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The Salem Witches Problem – Can they use game theory to survive?

March 29, 2017

Two suspected witches of Salem are subjected to a test by the Witchfinder General.

To ascertain whether they have magical powers of telepathy (They haven’t, by the way) they will be separated and seated at a table in the blue room (Suspect 1) and the yellow room (Suspect 2). They will be unable to see each other or communicate in any way.

Before being separated they are allowed a few private moments together.

After being separated, they are given a deck of cards each and asked to extract one card from the deck.

They are allowed to look at their chosen card if they wish, but what they must actually do is to name the colour of the card that the other suspect has drawn.

It is a standard deck of cards, so there is a 1 in 2 chance the chosen card is black, and the same that it is red.

The game will be repeated ten times, to reduce the chance that they will survive by simple good fortune.

If in any round they both correctly identify the colour of the other person’s card, then they will both die.

If both suspects are wrong, or one is wrong, in every round, then both are free to go.

There are two questions:

1.     What is the probability they will survive by chance?

2.     Is there a co-operative strategy they could agree on before being separated to guarantee they both survive?

Think about it: In any round, what is the chance that each suspect will correctly name the colour of the other suspect’s card? A half? A quarter? What about over ten successive rounds?

To survive, they must avoid this over ten rounds. Is there a way they can take chance out of it, and make sure that at least one of them names the wrong colour for the other suspect’s card, for ten rounds in a row.

If so, that is the door to freedom. Remember that they can secretly hatch a joint strategy and they either both survive or both die, so they can trust each other to stick to the plan, if there is one.

 

Spoiler Alert (The Solution)

In the first round, the chance that the suspect in the blue room will correctly name the colour of the other suspect’s card is ½. Similarly for the suspect in the yellow room.

These are independent events, so the probability of being condemned after first hands are dealt (i.e. both name the colour of the other suspect’s card correctly) = ½ x ½ = ¼.

So probability of surviving first hand = ¾

Probability of surviving 10 hands = (3/4)10 = 0.0563, i.e. 5.63%

But there is a strategy to ensure survival, if they can agree on it before.

 

Can you work it out?

The solution is for player 1 to guess the same colour as his own card, and player 2 to guess a different colour to his card. This way  they will always survive.

Thus:

Red Red gives Red Black – they survive.

Black Black gives Black Red – they survive.

Black Red gives Black Black – they survive.

Red Black gives Red Red – they survive.

To better conceal the strategy, they could also decide to alternate roles.

This is the optimal outcome in a game where the two players are able to co-ordinate a strategy in advance, and where trust is guaranteed because they both stand to gain by sticking to the strategy.

There are other scenarios in which the superior strategy from the point of view of one or both players is to defect from the strategy they would adopt if they were free to strike an enforceable deal. One such scenario is known as the Prisoner’s Dilemma problem. In this problem, the optimal strategy for each player, when a deal cannot be enforced, is to choose a strategy worse than they could reach co-operatively. This is in turn an example of a Nash equilibrium in which one or both players stand to gain by switching strategy from the current one.

These cases will be examined another time when we look at Game Theory and the important role of the Nash Equilibrium in it.

 

Further Reading and Links

https://selectnetworks.net

 

 

 

 

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