The Multi-Player Performance-Enhancing Drug Game

This paper extends classical work on economics of doping into a multi-player game setting. Apart from being among the first papers formally formulating and analysing a multi-player doping situation, we find interesting results related to different types of Nash-equilibria (NE). Based mainly on analytic results, we claim at least two different NE structures linked to the choice of prize functions. Linear prize functions provide NEs characterised by either everyone or nobody taking drugs, while non-linear prize functions lead to qualitatively different NEs with significantly more complex predictive characteristics.

1 The two-player performance-enhancing drug game as a special case In this section, we show how the multi-player doping game can be parametrised to reproduce the original two-player doping game of [1]; that is, we will check if the expanded model described in the Model set-up section of the main manuscript is able to regenerate the original performanceenhancing drug game in [1].
In order to mimic this model, we will need to assign correct values to the new set of parameters. It can be shown easily that the following parametrisation leads to this objective: n = 2, a 1 = a, a 2 = 0, r 1 = r 2 = r and c 1 = c 2 = c (S1) Given the parametric assumptions of Equation (S1), the variable d may take three values; 0, 1, or 2. If d = 0, nobody takes drugs and the expected payoff of players can be computed as: In the case of two drug takers (d = 2), the expected payoff of players takes the form: If d = 1, one player is "clean" while the other takes drugs. In this situation the payoff of the cheater becomes: while the payoff of the player that stays clean is: Now, comparing the results of (S2), (S3), (S4) and (S5) with the original pay-off structure of [1] a perfect correspondence is readily observed. (See Figure S1) Consequently, our extended multi player model is shown to represent the two-player case correctly.

Proofs of Theorems in the main manuscript
Let us start from the expected payoff of player i when he cheats with probability p i . Let D (i) denote the number of players taking doping, not counting player i itself (note that this is a random variable). The payoff function is then: is the average prize received by a player that cheated if there are d cheaters in total, andĀ d = n i=d+1 a i n−d is the average prize received by players that played fair. By applying these definitions, the expression can be simplified as: Taking the derivative: Since p i is constrained to the range [0; 1], the condition for a Nash equilibrium is that for each player, exactly one of the following three cases must hold: Note that if we expand the sum in the derivative and spell out the P(D (i) = d) probabilities exactly, we would find that the derivative of Π i w.r.t. p i depends only on p j where j = i, and their form is equivalent apart from the indices of the p i 's involved. Therefore, when we evaluate combinations of the three conditions outlined above, we only have to decide how many players will be treated according to the first, second or third conditions, without loss of generality.
Theorem 1. Given an instance of the n-player doping game, a common rc product for all players, and a linear prize structure (i.e. a common b = a 1 − a 2 ), the Nash-equilibria are as follows: 1. Everyone cheats when rc < n−1 2 b.
Proof. When the prize structure is linear, we know that a 1 − a 2 = a 3 − a 2 = . . . , and that a i = a 1 − (i − 1)b where b is defined as a 1 − a 2 . This implies that The general derivative in Equation (S6) is then simplified to: It is now clear that the derivatives are independent of i, i.e. they are the same for all the players. The derivatives are zero when rc = b n−1 2 , and since this constraint imposes no restriction on the probabilities, it follows that any pure or mixed strategy is a Nash equilibrium when rc = b n−1 2 and the prize structure is linear. rc < b n−1 2 implies a positive derivative for every player, leading to a situation where everyone cheats. Similarly, rc > b n−1 2 yields a negative derivative and everyone will play fair.
Theorem 2. Given an instance of the n-player doping game and a common rc product for all players, the sufficient and necessary condition for the "k players cheat and n − k players play fair" pure strategy to be a Nash-equilibrium is as follows: Proof. Without loss of generality, let us assume that players 1, 2, . . . , k cheat and the rest play fair. When we are in a Nash equilibrium and k players cheat with probability 1, this means that k players have no incentive for switching their strategy because their payoff function is locally maximal. Similarly, for the remaining n − k players, they also have no incentive to switch because their payoff function is locally maximal. If there were no limits on the range of p i (i.e. the probability of cheating), the equations were as follows: This means that the derivatives of the payoff function are zeros for everyone. However, since p i is bounded from above by 1 and from below by zero, the derivative of the payoff function of cheaters may actually be larger than or equal to zero, where being larger means that the player would try to cheat even more frequently if that were possible. A similar train of thought shows that the derivative of the payoff function of non-cheaters may actually be smaller than or equal to zero, meaning that they would try to play even more fairly if that were possible. This yields the following conditions to start out from: We use Equation (S6) to simplify them as follows: 1. For player i where i ≤ k, the player is cheating, and the number of other players who are cheating is k − 1 with certainty (since all p i probabilities are either zeros or ones).
2. For player i where i > k, the player is playing fair, and the number of other players who are cheating is k with certainty (since all p i probabilities are either zeros or ones).
We then assume a common rc product and move r i c i to the right hand side to obtain the final equation: This concludes our proof.
3 Nash-equilibria in the three-player case To describe the various Nash-equilibria we will find, we will use a simple encoding scheme where each letter describes the behaviour of one player. Each letter may be D (doping with certainty), F (plays fair with certainty), m (mixed strategy) or * (any pure or mixed strategy). For instance, DDF describes a strategy where two players cheat and the third plays fair all the time, while mmD is a strategy where two players play mixed strategies and the third cheats all the time. The order of letters does not matter, mmD is the same as mDm or Dmm. The difference between m and * is subtle but significant: when a player is marked by m, it means that the player will assume a mixed strategy with an exact probability for cheating, while * means that the player may take any pure or mixed strategy with any probabiity for cheating without affecting its own payoff function.
In the three-player case, we know the following: We also know that The derivative is then Note that the b i 's are just an alternate representation of the prize structure as b i is the difference between the ith and the i + 1th prize. The remaining derivatives are the same but with different indices for p i , r i and c i , but the indices for b i stay the same: This means that all the three players are according to condition 2. We have to decide whether all the derivatives are non-negative (given that all cheat, p 1 = p 2 = p 3 = 1): Therefore, this particular configuration (i.e. everyone cheats) is a Nash equilibrium if and only if 2b 2 +b 1 3 ≥ max i r i c i . Increasing r i c i above 2b 2 +b 1 3 for at least one player eradicates this configuration.
A special case is the case of linear prize structures where b 1 = b 2 = · · · = b n−1 = b, leading to b ≥ max r i c i . In other words, if the expected loss of player i when cheating is less than the difference between two consecutive places in the prize ladder, then the player will cheat.

