## Prisoner’s dilemma and game shows

Variation of prisoner’s dilemma is used in several games shows:

In Golden Balls, the rules are:

After five balls have been won, the contestants make one last decision to determine the final jackpot division. Each contestant chooses one of two final golden balls, one with “Split” printed on the cash background inside it, and one with “Steal” printed on the killer background inside it.

If both contestants choose a Split ball, the jackpot is split equally between them and they both go home with half the money they’ve won.

If one contestant chooses a Split ball and the other chooses a Steal ball, the Stealer goes home with all the money and the Splitter goes home empty-handed.

If both contestants choose Steal balls, they both go home empty-handed.

In Friend or Foe?, the rules are:

Both eliminated team members approach the “trust box” and make their arguments as to why their partner should vote “friend.” Both team members then secretly vote either “friend” or “foe.” If both team members vote friend, they split the trust fund evenly. If one member votes friend and the other votes foe, the team member who voted friend receives nothing and their teammate who voted foe wins the entire trust fund. However, if both team members vote foe, neither team member wins anything.

In Take it All, the rules are:

The final two contestants each select a random card from a deck of ten containing values ranging from $25,000 to $250,000. The values chosen are not revealed to either contestant upon selection. The two remaining contestants then separately and secretly decide to “keep mine” or “take it all”. If both contestants choose “keep mine”, each leaves the game with the prizes already accumulated earlier in the game, along with the cash value selected at the start of the Prize Fight. If one contestant chooses “keep mine” and the other chooses “take it all”, the contestant who chose “take it all” collects not only their prizes and cash award, but also the other contestant’s prizes and cash award, and the contestant who chose “keep mine” leaves empty-handed. However, if both players choose “take it all”, both players forfeit all prizes accumulated during the game as well as the cash awards, and both leave empty-handed

In Shafted, the rules are:

During the final round, the two remaining players stand behind podiums opposite each other; they are playing for the amount of money the leading player had at the end of the previous round. Each of the players is asked if they wish to “share” or to “shaft”. If both players decide to shaft, both players walk away empty-handed. If one decides to share and the other to shaft, then the person who shafted wins all the prize money; if they both decide to share the money is divided equally between the two players.

These games all uses one variation of the prisoner’s dilemma. In this variation of the prisoner’s dilemma, if one prisoner chooses to betray the other and confess, the other prisoner’s payoff remains the same no matter what choice is made.

The pay-offs are as follows

(a is a positive value, b is a positive value and b > a)

p2: do not betray the other | p2: betray the other | |

p1: do not betray the other | (p1: -a, p2: -a) | (p1: -b, p2: 0) |

p1: betray the other | (p1: 0, p2: -b) | (p1: -b, p2: -b) |

Analysis:

In this case, if prisoner 1 chooses to betray prisoner 2, prisoner 2’s choices are equal in terms of how many years he would be in prison,

If prisoner 1 chooses not to betray prisoner 2, prisoner 2 should betray prisoner 1 because in this way he will get less years in prison.

If prisoner 2 chooses to betray prisoner 1, prisoner 1’s choices are equal in terms of how many years he would be in prison,

If prisoner 2 chooses not to betray prisoner 1, prisoner 1 should betray prisoner 2 because in this way he will get less years in prison.

Mixed Strategy Nash Equilibrium

Prisoner 1 has possibility p to not betray prisoner 2

Prisoner 2 has possibility q to not betray prisoner 1

for prisoner 1:

-a*q – b*(1-q) = 0*q – b*(1-q)

-a*q = -b + b*q – b*q + b

a*q = 0

q = 0

for prisoner 2:

-a*p – b*(1-p) = 0*p – b*(1-p)

-a*p = -b + b*p – b*p + b

a*p = 0

p = 0

So both of them have more incentive to betray the other and confess.

Similarly in a game:

(a is a positive monetary value, b is also a positive monetary value and b > a)

p2: not steal | p2: steal | |

p1: not steal | (p1: a, p2: a) | (p1: 0, p2: b) |

p1: steal | (p1: b, p2: 0) | (p1: 0, p2: 0) |

Analysis:

In this case, if player 1 chooses to take play 2’s money, player 2’s choices are equal in terms of how many money he will get,

If player 1 chooses to just take his own share of the money, player 2 should choose to take player 1’s money because in this way he will get all the money.

If player 1 chooses to take play 2’s money, player 2’s choices are equal in terms of how many money he will get,

If player 1 chooses to just take his own share of the money, player 2 should choose to take player 1’s money because in this way he will get all the money.

Mixed Strategy Nash Equilibrium:

Player 1 has possibility p to not take player 2’s money

Player 2 has possibility q to not take player 1’s money

for player 1:

a*q + 0*(1-q) = b*q – 0*(1-q)

a*q = b*q

q = 0

for player 2:

a*p + 0*(1-p) = b*p – 0*(1-p)

a*p = b*p

p = 0

So both of them have more incentive to take the other’s money. Thus, in the game shows, people will be more likely to choose to steal the opponent’s money (or take all the money) instead of choosing to share the money with the opponent. And this will result in both of them not getting any money.

Source: https://en.wikipedia.org/wiki/Friend_or_Foe%3F_(TV_series)

https://en.wikipedia.org/wiki/Golden_Balls

https://en.wikipedia.org/wiki/Take_It_All_(game_show)

https://en.wikipedia.org/wiki/Shafted