M.O.P. (Chapter 16): Full-Street [0,1] No-Fold Game

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M.O.P. (Chapter 16): Full-Street [0,1] No-Fold Game

Example 16.1

[0,1] Game #4

"One-Full Street.  Neither Player May Fold.  There is one bet left. "

Why on earth does the optimal solution containing X betting all hands and Y calling with all hands? 

It seems dominated for X to bet with any hand worse than .5

How is it not in X's best interest to bet all hands which are in [0, .5] and then just check the rest and hope Y checks behind?


Edit:

I understand that Y can then "improve" their equity by betting all hands that X would be betting plus the top half of the X hands which X is checking, but this doesn't mean that the strategy of betting only those hands which have a positive (or zero) expectation of betting is not dominated.  I mean, this strategy they give as the "solution" for X is actually suggesting that X bets the worst possible hand when X knows that Y will call.

Why is the optimal strategy pair not just:

X bets [0,.5] and checks all other hands

Y bets [0,.75] and checks all other hands


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