and the minimum is achieved when $b=0$ so Tom can maximally exploit him (minimise $E(J)$) by never bluffing ($b=0$). Similarly you can work out that if John calls less often than this, so that $c-frac{1}{3}$ is negative, Tom can maximally exploit John by always bluffing ($b=1$). The only way that John can defend himself against exploitation is to use $c=frac{1}{3}$. This is his unexploitable calling (bluff catching) frequency. His bluff catches with K♠ and value calls with A♠ are in the appropriate ratio to make Tom indifferent, i.e. however often he bluffs, his win rate does not change.
We can also rearrange the expression for $E(J)= -E(T)$ to get
[ E(T) = frac{1}{2}cleft(frac{1}{3}-b
ight)+frac{1}{6}b. ]