Recall the game between R and D determined by the following data.
![\begin{array}{cc|ccc} \multicolumn{5}{c}{\text{Table 11/3/1}} \\[4pt] & & & R & \\ & & \alpha & \beta & \gamma \\ \hline & 1 & b & a & c \\ D & 2 & a & c & b \\ & 3 & c & b & a \end{array}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bcc%7Cccc%7D++%5Cmulticolumn%7B5%7D%7Bc%7D%7B%5Ctext%7BTable+11%2F3%2F1%7D%7D+%5C%5C%5B4pt%5D++%26+%26+%26+R+%26+%5C%5C++%26+%26+%5Calpha+%26+%5Cbeta+%26+%5Cgamma+%5C%5C++%5Chline++++%26+1+%26+b+%26+a+%26+c+%5C%5C++D+%26+2+%26+a+%26+c+%26+b+%5C%5C++++%26+3+%26+c+%26+b+%26+a++%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0&c=20201002)
Here is Ashby’s analysis of how it plays out.
Examination of the table soon shows that with this particular table R can win always. Whatever value D selects first, R can always select a Greek letter that will give the desired outcome. Thus if D selects 1, R selects β; if D selects 2, R selects α; and so on. In fact, if R acts according to the transformation
then he can always force the outcome to be a.
R‘s position, with this particular table, is peculiarly favourable, for not only can R always force a as the outcome, but he can as readily force, if desired, b or c as the outcome. R has, in fact, complete control of the outcome.
Reference
- Ashby, W.R. (1956), An Introduction to Cybernetics, Chapman and Hall, London, UK. Republished by Methuen and Company, London, UK, 1964. Online.
Inquiry Into Inquiry 
As a game, it becomes a little (but only a little) more interesting when R&D make their selections in ignorance of the other’s choice. From my brief glance, it looks like (under those constraints) a random choice matrix becomes the only one that “works” for both, where “works” means provides any chance of breaking even. But my intuitions on such things are demonstrably crappy, so I look forward to the next entry.
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