Donuts on a Stick – how to win the high-calorie challenge of the Taskmaster
‘Taskmaster’1 is a British television show, in which five contestants compete against each other over a series of episodes doing typically quite unusual tasks, such as dressing a coconut to look like a businessman. They collect points depending on how they perform and then win (often hilarious) prizes at the end of each show and the season.
Most of the time the competition is simply based on how the contestants each fare in their individual task. In such a setting, they can focus on doing the task to the best of their ability and do not need to worry about what others will do.
However, there are some exceptions as the show has introduced some tasks here and there where game theoretic considerations come into play, meaning that while conducting a task the participants also need to consider what others will be doing when contemplating their actions.
Donuts on a Stick
In our opinion, the most interesting case happened, when in Season 3 Episode 5 the participants played the game ‘Donuts on a Stick’. The rules were quite simple, yet tricky: The 5 participants were asked to secretly place between 1 and 5 donuts on a stick. The participant with the lowest unique number (i.e. the lowest number chosen by no other participant) of donuts wins. In case all chosen numbers of donuts are duplicated, then the game is played out again until there is a winner.
The first thing to note is that there is no so-called dominant strategy in this game. For example, placing 1 donut on your stick is a good idea if nobody else does, but if everyone else does you are better off choosing 2 donuts.
In the following we will focus on finding the so-called ‘Nash Equilibria’, which are combinations of strategies in which all of the participants choose the best available strategy given the chosen strategy of all the others.
Proceedings in the show
Let’s check out what happened in the show (game starts at 35:57 and lasts around 5 minutes):
In the first round, three of the five participants selected 1 donut and the remaining two selected 2 donuts, resulting in a tie round. This combination of strategies is not a Nash equilibrium, as each of the participants with 1 donut will have regretted not to have gone for e.g. 3 donuts instead and win the round.
Looking at the game, we can see that out of the 3125 possible combinations, 2052 are tie rounds and none of these will be a Nash equilibrium. The reason for this is that in a tie round at least one number of donuts has been chosen by three or five players, either of whom could profitably deviate by choosing a number of donuts not already chosen by somebody else.
In the second round, two participants selected 1 donut, and one each selected 2, 3 and 4 donuts respectively, resulting in a win for the participant with 2 donuts.
Quite remarkably, although only one player can be satisfied with the outcome, this combination of strategies is actually a Nash Equilibrium: None of the players – including the four players that lost this game – could have improved their result in this game by deviating from their choice given the choices of the others.
Is this the only Nash Equilibrium? No, actually we find that there are in total 1,3802 different combinations in this game that share the same characteristics as round two, i.e. where we end up with one person winning the game and no other players being able to improve upon their result of losing3.
Consequently, this leaves us with 1,540 combinations of one player winning the game and at least one other player regretting their choice, i.e. situations which are not a Nash Equilibrium.
Searching for a symmetric equilibrium
The Nash Equilibria in the pure strategies mentioned in the previous section are not very stable in the sense that only one player is happy with the outcome. In the following example, we are going to extend the analysis to mixed strategies in which players choose a probability distribution over the pure strategies and will be looking for a symmetrical Nash Equilibrium, i.e. an equilibrium where all players choose the same mixed strategy.
Given the game with 5 players is actually quite complex, we will be modelling an n-player game in which players can choose between 1 to n donuts and see whether we learn anything from a simpler scenario. In this n-player game, the winner receives n points, the losers receive 0 points and we have modelled the tie round as everybody receiving 1 point (giving an equal chance to eventually win the repeat game). We will denote the symmetrical Nash Equilibrium strategy in this game by (x1*, … , xn*) where xi*∈[0,1] represents the probability of choosing i donuts.