Case 2: two players cheat, one player plays fair (DDF)
Without loss of generality, we can assume that players 1 and 2 cheat and player 3 plays fair. Thus, p 1 + p 2 = 2, p 1 + p 3 = 1 and p 2 + p 3 = 1. The equations to check are: Here it becomes easier to assume that r i c i is independent of i, yielding: The above equation may hold only when b 1 ≥ b 2 ; in other words, when the prize structure is convex or linear. If we assume the case of linear prize structure (i.e. b 1 = b 2 = b), we obtain b ≥ rc ≥ b, which can happen only if rc = b. We will see later that rc = b collapses the game into a configuration where all the strategies are equivalent anyway.

Case 3: one player cheats, two play fair (DFF)
Again, without loss of generality, we can assume that player 1 cheats and players 2 and 3 play fair. Thus, p 1 + p 2 = 1, p 1 + p 3 = 1 and p 2 + p 3 = 0. The equations are: Assuming r i c i = rc: Note that this can hold only when b 1 ≤ b 2 ; in other words, when the prize structure is concave or linear.

Case 4: everyone plays fair (FFF)
In this case, we have p 1 = p 2 = p 3 = 0, which leads to Therefore, this particular configuration (i.e. everyone plays fair) is a Nash equilibrium if and only if 2b 1 +b 2 3 ≤ min r i c i . Decreasing r i c i below 2b 1 +b 2 3 for at least one player eradicates this configuration.
A special case is the case of linear prize structures which leads to b ≤ min r i c i . In other words, if the expected loss of player i when cheating is greater than the difference between two consecutive places in the prize ladder, then the player will play fair.
Let us take a step back now and examine what we have observed so far. The presence or absence of various pure Nash-equilibria depend on where the value of the rc product falls compared to 2b 1 +b 2 3 , b 1 +b 2 2 and b 1 +2b 2 3 . Furthermore, note that the thresholds are also equal to A 1 −Ā 0 , A 2 −Ā 1 and A 3 −Ā 2 . We will see that these thresholds will also determine the behaviour of the model when we allow for mixed strategies as well.

