Equilibrium strategy for half-street [0,1] poker game, no-limit (from "The mathematics of poker") I have been trying to understand Example 14.3 from p154 of Bill Chen's and Jerrod Ankenman's book The Mathematics of Poker without much success.  In this section they are analyzing what they refer to as [0,1] Game #3, and I think their argument has some flaw to it.  However, I have reason to believe that their answer is correct, so it could just be a case of poor wording. 
The rules of this game are as follows.  The game is one "half street", meaning that only one player can bet.  Betting is no-limit, and both players have arbitrarily large stacks.  This means that a bet size can be any number on $(0,\infty)$.  The pot starts out with 1 unit in it.
The first player, player X, is forced to check "in the dark" (without looking at his holdings).  Player Y can either bet some amount or check.  If he checks, X must check back, and they show down hands to determine the winner.  If Y bets, X can only call or fold. No raising.  This is the "half-street" condition. X has no aggressive options.
Chen and Ankenman  start the analysis by finding X's calling threshold, $x(s)$, when facing a bet of size $s$.  This is the worst hand with which X is willing to call a bet of that size.  In particular, X won't call a bet of any size with a hand worse than $x(0) \equiv x_0$ (a value that is not yet known at this stage of the argument).
Then they reason that X's calling range ($x_0$ or better) consists only of hands that beat all of Y's bluffs, which makes sense.  They continue,

...X will choose his function $x(t)$ such that if Y's hand $y$ is weaker than $x_0$, then Y is indifferent to bluffing an amount $s$.  In other words, X will call a bet of size $t$ with exactly enough hands that Y is indifferent between betting $t$ or checking for any value of $t$.  So $t_1$, $t_2$, $t_3$ and so on, all have equity equal to checking.  Hence Y is indifferent to betting any $s$ value.

But this can't be right as stated.  If Y is indifferent as to bet size for each hand $y$ in his bluffing range, he can't also be indifferent between bluffing and checking.  For all of his bluff hands except the cutoff (the best hand he will bluff with), he definitely prefers bluffing to checking.  Even bluffing an infinitesimal amount gets X to fold out a range of better hands (hands worse than $x_0$ but better than $y$.  
The authors go on to derive $x(s) = x_0/(1+s)$.  When you put this back into the expected gain from bluffing amount $s$ with hand $y$, namely $y-x(s)-sx(s)$, you get $y-x_0$.  This doesn't agree with the idea that Y is indifferent between bluffing and checking, which has expected gain of zero (above the value you get from merely showing down hands to determine the winner).  Note that the above formulas follow the counterintuitive convention that 0 is the best hand and 1 is the worst.  This makes formulas simpler.
So, what were they really trying to say in the paragraph quoted above?  Because it doesn't make sense to me.
Their math must be right, though, even if the wording isn't.  Their answer looks a lot like the one Newman gave in his 1953 paper, "A Model for Real Poker", which was probably the first ever analysis of this game.  He did not show how he derived it, unfortunately.  He merely verified that it satisfied the minimax condition.
Thanks in advance!
 A: I just encountered this, and you're right: Y prefers bluffing to checking for all bluffs below the threshold. It's pretty simple to prove this: for y_a and y_b below the bluffing threshold with y_a > y_b, the EV of bluffing a given amount s is equal (since you win vs all x > x(s), which fold, and lose vs all x < x(s), which call) but the EV of checking y_a is less than the EV of checking y_b, since y_a is weaker - so if EV(check y_a) = EV(bluff y_a), then it's not possible that EV(check y_b) = EV(bluff y_b).
As you note, the indifference must occur at the bluffing threshold, which the book calls y'(0). The indifference calculation is similar to the one on page 152: when we bluff here, we risk losing an additional s units x(s) of the time and gain the 1 unit in the pot an additional (y'(0) - x(s)) of the time, so for indifference we need:
-s * x(s) + 1 * (y'(0) - x(s)) = 0
So therefore x(s) = y'(0) / (1 + s), and we can plug in s = 0 to get x(0) = y'(0) and recover formula 14.2:
x(s) = x_0 / (1 + s)
This also explains the claim at the top of page 154 that was presented without justification: "Since the bluffing region begins immediately after x_0, ...". This is the part that tripped me up, but it falls out easily when formula 14.2 is solved the "right" way as above.
