# Straffin, P. D. (1993): Game Theory and Strategy. How is the security level on p. 103 calculated?

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):

• (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. – Stephan Kolassa Jul 17 '18 at 9:36
• 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. – Joram Schito Jul 17 '18 at 10:24
• 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. – whuber Jul 17 '18 at 11:13
• OK, thank you! I will try it also on the SE economics site. – Joram Schito Jul 17 '18 at 11:39
• – Giskard Jul 17 '18 at 13:01

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.

• 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? – Joram Schito Jul 25 '18 at 7:41

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.