Mixed strategies 3.2.1 Case 5: everyone plays a mixed strategy (mmm)
In this configuration, all the partial derivatives must be equal to zero: the prize structure is linear), the first term is zero and we recover cases 1 and 4, respectively. Let us therefore assume that b 1 = b 2 . This leads to: Again, assuming that r i c i = rc, we obtain We also have to ensure that 0 ≤ p i ≤ 1, otherwise this solution would not exist. Therefore, this Nash equilibrium occurs if and only if where we notice the same thresholds again as the ones we have seen for the pure case.

Case 6: one player cheats with certainty, others play a mixed strategy (Dmm)
Without loss of generality, assume that player 1 cheats with certainty. The equations are: Again, when b 1 = b 2 , the conditions recover case 3, so let us assume that b 1 = b 2 . We then have to distinguish two cases.
The first case is when b 1 > b 2 , i.e. a convex prize structure. In this case, b 2 − b 1 < 0, so the relation will turn in the first equation as we divide by a negative number: Assuming a common rc, this means that , but since we also have to satisfy the first equation, this will happen only if We also have to ensure that p 2 and p 3 are between 0 and 1: The other case is when b 2 > b 1 , i.e. the prize structure is concave. In this case, the relation mark does not turn, but since p 2 and p 3 were determined from the second and third equations that do not contain the relation mark, the solution will be the same. Substituting p 2 and p 3 back into the first equation yields rc ≥ b 1 +2b 2 3 . Therefore, this Nash equilibrium will appear if and only if 3.2.3 Case 7: one player plays fair with certainty, others play a mixed strategy (Fmm) Without loss of generality, assume that player 1 plays fair with certainty. The equations are: With a common rc, the second and third equations yield , and to satisfy the first one, we also need rc ≤ 2b 1 +b 2 3 . Since p 2 and p 3 must be between zero and one, the complete set of conditions is: 3.2.4 Case 8: two players cheat with certainty, one plays a mixed strategy (DDm) Without loss of generality, assume that players 1 and 2 cheat with certainty. The equations are: Assuming a common rc, the third equation clearly places a constraint on the occurrence of this NE -it will occur only for a particular value of rc. In that case, the first two equations will prescribe an upper or a lower bound on p 3 , depending on the sign of , but since we know rc exactly from the third equation, this yields p 3 ≥ 1. Since p 3 is a probability, the only allowed value is p 3 = 1, which is in fact case 1 (three players cheat with certainty). Thus, case 8 does not exist for concave prize structures.
When b 2 < b 1 , there will be a trivial upper bound on p 3 , i.e. p 3 ≤ 1, which is not a restriction on p 3 , meaning that case 8 is better marked as the DD * case.
3.2.5 Case 9: two players play fair with certainty, one plays a mixed strategy (FFm) Without loss of generality, assume that players 1 and 2 play fair with certainty. The equations are: Similarly to case 8, case 9 will either collapse to case 4 (FFF) when b 2 > b 1 , or will collapse to FF * if b 2 < b 1 .
3.2.6 Case 10: one player plays fair, one player cheats, the third is mixed (DFm) Without loss of generality, assume that player 1 plays fair, player 2 cheats and player 3 is mixed. The equations are: Let us assume a common rc: the only solution is p 3 = 0, collapsing this case into case 2 (DDF). If b 1 > b 2 , any p 4 satisfies the equations, making it equivalent to DF * .

Summary of the three-player case
All the conditions we have obtained for rc depended on three thresholds: . These are also equal to A 1 −Ā 0 , A 2 −Ā 1 and A 3 −Ā 2 . The behaviour of the model will depend on the relation of rc to these thresholds. First, let us assume that b 1 < b 2 , i.e. the prize structure is concave. In this case, we have the following: rc Cases of Nash-equilibria Between A 3 −Ā 2 and A 2 −Ā 1 DDF, Dmm, mmm Exactly A 2 −Ā 1 DDF, DFF, mmm with p = 0.5 Between A 2 −Ā 1 and A 1 −Ā 0 DFF, Fmm, mmm When the prize structure is convex, b 1 is larger than b 2 , leading to the following cases: rc Cases of Nash-equilibria and we obtain: rc Cases of Nash-equilibria Finally, we present an alternative representation of the the different types of Nash equilibria on Figure S2. On this figure, each Nash equilibrium is represented by a box and Nash equilibria corresponding to the same rc value are arranged in the same column. The value of the rc product increases from left to right. Arrows denote the "transition" of one type of Nash equilibrium into another (for instance, a DDD type Nash equilibrium into Dmm) as the value of rc changes. The upper panel contains the case of concave prize functions while the lower panel contains the convex case.

