[Original review, Jun 21 2021]
I don't know enough game theory to be able to give a competent opinion on this book, but like everything from Binmore it's well-written, full of provocative thoughts, and based on a phenomenal amount of reading. The central idea is characteristically imaginative: maybe there are types of game strategies that no one has even considered?
To give some background, Binmore spends quite a lot of time here discussing a toy game he calls micro-poker. This is played between two people we'll refer to as Q and JK, using a deck of three cards which contains a Jack, a Queen and a King. At the start of the game, Q gets the Queen face-up, and JK gets either the Jack or the King face-down, with a 50% chance of receiving either one. JK can look at his card but Q doesn't know what it is.
Now JK has to decide whether to check or to raise. If he checks, he loses one dollar. If he raises, Q has to decide whether to fold or to see. If he folds, he loses one dollar. If he sees, the player with the higher card wins two dollars and the other one loses two dollars. What are the best strategies for each player?
Evidently JK will raise if he has the King, but what does he do if he's dealt the Jack? If he always raises, then after a while Q will notice this and always call. Then JK will win when he has the King and lose when he has the Jack, so he'll break even. Similarly, if he never raises with the Jack, then Q will again notice after a while that JK's only raising half the time. He'll stop seeing altogether, and again he'll break even.
As any fule kno, both players should choose mixed strategies. If JK has the Jack, he should raise with some probability x, and if he raises then Q should see with some probability y, using some kind of method (a "randomising box") to choose the number in a way that can't be predicted. The question is what the optimal values are for the probabilities. It's fun to use a little algebra to work it out! JK's payoff is:
0.5(2y + (1-y)) + 0.5(-(1-x) + x(-2y + (1-y))) =
0.5(2y + (1-y) + -(1-x) + x(-2y + (1-y))) =
0.5(y + x + x(-3y + 1)) =
0.5(y + 2x - 3xy) [P]
If JK chooses x = 1/3, the formula for the payoff P reduces to
0.5(y + 2*1/3 - 3*1/3*y) = 1/3
which doesn't depend on y. So whatever strategy Q chooses, JK will come out ahead in the long run, gaining an average of 33 cents per game. Similarly, if Q chooses y = 2/3, he can make sure that X doesn't do any better than this, and the two choices form a Nash equilibrium.
I though all games were in principle similar, just more complicated. But Binmore says, suppose you could make choices that don't have an associated probability? He goes to some lengths to specify exactly what this means (there is quite a lot of nontrivial mathematics), but at the end of the day he has enough machinery to be able to define what he calls a "totally muddling box". This takes the concept of a randomising box a step further: a totally muddling box produces numbers randomly, and you can't assign a probability to any given thing happening. He shows you a simple game where he says a totally muddling box can give you a better strategy than any ordinary randomising box.
And then, just when I thought things had got really interesting, the book stops! Damn. Why?! Did I miss something? Couldn't he think of a suitable way to continue? Had he made his point and felt no need to do any more? Those mathematicians. Sometimes they're too smart for their own good.
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[Update, Jun 23 2021]
I showed the micro-poker problem to Not and another strong bridge-player we know, and they both gave the same answer after only a few seconds: if they were playing JK and were dealt the Jack, they'd raise all the time.
I pointed out that this is suboptimal according to the game-theoretical analysis above, but Not, a practical card-player with vast experience, was unimpressed. If we set x = 1 (i.e. JK raises all the time on the Jack), the value for the payoff formula P reduces to y - 1. If Q sees at their game-theoretical optimal rate of 2/3, JK will do as well raising all the time as they would with the game-theoretical optimal rate of raising one-third of the time. For JK to do worse by raising all the time, Q has to realise that they should be seeing more often than two-thirds of the time. It could take them a while to figure out that they need to adopt this seemingly risky strategy. As Binmore points out several times in the book, most people are instinctively risk-averse. The bottom line is that raising all the time can't do worse than break even, is likely to do as well as the theoretically optimal strategy, and gives the opponent plenty of chances to go wrong.
I'm reminded of a famous quote from Karpov, one of the world's great pragmatic chessplayers. A journalist asked Karpov what he'd do given a choice between a brilliant but complicated sacrificial combination which might lead to a forced win, and an endgame with a tiny advantage.
"The endgame, without even thinking," said Karpov.