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Could somebody explain how Straffin calculated the security levels for Rose and Colin on p. 103? As far as I understood, Colin's security level is 6 because it is the minimum of the strategy Colin A because Colin A dominates Colin B. If both, Colin A and Colin B were considered, Rose should follow a mixed strategy of 8/9 for Rose A and 1/9 for Rose B, which leads to a security level of 20/9, but only if Colin A was considered. I don't understand, how Straffin came to the security level of 10/3 for Rose and why the strategy BA should be played at least 1/3 of the time.

Here are the two pages that describe the problem (I apologize for the bad resolution): enter image description here

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  • $\begingroup$ (1) This is impossible to answer for anyone who doesn't have the book at hand. Please add more information. (2) This sounds like straightforward game theory, not statistics. $\endgroup$ – Stephan Kolassa Jul 17 '18 at 9:36
  • $\begingroup$ Thank you for your comment. I deleted the statistics tag and I uploaded the two pages, in which the left page describes the problem and the right page describes the results. $\endgroup$ – Joram Schito Jul 17 '18 at 10:24
  • $\begingroup$ Although I view game theory as one theoretical underpinnning of statistical theory, I suspect most users of this site will perceive this topic as peripheral at best. You might get better responses on the SE economics site. $\endgroup$ – whuber Jul 17 '18 at 11:13
  • $\begingroup$ OK, thank you! I will try it also on the SE economics site. $\endgroup$ – Joram Schito Jul 17 '18 at 11:39
  • $\begingroup$ Possible duplicate of Straffin, P. D. (1993): Game Theory and Strategy. How is the security level on p. 103 calculated? $\endgroup$ – Giskard Jul 17 '18 at 13:01
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Suppose Colin plays A with probability c and B with probability 1-c, while Rose plays A with probability r and B with probability 1-r. Then Rose's payoff is

$2rc+4(1-r)c+10r(1-c)+0(1-r)(1-c)$

Rose's security level is calculated by taking the max-mini payoff: that is, for each value of r, find the minimum payoff over all c, then find the maximum such value.

So, first, we can simplify the above expression:

$2rc+4c-4rc+10r-10rc = 4c+10r-12rc$

As a function of c, this is a line with intercept $10r$ and slope $4-12r$.

Case I: $r = \frac13$, this line is a constant $10r$. Plug $r =\frac13$ in and you get $\frac{10}3$

Case II: $r > \frac13$, then it's negatively sloped, and the minimum occurs at $c = 1$, with a payoff of $4-2r$. Since $r > \frac13$, we have that the payoff is less than $4-2/3 = \frac{10}3$.

Case II: $r < \frac13$, then it's positively sloped, and the minimum occurs at $c = 0$, with a payoff of $10r$. Since $r < \frac13$, the payoff is less than $\frac{10}3$.

In other words, if Rose plays A $\frac13$ of the time, then she gets an average of $\frac{10}3$ regardless of what Colin does. If she plays A less than $\frac13$ of the time, then if Colin plays B, then she gets 10 less than $\frac13$ of the time and 0 the rest of the time, getting her less than $\frac{10}3$. If she plays A more than $\frac13$ of the time, then if Colin plays A, she gets 2 more than $\frac13$ of the time and 4 the rest of the time, getting her less than $\frac{10}3$.

So Rose can guarantee herself $\frac{10}3$ by playing A $\frac13$ of the time, and every other strategy has a possibility of giving her less than that. Thus, $\frac{10}3$ is her security level.

The reason BA should be played at least $\frac13$ of the time is that if BA is played $\frac13$ of the time and AB is played $\frac23$ of the time, then Colin's payoff is $8*\frac13+ 5*\frac23 = \frac83 + \frac{10}3 = \frac{18}3=6$, which is his security level. If BA is played less than $\frac13$ of the time, then his payoff goes below his security level, so he would be better off not cooperating.

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  • $\begingroup$ Thank you, @Acccumulation, for your reply. I understood that Rose’s payoff will be constantly $\frac{1}{3}$ if Rose plays A $\frac{1}{3}$ of the time, no matter what Colin does. By using Geogebra, I also understood that Rose’s payoff is minimal in case II if $c=1$ and in case III if $c=0$. How did you came to this conclusion? Just by trying or by using calculus? Did I understand you correct that you determined Rose’s minimum payoff in terms of “worst case scenario” what happened under the condition that Colin would do the worst? $\endgroup$ – Joram Schito Jul 25 '18 at 7:41
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Based on @Acccumulation 's answer, I try to put the reason, why BA should be played at least $\frac{1}{3}$ of the time, in other words. If Colin would not cooperate, he could always play Colin A because Colin A dominates Colin B. Therefore, he was granted a payoff of $6$ for sure. Now, consider the cooperation cases AB and BA: For AB, Colin would get a payoff of $5$, whereas he would get a payoff of $8$ for BA.

The question now is: How large should the ratio of playing BA at least be in order to guarantee Colin to cooperate by playing a mixture of AB and BA and not to play only Colin A?

To answer this, put this into a formula by determining $x$ as the minimum occurrence of BA with payoff $8$ for Colin and $1-x$ as the remaining occurrence of AB with payoff $5$. Remember that $6$ is the minimum payoff Colin would get if he played only the dominant strategy Colin A: $$x\cdot 8+(1-x)\cdot5 \geq 6$$ $$3\cdot x+5 \geq 6$$ $$x \geq \frac{1}{3}$$ Thus, BA should be played at least $\frac{1}{3}$ of the time to motivate Colin to cooperate.

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