Looking at different cases for n we find the following:
- n = 1: (x1*)=(1)
- n = 2: (x1*, x2*),=(1,0)
Both players will always select 1 donut and will always tie
- n = 3: (x1*, x2*, x3*),=(1/2, 1/4, 1/4)4
If only 1 and 2 donuts were part of the equilibrium strategy, then they would need to be played with the same probability of 50% (otherwise it would be profitable to always choose the option with the lower probability to avoid ties).
However (1/2, 1/2, 0), if selected by all players, is no equilibrium, as a player could profitably switch to selecting 3 donuts, which wins the game 50% of the time.
That is why in equilibrium all three pure strategies must be played to a certain percentage in the equilibrium strategy.
- n = 4: (x1*, x2*, x3*, x4*),=(1/2, 1/2, 0, 0)4
Quite interestingly, we add one player and one additional option (selecting 4 donuts) and as a result in the symmetric equilibrium one option less is selected!
The reason this happens is the fact that with three other players playing the equilibrium strategy the probability of a Three-way-tie between them is only 25%, which means that deviating by choosing e.g. 3 donuts does not improve a player’s expected outcome (it gives them the same outcome as in equilibrium, i.e. 1 point).
Similar to the argument given above, 1 and 2 donuts are selected with the same probability as otherwise it would be profitable to always go for the option selected with lower probability in order to try to avoid a tie.
- n = 5: (x1*, x2*, x3*, x4*, x5*),≈(37%, 34%, 20%, 8%, 1%)5
Amazingly enough, when one player is added, we go from two out of four strategies being played, to all five of them being included!
We think this is particularly counter-intuitive with respect to the choice of 5 donuts, which on first glance looks like an awful choice. However there is some method in the madness in that going for that choice gives the highest chance of avoiding a tie and gives the chance to clinch the win in some scenarios where everyone else is tied on the more preferred numbers.
What does make sense intuitively, however, is that the probabilities are decreasing with an increasing number of donuts, accounting for the fact that there are more cases of you winning with a lower amount of donuts, if this number is unique.
Unfortunately, we do not see a clear pattern evolving for the n-player game, and given we also have a day job to attend, we will stop the analysis at this point. But if somebody wants to give this a go, we’d be very curious to hear the answers to the following:
- What happens in the case of n→∞?
- For n=1, 3 and 5 all pure strategies were included in the symmetric equilibrium strategy. Is that always the case when there is an uneven number of players?
Changing the Game
The show is known for its participants trying to find loopholes in the task descriptions to gain an advantage. We wonder why nobody tried to change the game in the following manner: A participant could have taken out four donuts from their box and gift them to the audience clearly visible for the others. This means that they would have committed to choosing 1 donut on their stick, which prevents others from also choosing 1 donut as they know they cannot win with this choice. If nobody else chooses 1 donut, then our bold player will win the game.
This strategy is however not risk-free, as there is a good chance that some of the others would not have liked such behaviour to be rewarded and would have sacrificed themselves for the group by choosing 1 donut too, even if they do not personally benefit from that choice.
It certainly would have been interesting to see how that action would have played out!
1 All episodes are available on their Taskmaster Youtube Channel
2 Readers who enjoy combinatorics are free to recreate those calculations
3 Note that in this argument we remain within the scope of the game and abstract away from the concrete show, where players could theoretically have considerations like they’d rather want player A to win this game than player B because in the long-term series standings they consider B a more severe competitor than A. In the show that could have actually been a valid point given this game was not only the last game of the episode, but as well the last game of the season and eventually decided the season champion!
4 The stated equilibria were found using a combination of logic, combinatorics and polynomial equations. We have omitted them from the post for the sake of brevity, but are happy to provide details for those interested
5 To be fair, we only got an approximate solution (rounded above for ease of readability), which could be considered ‘close enough’ for practical reasons as in that solution the expected value of a player choosing either option is 1.000000 points when rounding to the sixth decimal. We tried with leading mathematical software to find an exact solution, however it looks like the series of polynomial equations of degree 4 to be solved goes beyond today’s possibilities. If any reader wants to give it a go, please feel free to get in touch with us!