As promised, here's the solution of the AKQJ game. Remember the deck contains just A, K, Q and J, the pot contains $2, players X and Y get one card each, X checks, then Y can either bet $1 or check. If Y bets, X can either call or fold. It turns out (details below for those interested) that there is a family of optimal solutions for X. Y bluffs 1/3 of his Js, checks with K and Q and bets A, but X calls with A and a proportion cK of his Ks and (1-cK) of his Qs, and folds J. The only restriction is that cK >= 1/3. This makes Y a profit of $1/12 per hand for any choice of cK, and neither player can exploit the other.
Remember that the solution of the AKQ game was for Y to bluff 1/3 of his Qs, check with K and bet with A, and X needed to call with A, call with 1/3 of his Ks and fold Q. This makes Y a profit of $1/18 per hand.
So we can see that the AKQJ game is more profitable for Y than the AKQ game. It's very interesting that X has a range of optimal strategies. He can even choose a nonmixed strategy if he wants to (call all Ks, fold all Qs)
There are two obvious questions now.
i) Can this analysis be generalized to a deck with N cards? We've done N=3 and N=4. This looks very hard to me, but I may be making the analysis harder than it should be.
ii) What happens in the limit N -> infinity? This seems like a silly question, but really, this should converge to another game analysed in Chen and Ankeman's book, where instead of discrete cards, each player has a random number (x and y) drawn from the uniform distribution on [0,1], with 0 being (perversely) the best hand and 1 the worst. The solution is indeed that Y should bet his best hands (0<y<y1), bluff his weakest hands (y2<y<1) and check his middling hands (y1<y<y2), whilst x should call with his best hands (0<x<x1) and fold the rest. The key point is that y2>x1>y1. X must bluff in order to get value from his best hands, making Y have some bluff catchers in his range. I can't remember the values of x1, y1 and y2, but they're in C and A's book.
Maybe I'll let one of my students take it from here! ♠_evil:
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Analysis of the AKQJ game
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We are looking for the optimal values of Y's betting frequency with J, Q and K (bJ, bQ, bK) and X's calling frequency with Q and K (cQ and cK). Y always bets A and X always calls with A and folds J.
With X's hand given first and Y second, X's additional profit from the betting is:
JQ, JK, JA, zero since X folds
QJ, X wins an extra $1 if Y bluffs and he calls, but loses $2 if he folds, bJ(cQ-2(1-cQ))
QK, X loses an extra $1 if Y bets and he calls, -bK cQ
QA, X loses an extra $1 is he calls, -cQ
KJ, bJ(cK-2(1-cK))
KQ, bQ(cK-2(1-cK))
KA, -cK
AJ, bJ
AQ, bQ
AK, bK
Adding this all together, X's profit is
P = ((3bJ-bK-1) cQ + (3bJ+3bQ-1)cK -3bJ-bQ+bK)/12 = (3(cQ+cK-1)bJ + (3cK-1)bQ + (1-cQ)bK -cQ -cK)/12
This is rather more complicated than the AKQ game, not least because there are five unknowns instead of two! I sorted this out by going through the various cases systematically, but I can't help thinking that those clever game theorists have a better way of doing this!
Here goes! The optimal solution must satisfy the following five conditions
1) cQ+cK=1, or cQ+cK>1 and bJ=0, or cQ+cK<1 and bJ=1,
2) cK = 1/3, or cK>1/3 and bQ=0, or cK<1/3 and bQ=1,
3) cQ = 1, or cQ<1 and bK=0,
4) 3bJ-bK=1, or 3bJ-bK>1 and cQ=1, or 3bJ-bK<1 and cQ=0,
5) bJ+bQ=1/3 or bJ+bQ>1/3 and cK=1, or bJ+bQ<1/3 and cK=0.
Actually, the poker interpretations of these conditions are clear:
1) if X calls too much/too little, Y never/always bluffs J
2) if X calls too much/too little with K, Y never/always bluffs Q
3) if X calls too little with Q, Y never bets K
5) if Y bluffs too much/too little, X always/never calls with K
4) is similar, but kind of mixed up.
Now consider the three possibilities in case 1).
i) bJ=1, then 4) => cQ=1, 5) => cK=1, 2) => bQ=0 and 1) => bJ=0, which is a contradiction.
ii) bJ = 0, then 4) => cQ=0, 3) =>bK = 0. Then 2) and 5) become
2) cK = 1/3, or cK>1/3 and bQ=0, or cK<1/3 and bQ = 1
5) bQ = 1/3, or bQ>1/3 and cK=1, or bQ<1/3 and cK = 0
The only consistent solution of these is bQ = cK = 1/3. Then cQ+cK = 1/3 and 1) => bJ = 1, which is a contradiction.
iii) cQ+cK = 1, then we must consider two subcases from 3)
a) cQ = 1 => cK = 0, then 2) =>bQ = 1, => bJ + bQ = bJ+1 >1/3, so that 5) => cK = 1, which is a contradiction.
b) bK = 0, cQ<1. Next there are three subcases arising from 4)
I) bJ>1/3 => cQ = 1, a contradiction
II) bJ<1/3 => cQ = 0 => cK=1, then 2) => bQ = 0 and 5) => bJ + bQ > 1/3, a contradiction.
III) bJ = 1/3, then 5) => two subcases
A) bQ>0 and cK=1, when 2) gives bQ = 0, a contradiction
B) bQ = 0, and 2) => cK>1/3. And this is it. We can satisfy all the conditions provided that
bJ = 1/3, bQ = bK = 0, cQ+cK = 1, cK>1/3.
There MUST be an easier way to do this. There's no chance of this working for an N card deck, or at least I wouldn't have thought so. I may have to go away and learn some more game theory. It would be nice if this was new and I had to go and invent some new maths, but I doubt it! Or of course, I might have have made a complete bollocks of the whole thing. Any one fancy checking? 
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I once had an argument with my college english teacher - I was trying to prove to her that homeworks exists, just like waters or fishes - since I was referring to multiple different types of homework which would require certain distinction in the word itself, which in singular form obviously wouldn't fit that role quite well.