Nash-equilibria in the four-player case
In the four-player case, we know the following: We also know that The derivative of the payoff function Π 1 with respect to p 1 is then We have to decide whether all the derivatives are non-negative (given that all cheat, p 1 = p 2 = p 3 = p 4 = 1): Therefore, this particular configuration is a Nash equilibrium if and only if b 1 +2b 2 +3b 3 4 ≥ max i r i c i ; note that the left hand side is equal to A 4 −Ā 3 , which echoes the rule we have seen for the three player case, where the threshold was A 3 −Ā 2 .

Case 3: two players cheat, two play fair (DDFF)
Without loss of generality, we can assume that players 1, 2 cheat and players 3 and 4 play fair.
The equations to check are: Assuming a common rc yields: 3b 1 + 4b 2 + 2b 3 6 ≥ rc ≥ 2b 1 + 4b 2 + 3b 3 6 where the left hand side is equal to A 2 −Ā 1 and the right hand side is equal to A 3 −Ā 2 . For linear prize structures, this will collapse into rc = 3 2 b as above.

Case 4: one player cheats, three play fair (DFFF)
Without loss of generality, we can assume that player 1 cheats and players 2, 3 and 4 play fair. The equations to check are: Assuming a common rc yields: where the left hand side is equal to A 1 −Ā 0 and the right hand side is equal to A 2 −Ā 1 . For linear prize structures, this will collapse into rc = 3 2 b as above.

Case 5: everyone plays fair (FFFF)
We have to decide whether all the derivatives are negative (given that all play fair, p 1 = p 2 = p 3 = p 4 = 0): Therefore, this particular configuration is a Nash equilibrium if and only if 3b 1 +2b 2 +b 3 4 ≤ min i r i c i ; note that the left hand side is equal to A 1 −Ā 0 , which echoes the rule we have seen in case 4 of the three-player case.
Let us take a step back now again and examine what we have observed so far. The presence or absence of various pure Nash-equilibria depend on where the value of the rc product falls compared to 3b 1 +2b 2 +b 3 4 , 3b 1 +4b 2 +2b 3 6 , 2b 1 +4b 2 +3b 3 6 and b 1 +2b 2 +3b 3 4 . Furthermore, note that the thresholds are also equal to A 1 −Ā 0 , A 2 −Ā 1 , A 3 −Ā 2 and A 4 −Ā 3 . We will encounter these thresholds later when we allow for mixed strategies as well.

