I got this question, or something like it, while presenting the example game theory solution in my mathematical poker talk. The solution to that game is that player X will check-call with hands worse than 4/9, check-raise with hands better than 8/9, and bet the hands in between. If X bets, Y will raise with 2/3 or better and call otherwise. If X checks, Y will bet with 1/3 or better and check otherwise.
This is one of the games in The Mathematics of Poker, except, as with all the [0,1] games in that book, Chen and Ankenman define lower hands to be better, so 0 is the best hand and 1 is the worst. Also, there is a typo in one of the indifference charts for this game, but it does not affect the resultant indifference equation.
The answer I gave during the talk was correct: yes, they do really play pure strategies here. I hadn't reached the point in the talk where I explain pure and mixed strategies, so I used this opportunity to introduce those concepts. A pure strategy would be if the players always play the same way given any particular hand. This is true everywhere except for the indifference points. As an example, I pointed out that in this game, if X held the hand 1/3, he always checks and then calls. An example of a mixed strategy would be if X sometimes check-called when holding 1/3 but sometimes chose another play (such as betting) when holding this hand.
That's what I said during the talk. Let me expand on it a bit.
In Texas Holdem and other common poker games, there are only a few "indifference point" hands where mixed strategies make sense (like X with 4/9 or 8/9). The rest of the possible hands require pure strategies. This is complicated a bit by the fact that real poker games are dynamic, with hands changing value as new cards are dealt, but I think the principle still holds that most hands have one play that has greater EV associated with it than the other available plays. This is why I so often argue against randomizing your play and instead advocate just choosing which hands are your indifference points. This wouldn't have been appropriate to address during the talk, since I didn't want to discuss any real poker games such as Texas Holdem; many people in the audience weren't familiar with the rules of those games.
One thing I wish I did add during the talk was an explanation of why, for this particular [0,1] game, mixed strategies were not appropriate. It would have been very difficult to succinctly prove the point without preparation, but I could have at least elucidated the general principle that was at play: players in a Nash equilibrium will never make a play that has less EV than another available play in any situation, nor will they make a play that has the same EV if that play can then be exploited by an opponent.
With the hand of 7/9, for example, player X has the same EV whether he check-raises or bets. However, this is not his "indifference point" because if he check-raises here instead of betting like he "should" in the Nash equilibrium, he allows player Y to change his strategy and improve his (Y's) EV. This violates the definition of Nash equilibrium (in which no player can change his strategy to improve his EV). I think the more fundamental point that the professor was getting at with this question may have been this: isn't it worth mixing up your strategy with every hand in order to avoid becoming predictable to your opponent? The answer to this is no. I'm not sure why this answer meets with such resistance among poker players (necessitating my repeated refutations), but it really does seem to offend most players' basic idea of what good poker looks like. I guess they think that their opponents will be able to figure out what they are holding if they play their hand the "right" way every time. However, this problem of predictability is completely solved by simply playing other hands the same way. Player Y cannot know which of the hands in the range [0,4/9] and [8/9,1] X has checked with. [Edit.] No need to sacrifice any EV trying to throw your opponent off any further. Sacrificing EV is, by definition, always the wrong play unless you gain it back somehow. In this case you gain nothing back.
Four more points about randomizing your play...
First, I'm not saying that X should always play the hand "1/3" the same way against all opponents, only that he should play it the same way against player Y, who plays optimally. Against other players, X might maximize his EV by betting with the hand 1/3. In any given situation, X should not be randomizing his play between two or more actions; he should be choosing the single action that seems to maximize his EV based on all the information that is available to him.
Second, the very fact that X sometimes plays this hand differently based on his opponents and other situational factors serves as a sort of pseudo-randomizing of X's play. That is to say, to other players, X's strategy will look random even though it's not. X knows the reason he is playing differently, so he is not "randomizing" his play; he is always maximizing his EV for a given situation. X's opponents, unable to completely discern all the factors X has taken into account before making his play, will view his play as being random. Even if you believe you need to make your play look random to your opponents (which I think is wrong), this should already be accomplished because the situational factors are just to complex for all players to evaluate in the same way.
Third, your play will be randomized because you will make mistakes. No need to add extra "intentional mistakes" when you are playing a game that is hard enough already!
Fourth, your opponents are not likely to be paying close enough attention to take advantage of your predictability anyway, and if they try they will likely do it wrong.
I'm writing this from a hotel in Baltimore; tomorrow begins the recruitment weekend for the Johns Hopkins Biostatistics department.
Edit (3/1/12). This originally said "[0, 1/3] and [2/3, 1]". I fixed it to reflect X's actually check-calling and check-raising ranges of [0, 4/9] and [8/9, 1].