0
$\begingroup$

I am considering the iterated prisoners dilemma. In particular, suppose after each game, the game is played again with probability $p$. Additionally, assume it is known that my opponent will confess on every turn once I have confessed once, but will stay mute until so.

Just to reiterate the rules, if A confesses and B does as well, both get $5$ years in jail. If one is mute but the other confesses, the one who was mute gets $10$ years in jail. If both remain mute, both get 1 year in jail.

So far, I've calculated that following: The expected payoff is,
$$5 - \frac5{(1-p)}$$ if one always confesses. Additionally, if I always stayed mute it would be $$\frac{−1}{(1 − p)}$$ Thus, the latter strategy is better if $p > 1/5$.

However, there are some pieces here I do not understand. For example, let $Y_r = 1$ if there are at least $r$ rounds of the game (and $0$ otherwise). I am expected to find the value of $E[Y_r]$ as a function of $p$ and $r$.

From intuition, it is clear that $E[Y_1] = 1$ and $E[Y_2] = p$. Since the stopping condition is geometric with $\frac{1}{1-p}$, I am unsure on how to implement this into getting the expected number of rounds played.

Additionally, I am trying to calculate the expected payoff I would receive if my strategy was to stay mute on the first game and then confess on any subsequent game. I was thinking to obtain this payoff as a linear combination of $Y_1 , Y_2 , Y_3 , \ldots ,$ and use linearity of expectation to obtain my equation.

How can I properly calculate $Y_r$ and the expected payout should I remain mute on the first game and confess on any subsequent game?

$\endgroup$
1
$\begingroup$

Denote by $u^t_i$ the payoff obtained by Player~$i$ in round~$t$, and assume zero payoffs once the game stops. Since it is not mentioned, I ignore the possibility of a discount factor and let $\delta=1$.

The (random) sum of payoffs to $i$ is $$\sum_{t=0}^{+\infty} u^t_i Y_t$$ where $Y_t$ has a bernoullian distribution with $P(Y_t = 1) = p^t$ because the probability of a round is $p$ and thus the probability of at least $t$ rounds is $p^t$.

Therefore, the expected payoffs to $i$ is $$\sum_{t=0}^{+\infty} u^t_i E(Y_t) = \sum_{t=0}^{+\infty} u^t_i p^t$$

Now you can apply this formula to specific cases. F.i., if you stay mute in the first period and then always confess, you get $$-1 + \sum_{t=1}^{+\infty} (-5) p^t = -1 + \frac{5p}{1-p}$$ because $$\sum_{t=1}^{+\infty} p^t = \sum_{t=0}^{+\infty} p^t - 1 = \frac{1}{1-p} - 1 = \frac{p}{1-p}$$

Likewise, you can rationalise the other payoffs presented in your question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.