Mixed strategies 4.2.1 Case 6: everyone plays a mixed strategy (mmmm)
In this configuration, all the partial derivatives must be equal to zero. Let's see where this leads: and so on for r 2 c 2 , r 3 c 3 and r 4 c 4 . Here we assume a common rc again. For sake of simplicity, let us denote b 3 − 2b 2 + b 1 with C 1 , b 3 + 2b 2 − 3b 1 with C 2 and 3b 1 + 2b 2 + b 3 with C 3 : (p 2 p 3 + p 2 p 4 + p 3 p 4 )C 1 + (p 2 + p 3 + p 4 )C 2 = 12rc − 3C 3 (p 1 p 3 + p 1 p 4 + p 3 p 4 )C 1 + (p 1 + p 3 + p 4 )C 2 = 12rc − 3C 3 (p 1 p 2 + p 1 p 4 + p 2 p 4 )C 1 + (p 1 + p 2 + p 4 )C 2 = 12rc − 3C 3 (p 1 p 2 + p 1 p 3 + p 2 p 3 )C 1 + (p 1 + p 2 + p 3 )C 2 = 12rc − 3C 3 Because of the common rc, all the players are equivalent and all the equations are symmetric so we have no reason to assume that p 1 = p 2 or p 2 = p 3 or p 3 = p 4 . (Proof by contradiction: suppose that p 1 = p 2 ; in this case, the left hand side of the first two equations are different but the right hand side are the same, so they cannot be both satisfied). Switching to a common p gives us a single quadratic equation: C 1 p 2 + C 2 p + C 3 − 4rc = 0 and the two solutions are: However, since we also know that p must be between 0 and 1, we can bound rc from above and below to ensure that. Let us assume that p = 0; in this case, rc must be exactly C 3 /4 = 3b 1 +2b 2 +b 3 4 = A 1 −Ā 0 . Let us now assume that p = 1; in this case, rc must be C 1 +C 2 +C 3 It can easily be shown that any rc between these two extremes will yield a solution for p between 0 and 1, and that there will be only one such solution because the other solution of the quadratic equation will fall outside the allowed region.
When we compare this case with case 5 (mmm) of the three-player case, we can see that the rc region is similar as it fills the space on the rc axis between the "everyone cheats" and the "everyone plays fair" region. However, a significant difference is that in the three-player case, the actual value of p was a linear function of rc, while in this case it depends on √ rc. We conjecture that in the general n-player case, the value of p will depend on (rc) 1/(n−2) .
Finally, we can evaluate what happens when the prize structure is linear. In this case, C 1 = C 2 = 0 (since all the b's are equal), so the equation is not quadratic any more; we obtain 3 2 b = rc 4.2.2 Case 7: three players cheat with certainty, one plays a mixed strategy (DDDm) Without loss of generality, assume that players 1, 2 and 3 cheat with certainty. The equations are: The second equation gives us a condition that must hold irrespectively of the actual value of p 4 ; it dictates that b 1 +2b 2 +3b 3 4 = A 4 −Ā 3 = rc. When this holds, the first equation is simplified to (p 4 − 1)(2C 1 + C 2 ) ≥ 0 and since p 4 < 1 (note that p 4 plays a mixed strategy; we have already treated the case of p 4 = 1 in case 1), it follows that we need 3b 3 − 2b 2 − b 1 ≤ 0. Therefore, this Nash equilibrium arises if the following two conditions both hold at the same time: If both of the above conditions hold, any p 4 will satisfy the equations (including p 4 = 0 or p 4 = 1, which were already shown to be Nash-equilibria in Cases 1 and 2). Since any p 4 is suitable, this strategy then becomes DDD * .
For convex prize functions (b 1 ≥ b 2 ≥ b 3 ), the second condition trivially holds and all we need is rc = A 4 −Ā 3 . For linear prize functions, rc becomes 3 2 b. For concave prize functions (b 1 ≤ b 2 ≤ b 3 ), the second condition will never hold (unless b 1 = b 2 = b 3 ) and this case will not be a Nash-equilibrium. 4.2.3 Case 8: two players cheat with certainty, one plays fair with certainty, one plays a mixed strategy (DDFm) Without loss of generality, assume that players 1 and 2 cheat with certainty, and player 3 plays fair. The equations are: Again, the third equation will give us a necessary condition for this configuration to be a Nash-equilibrium: rc = A 3 −Ā 2 . When this holds, the remaining two equations are as follows: The implications are that C 1 + C 2 ≤ 0 and 2C 1 + C 2 ≤ 0. Substituting C 1 and C 2 back yields the following two conditions for this Nash-equilibrium, both of which must be satisfied: ).
If the above conditions both hold, any p 4 will satisfy the equations, including p 4 = 0 or p 4 = 1, which were already treated in Cases 2 and 3. Since any p 4 is suitable, this strategy then becomes DDF * .
For convex prize functions (b 1 ≥ b 2 ≥ b 3 ), the second condition trivially holds and all we need is rc = A 3 −Ā 2 . For linear prize functions, rc becomes 3 2 b. For concave prize functions (b 1 ≤ b 2 ≤ b 3 ), the second condition will never hold (unless b 1 = b 2 = b 3 ) and this case will not be a Nash-equilibrium.
4.2.4 Case 9: one player cheats with certainty, two play fair with certainty, one plays a mixed strategy (DFFm) Without loss of generality, assume that player 1 cheats with certainty, and players 2 and 3 play fair. The equations are: Again, the third equation will give us a necessary condition for this configuration to be a Nash-equilibrium: rc = A 2 −Ā 1 . When this holds, the remaining two equations are as follows: The first equation implies that C 2 ≤ 0 (since otherwise p 4 would have to be larger than 1, which is impossible). This is equivalent to b 3 + 2b 2 ≤ 3b 1 . The second equation implies that C 1 + C 2 ≤ 0 (again, otherwise p 4 would have to be negative, which is impossible). This is equivalent to b 3 ≤ b 1 . Therefore, this Nash equilibrium arises if the following two conditions hold: ).
If the above conditions both hold, any p 4 will satisfy the equations, including p 4 = 0 or p 4 = 1, which were already treated in Cases 3 and 4. Since any p 4 is suitable, this strategy then becomes DFF * .
