n)q^n = (2-q^n)q^n.$$ \noindent (ii) $A$ scores some $m\geq 2$, and then team $B$ scores at least $(m-1).$ The collective chance of this is $$(1-q^n)^mq^n(1-q^{n-1})^{m-1}.$$ Overall, $Pr(B)$ is the sum of all these quantities. This might appear rather complicated, but a calculation leads to the neat answer $$Pr(B)=\frac{q(1+q^{n-1}-q^{2n-1})}{1+q-q^n}.$$
(See here for a hint.)
In our model, if we can choose values of $q$ to make the expression for $Pr(B)$ equal to $3/5$ when $n=3$, and to $4/7$ when $n=4$, the game would be fair! With five players, $q=0.636..$ achieves that goal, with seven players, it's $q=0.674..$. This means that the chance $p$ of answering a question successfully is $p=1-q \approx 0.36$ with five players and $p=1-q \approx 0.33$ with seven players.Conclusion
Our analysis suggests that we can make the game fair by:
- Giving the smaller team a one-point lead,
- Handing victory to the smaller team if there is a tie, and
- Devising questions so that the chance of success for a random player is around 36% when five players remain, or 33% when seven players remain.
With modest cash awards, questions shouldn't be too hard.
These values are small enough to make it very likely that a team collectively fails on some question well before their minute is up, so ignoring the time limit is justified.
But it also means that the questions are hard, with roughly only a one in three chance of getting a single question right. Setting questions with that degree of difficulty would change the game beyond recognition! With only modest cash prizes on offer, contestants should not be discouraged or humiliated, they should face relatively straightforward questions; perhaps both teams might well expect to collectively answer all the questions offered during their one minute, most of the time.
The crucial point is that fairness demands that the winning chance of the smaller team be either $3/5$ or $4/7$. So if the type of questions asked remains as now, our suggestion is to give the smaller team a one-point start and victory if the scores are tied. Worth a try?About the author
John Haigh teaches mathematics at the University of Sussex. He is addicted to some (but not all) TV game shows, and finds extra pleasure in exploring their mathematical properties.