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The Game of Death is a typical game that teenagers play in South Korea. We play “The Game of Death” with N people. If you are able to avoid saying number 0 up to the end of the game, you win. We’ll explore the winning strategy of a team game in The Game of Death using the pigeonhole principle and concept of a prime number. In the team game, the starting team wins the game if the starter says the prime number bigger than N+1. Moreover, there is no obvious winning strategy for the non-starting team, but the optimal solution exists.
It was interesting to apply mathematical concepts to simple games.
We play “The Game of Death” with N people. If you are able to avoid saying number 0 up to the end of the game, you win. Before the game is started, each person designates and points out another person to take turns after you. When you get pointed out, then you should point out the person you designated formerly, and you cannot change the person later.
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The game starts as the first player announces one natural number K and then points out the person the first player had formerly designated. Then the next person pointed out by the first player shouts the number K - 1 and then points out the third person that the second player formerly designated. The game goes on as the next person keeps shouting one smaller number than the previous number. The person who gets pointed out by the player who shouts number 1 has to say number 0 and loses the game in the end.
In a team game, the whole team loses the game if one of the teammates said 0.
For example, we can look at the following case.
In this case, N=6 and K=14.
So Player 4 will lose this game.
We’ll explore the winning strategy of team games by mathematically proving it. Winning strategy in individual game exists, so we’ll focus on winning strategy for the starter in the team game, and the optimal winning strategy for the non-starter in the team game.
When Team A starts the game, the first player is represented by A1, and A1’s teammates are represented by A2, A3, and so on. Their opposition is represented by B1, B2, and so on. There should be the same number of Team A and Team B. Trivial winning strategies such as A1 announcing 1 and nominating B1, or A1 announcing the number that makes no chances for Team B to nominate will not be considered. For example, in the following case, N=4 and K = 4, and the cycle is A1→A2→A3→A4→B1.
In a team game, the key point is to involve fewer people in the cycle since if there are more teammates involved in a cycle, since it increases the possibility to call 0 and lose the game in a cycle.We’ll focus on 4:4 team game to obtain the winning strategy. As a strategy, suppose A1, A2, A3, and A4 all nominate B1 in their turn. When the game starts, A1 nominates B1 and announces K. This can be divided into two cases: B1 nominates B2, or B1 nominates A(A represents any people in Team A, because the person who gets nominated is insignificant as Team A all nominates B1).
(Case 1) B1 nominates B2
For convenience, assume the next person B1 nominates in Team B is B2. B2 will not nominate B1 because They will nominate each other for the rest of the game and lose the game. This case can be divided into two different cases by whom B2 nominates: B2 nominates B3, or B2 nominates A.
(Case 1-1) B2 nominates B3
For convenience, assume the next person B2 nominates in Team B is B3. B3 will not nominate B1 nor B2 because it will then cause a cycle in their team and subsequently lose the game. This case can be divided into two different cases: B3 nominates B4, or B3 nominates A.
(Case 1-1-1) B3 nominates B4
Similar to B3, B4 will not nominate a person in Team B, and B4 should nominate person in Team A. Then the following cycle recurs: A1→B1→B2→B→B4→A→B1→B2… From now on, in order to avoid confusion, arrows in the diagram will be omitted if the player didn’t have a chance. In this case, there are 5 people in the cycle, so if K is not multiple of 5, Team A wins the game.
(Case 1-1-2) B3 nominates A
Following cycle recurs: A1→B1→B2→B3→A→B1→B2→B3→A… In this case, there are 4 people in a cycle, so if K is not multiple of 4, team A wins the game.
(Case 1-2) B2 nominates A
Following cycle recurs: A1→B1→B2→A→B1→B2→A… There are 3 people in the cycle, so if K is not multiple of 3, Team A wins the game.
(Case 2) B1 nominates A
The following cycle would recurs: A1→B→A→B→A… There are 2 people in the cycle, so if K is not multiple of 2, Team A wins the game. In summary, K ≠ 2m, 3m, 4m, 5m (m is an arbitrary natural number). There is a sufficient amount of numbers that are meeting those conditions. For example, K can be 7, 11, 13, 17, 19, 23.
Since Team A can make P, the number of people in the cycle, from 2 (one A and one B) to N + 1 (one A and N B), general formula can be derived. K should not be a multiple of 2, 3, 4, …., N - 1, N, and N + 1. Similar to the individual game, the least common multiple of numbers from 2 to N + 1 plus 1 or (N + 1)! + 1 is possible. However, since these numbers are all hard to find and prime numbers (a whole number greater than 1 and impossible to be expressed as a product of two numbers that are less than the number) greater than N + 1 are a subset of these numbers, a prime number greater than N + 1 is enough. Therefore, if K is a prime number greater than N + 1, Team A wins the game.
However, what if Team A is not always starting? In this case, assume that Team B doesn’t know the winning strategy since if they know it, then they will definitely win the game. So Team B doesn’t call prime numbers or doesn’t nominate one single person like all team B people nominating A1. This case can be divided into two cases: B1 nomintates B2 or A1.
(Case 1) B1 nominates B2
This case can be divided into two cases: B2 nominates B3 or A1.
(Case 1-1) B2 nominates B3
This case can be divided into two cases: B3 nominates B4 or A1.
(Case 1-1-1) B3 nominate B4
B4 wouldn’t nominate a person in Team B because the cycle will recur in their team. B4 should nominate A1 (for convenience, assume that the first person to get nominated in Team A is A1). A1 can create a cycle at this stage. If the number of people in a cycle is P, the person who is continuously saying multiple of P loses because he will eventually say 0 as well. In a 4:4 game, there can be 7 different Ps: 2, 3, 4, 5, 6, 7, 8. (P = 2) To make P = 2, A1 should nominate B4 right back.