For convex prize functions (b 1 ≥ b 2 ≥ b 3 ), the second condition trivially holds and all we need is rc = A 2 −Ā 1 . For linear prize functions, rc becomes 3 2 b. For concave prize functions (b 1 ≤ b 2 ≤ b 3 ), the second condition will never hold (unless b 1 = b 2 = b 3 ) and this case will not be a Nash-equilibrium. 4.2.5 Case 10: three players play fair with certainty, one plays a mixed strategy (FFFm) Without loss of generality, assume that players 1, 2 and 3 play fair with certainty. The equations are: The second equation says that this is a Nash-equilibrium only if rc = 3b 1 +2b 2 +b 3 4 = A 1 −Ā 0 . The first equation dictates that in this case, C 2 must be non-positive (because p 2 is non-negative and the right hand side is zero). Therefore, this Nash equilibrium arises if the following two conditions hold: If both the above conditions hold, any p 4 will satisfy the equations, including p 4 = 0 or p 4 = 1, which were already treated in Cases 4 and 5. Since any p 4 is suitable, this strategy then becomes FFF * .
For convex prize functions (b 1 ≥ b 2 ≥ b 3 ), the second condition trivially holds and all we need is rc = A 1 −Ā 0 . For linear prize functions, rc becomes 3 2 b. For concave prize functions (b 1 ≤ b 2 ≤ b 3 ), the second condition will never hold (unless b 1 = b 2 = b 3 ) and this case will not be a Nash-equilibrium. 4.2.6 Case 11: two players cheat with certainty, two play a mixed strategy (DDmm) Without loss of generality, assume that players 1 and 2 cheat with certainty. The equations are: Again, the last two equations are symmetric w.r.t. p 3 and p 4 so we can replace them with a common p. The first two equations collapse into one: Let us work out the second equation first: In order for the first equation to be true, we also need: At this stage, we must distinguish between three sub-cases: We also have to ensure that the calculated p is between 0 and 1. p > 0 is satisfied when rc > A 3 −Ā 2 and 3b 3 > 2b 2 + b 1 or rc < A 3 −Ā 2 and 3b 3 < 2b 2 + b 1 while p < 1 is satisfied when Let us start with the second equation: The other equation yields two roots: Now, if C 1 < 0, the equation is satisfied if p is between p 1 and p 2 , and if C 1 > 0, the equation is satisfied if p is less than min(p 1 , p 2 ) or greater than max(p 1 , p 2 ). Let's cover all the four cases: • If C 1 < 0 and C 2 < 0, p 2 is negative, so every p between 0 and 1 satisfies the equation, including p = 12rc−6b 1 −8b 2 −4b 3 2b 3 −2b 1 that we already determined.
• If C 1 < 0 and C 2 > 0, p 2 is positive. If C 2 < −C 1 , we have p > − C 2 C 1 , otherwise there is no valid p which gives a solution.
• If C 1 > 0 and C 2 < 0, p 2 is positive. If C 2 < −C 1 , we have p < − C 2 C 1 , otherwise there is no valid p which gives a solution.
• If C 1 > 0 and C 2 > 0, p 2 is negative, which excludes any valid p between 0 and 1 from being a solution.
To sum it all up, this Nash-equilibrium will arise if the following conditions all hold at the same time: 4.2.8 Case 13: two players play fair with certainty, two play a mixed strategy (FFmm) Without loss of generality, assume that players 1 and 2 play fair with certainty. The equations are: Again, the last two equations are symmetric w.r.t. p 3 and p 4 so we can replace them with a common p. The first two equations collapse into one: Let us work out the second equation first: In order for the first equation to be true, we also need: At this stage, we must distinguish between three sub-cases: We also have to ensure that the calculated p is between 0 and 1. p > 0 is satisfied when Note that we do not consider p = 0 or p = 1 because they have already been treated in cases 3 and 5. To sum it all up, this Nash-equilibrium will arise if the following conditions all hold at the same time: 4.2.9 Case 14: one player cheats with certainty, others play a mixed strategy (Dmmm) Without loss of generality, assume that player 1 cheats with certainty. The equations are: Again, the last three equations are symmetric w.r.t. p 2 , p 3 and p 4 so we can replace them with a common p: Both quadratic equations have two roots; let us work out the roots of the second equation first: Since p must lie beween 0 and 1, we can also use the second equation to bound rc from above and below. Let us first assume that p = 0; in this case, rc = C 2 +3C 3 12 = A 2 −Ā 1 . Similarly, when p = 1, rc becomes C 1 +C 2 +C 3 4 = A 4 −Ā 3 . Therefore, it must be ensured that rc is between A 2 −Ā 1 and A 4 −Ā 3 in order for this Nash-equilibrium to arise.
The first equation also gives us some further necessary conditions on rc; the roots here are similar to the roots of Case 6 (note that the equation is the same): When C 1 < 0, p must be between p 1 and p 2 ; when C 1 > 0, p must be less than min(p 1 , p 2 ) or greater than max(p 1 , p 2 ).
4.2.10 Case 15: one player plays fair with certainty, others play a mixed strategy (Fmmm) Without loss of generality, assume that player 1 plays fair with certainty. The equations are: (p 2 p 3 + p 2 p 4 + p 3 p 4 )C 1 + (p 2 + p 3 + p 4 )C 2 ≤ 12rc − 3C 3 p 3 p 4 C 1 + (p 3 + p 4 )C 2 = 12rc − 3C 3 p 2 p 4 C 1 + (p 2 + p 4 )C 2 = 12rc − 3C 3 p 2 p 3 C 1 + (p 2 + p 3 )C 2 = 12rc − 3C 3 Again, the last three equations are symmetric w.r.t. p 2 , p 3 and p 4 so we can replace them with a common p: Both quadratic equations have two roots; let us work out the roots of the second equation first: Since p must lie beween 0 and 1, we can also use the second equation to bound rc from above and below. Let us first assume that p = 0; in this case, rc = C 3 4 = A 1 −Ā 0 . Similarly, when p = 1, rc becomes C 1 +2C 2 +3C 3 12 = A 3 −Ā 2 . Therefore, it must be ensured that rc is between A 1 −Ā 0 and A 3 −Ā 2 in order for this Nash-equilibrium to arise.
The first equation also gives us some further necessary conditions on rc; the roots here are similar to the roots of Case 6 (note that the equation is the same): When C 1 > 0, p must be between p 1 and p 2 ; when C 1 < 0, p must be less than min(p 1 , p 2 ) or greater than max(p 1 , p 2 ).

