The Poisoned Cup, revisited

[Yoruichi-san]

Many games, in fact, probably most games, do not have an optimal solution or Nash equilibrium, that is, if a player chooses one strategy, his opponent(s) always can do better for themselves by changing to a different strategy.

One such example is the infamous "which is the poisoned cup?" scenario, in which, for example, if the poisoner picks the strategy "he thinks I'm going to put it in his cup, so I'll put it in mine", then the poisonee is incentivized to move to the strategy "he thinks I think he's going to put it in my cup, so he'll put it in his, so I'll pick mine," and the poisoner then is incentivized to change to "he thinks I think he thinks..." and so on and so forth.

However, there are strategies that, once adopted, force a local equilibrium, that is, the players involved have no incentive to change strategies.

Now let's make this slightly more interesting...let's say that one cup contains water and the other coffee, which, due to its antioxidant properties *cough* makes the poison 20% less effective, that is, it will only kill you 80% of the time. Find equilibrium strategies for this scenario.

Edited June 8, 2012 by Yoruichi-san

[WitchOfSecrets]

I don't know microecon stuff well, but my intuition says that the equilibrium strategies for the original poisoned cup scenario are:

both choose at random. There's no reason to switch to a different strategy if your opponent's choosing randomly.

Let's see if I can start from that assumption and see how adding coffee switches the equilibrium...

[WitchOfSecrets]

Here we go.

Let's assume that we're going for two strategies where each person has a certain probability of choosing the coffee, and one-minus-that probability of choosing the water. Let's also say that the cost of being poisoned is "C" and benefit of poisoning someone else is "B".

Call the probability that the poisoner poisons the coffee X. The probability that they poison the water is 1-X.

Call the probability that the poisonee drinks the coffee Y, and the probability that they don't 1-Y.

The expected gain the poisoner has for a given strategy is:G = .8X*Y*B + (1-X)(1-Y)*B

The expected loss that the poisonee has for a given strategy is : L = -.8X*Y*C - (1-X)(1-Y)*C = - .8CXY -1C + XC + YC- XYC

Now, the poisonee can change Y, but they have no control over X. We can express the effect of changing their probability of drinking coffee by taking the derivative dL/dY and treating X as a constant. So dL/dY = -.8CX + C - CX = C(-.8X + 1 - X) = C(-1.8X + 1)

To find an equilibrium point, I'm thinking we have to set this derivative to zero - that's a place where changing Y doesn't alter L. So 0 = -1.8X + 1. X = 5/9. So if the poisoner chooses coffee 5/9 of the time, any strategy works equally well for the poisonee.. Now let's see what the poisonee does. L = -G if we set B and C to 1 for convenience, so their derivatives should be zero at the same value, I think.

I think that my summary was incomprehensible! But I believe that the poisonee should ALSO choose coffee 5/9 of the time.

This results in the poisonee dying 4/9 of the time.

Edited June 9, 2012 by WitchOfSecrets

[Anza Power]

Poison both cups.

[Half Fast]

Not too good at reckoning so I'll put it in the coffee, it'll probably kill them and might not kill me depending on who gets it, providing putting it in neither is not an option.

[Yoruichi-san]

Here we go.

Let's assume that we're going for two strategies where each person has a certain probability of choosing the coffee, and one-minus-that probability of choosing the water. Let's also say that the cost of being poisoned is "C" and benefit of poisoning someone else is "B".

Call the probability that the poisoner poisons the coffee X. The probability that they poison the water is 1-X.

Call the probability that the poisonee drinks the coffee Y, and the probability that they don't 1-Y.

The expected gain the poisoner has for a given strategy is:G = .8X*Y*B + (1-X)(1-Y)*B

The expected loss that the poisonee has for a given strategy is : L = -.8X*Y*C - (1-X)(1-Y)*C = - .8CXY -1C + XC + YC- XYC

Now, the poisonee can change Y, but they have no control over X. We can express the effect of changing their probability of drinking coffee by taking the derivative dL/dY and treating X as a constant. So dL/dY = -.8CX + C - CX = C(-.8X + 1 - X) = C(-1.8X + 1)

To find an equilibrium point, I'm thinking we have to set this derivative to zero - that's a place where changing Y doesn't alter L. So 0 = -1.8X + 1. X = 5/9. So if the poisoner chooses coffee 5/9 of the time, any strategy works equally well for the poisonee.. Now let's see what the poisonee does. L = -G if we set B and C to 1 for convenience, so their derivatives should be zero at the same value, I think.

I think that my summary was incomprehensible! But I believe that the poisonee should ALSO choose coffee 5/9 of the time.

This results in the poisonee dying 4/9 of the time.

Very good! You even did the full analysis with calculus...hmm...I'm going to have to figure out a special symbol to give out...maybe the ITG double star?

Even though you might not know the genre if you're unfamiliar with game theory, what you're doing is constructing a probabilistic strategy, and the easy way to calculate the equilibrium point is found is to choose P so that the expectation values are equal, i.e. P*(1)=(1-P)(.8), solve for P.

That was what I was expecting (no diddiddiddiddid(114)diddiddiddiddid) , but you went one better by proving it is the equilibrium point. Excellent work.