Then the cycle becomes B1→B2→B3→B4→A1→B4→A1→B4…. To win the game, A1 needs to avoid saying multiple of P. It means that A1 should say 2m+1 every cycle. If B1 says 2m+1, then B1 says (2m+1) → B2 says (2m) → B3 says (2m+1) → B4 says (2m) → A1 says (2m+1). Thus, Team A wins the game. We’ll call this backtracking, “an algorithm which is allowed to run forward until a dead end is reached, at which point previous steps are retraced and the algorithm is allowed to run forward again” [1].
(P = 3) To make P = 3, A1 should nominate B3.
Then the cycle becomes B1→B2→B3→B4→A1→B3→B4→A1→B3→B4→A1→B3….
To win the game, A1 needs to avoid saying multiple of P. This means that B1, B2, B3, or B4 should say multiple of 3 every cycle, and A1 should say 3m+1 or 3m+2 every cycle. When it is backtracked, Team A wins the game if B1 calls 3m+2 or 3m. Again, A1 should not nominate A2 to make a cycle of three because the number that A1 says will not change. A1 will always say 3m even A1 nominates A2 and A2 nominates B1 to make P = 3. In addition, A2 also needs to avoid saying multiple of P.
(P = 4) To make P = 4, A1 should nominate B2. Then the cycle becomes B1→B2→B3→B4→A1→B2→B3→B4→A1→B2…. So B1, B2, B3, or B4 should say multiple of 4 every cycle. When it is backtracked, Team A wins if B1 says 4m+1, 4m+2, or 4m+3. (P = 5) To make P = 5, A1 should nominate B1. Then the cycle becomes B1→B2→B3→B4→A1→B1→B2→B3→B4→A1→B1…. So B1, B2, B3, or B4 should say multiple of 5 every cycle. When it is backtracked, Team A wins if B1 says 5m, 5m+1, 5m+2, or 5m+3.
(P=6) If P exceeds 5, which is the number including oneself and all the Team B people, then A1 should nominate someone in their team. In this case, A1 should nominate A2, and A2 should nominate B1 to make P=6.
Then the cycle becomes
B1→B2→B3→B4→A1→A2→B1→B2→B3→B4→A1→A2→B1…. So B1, B2, B3, or B4 should say multiple of 6 every cycle. When it is backtracked, Team A wins if B1 says 6m,
6m+1, 6m+2, and 6m+3.
(P = 7) To make P = 7, A3 should nominate B1.
The cycle becomes
B1→B2→B3→B4→A1→A2→A3→B1→B2→B3→B4→A1→A2→A3→B1.... So B1, B2, B3, or B4 should say multiple of 7 every cycle. When it is backtracked, Team A wins the game if B1 says 7m, 7m+1, 7m+2, or 7m+3.
(P=8) To make P = 8, A4 should nominate B1. Then the cycle becomes
B1→B2→B3→B4→A1→A2→A3→A4→B1→B2→B3→B4→A1→A2→A3→A4→B1…. A1, A2, A3, and A4 need to avoid saying multiple of P. Team A win this game if B1, B2, B3, or B4 says multiple of P. When it is backtracked, Team A wins the game if B1 says 8m, 8m+1, 8m+2 or 8m+3.
At this stage, we can derive a general formula. Factors affecting the result are the number of people, N, the number first player has announced, K, the number of people that spoke before passing to Team A, S, and the number of people in a cycle Team A can create, P. This whole scenario when Team B starts can be divided into two cases: only one person is involved in a cycle from Team A, or more than one person is involved in a cycle from Team A. (Case 1) Only one person is involved in a cycle from Team A.
When A1 is trying to make a cycle that has P people, A1 can make a value of P from 2 (nominating BS) to S + 1 (nominating B1). The number of people who spoke but not involved in a cycle is S - (P - 1). A1 needs to avoid saying multiple of P, so K - S should not be multiple of P. Thus, A1 should find P that is not a factor of K – S, and nominate B(S - (P -1 + 1)), to win the game. (Case 2) More than one person is involved in a cycle from Team A
When A1 tries to make a cycle that has P people, A1 can make a value of P from 3 to N + S. In this case, to minimize the number of people in Team A in the cycle, S people in Team B are all involved in a cycle. Thus, the remainder of K divided by P should be smaller than S for one of the B1, B2, … BS to say multiple of P in every cycle. Since this winning strategy is obtained, it is trivial and repetitive to do the rest of the cases.
We explored winning strategies of “The Game of Death” in team games. In team games, if the game is N:N and Team A is starting, A1 should call a prime number that is bigger than N+1.
However, when Team B starts the game, there is no absolute guarantee to win the game this time. When Team B passes to Team A on Sth person and K-S is not multiple of P (2≤P≤S+1), or the remainder of K divided by P is equal or smaller than S (S+2≤P≤2N), then Team A can win the game. The number of people is N, the number first player has announced is K, the number of people that spoke before passing to Team A is S, and the number of people in a cycle Team A can create is P.
It was interesting to apply math in playing a simple fun game. This kind of calculation using number theory can be utilized in different places, not restricted to winning strategy. For example, this is applicable to making cookies. Leftover cookie dough is used again to minimize the waste in the dough and make extra cookies. Moreover, when considering many different types of cases, such as making an optimal decision or investment, this kind of calculation can be applied as well.
A Study on the Winning Strategy of “The Game of Death”. (2024, Feb 21). Retrieved from https://studymoose.com/document/a-study-on-the-winning-strategy-of-the-game-of-death
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