Summary of the four-player case
All the conditions we have obtained for rc depended on four thresholds: The behaviour of the model will primarily depend on the relation of rc to these thresholds as each region allows only specific types of Nash equilibria. Note that some of the cases as discussed above also depend on other criteria that are mostly related to the b i variables that define the shape of the prize function.
In the three-player case that we discussed before, prize functions were categorized into three classes (concave, convex and linear ones) depending on the relation of b 1 to b 2 . In the four-player case, we have b 1 , b 2 and b 3 , therefore the classification is not so straightforward. However, we can say that "realistic" prize functions are typically convex (i.e. b 1 ≥ b 2 ≥ b 3 ), and we can also evaluate linear (b 1 = b 2 = b 3 ) and concave (b 1 ≤ b 2 ≤ b 3 ) prize functions to complement our analysis and to make it comparable with the three-player case.
Let us start with the concave case: , the table is simplified to the following as all the thresholds collapse into a single one: rc Cases of Nash-equilibria Less than 3 2 b DDDD Exactly 3 2 b * * * * Greater than 3 2 b FFFF Finally, we present an alternative representation of the the different types of Nash equilibria on Figure S3. On this figure, each Nash equilibrium is represented by a box and Nash equilibria corresponding to the same rc value are arranged in the same column. The value of the rc product increases from left to right. Arrows denote the "transition" of one type of Nash equilibrium into another (for instance, a DDDD type Nash equilibrium into Dmmm) as the value of rc changes. The upper panel contains the case of concave prize functions while the lower panel contains the convex case.  Figure S1: The original pay-off matrix from [1].  Figure S2: Nash equilibria in the three-player doping game as a function of the value of rc. Each Nash equilibrium is represented by a box. Nash equilibria with the same rc product are in the same column. rc increases from left to right. An arrow points from one Nash equilibrium into another if changing rc transforms a Nash equilibrium into another one.   Figure S3: Nash equilibria in the four-player doping game as a function of the value of rc. Each Nash equilibrium is represented by a box. Nash equilibria with the same rc product are in the same column. rc increases from left to right. An arrow points from one Nash equilibrium into another if changing rc transforms a Nash equilibrium into another one.