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

Case II: $r &lt; \frac13$, then it's positively sloped, and the minimum occurs at $c = 0$, with a payoff of $10r$. Since $r &lt; \frac13$, 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.