So this question is only to those who have a book;
example 11.1, chapter 11,part III
could someone explain me this "clairvoyance game"? What does it have to do with clairvoyance? Which player acts first? I also think that this payoff matrix is wrong- player x should be on player y place, and bet with nuts when the result is check-fold does have some value too right? Isn't it +P? And could someone get me the whole ev equation for <Y>, with bluff ev too (i know it is 0)?

It's called a "clairvoyance game" because Y knows what X's cards are. One important point they fail to mention is that X knows that Y is clairvoyant. If X didn't know that, he should call as long as the pot odds he's getting are better than the odds his hand is best. Y would either always bluff or never bluff depending on whether X's calling frequency gave him +EV. If X thought he would win 50% of the time, he should always call since he's getting better than even money, and Y should never bluff. Both this example and the text also assume that Y knows whether X knows that Y is clairvoyant, and X knows that, etc. Clairvoyance is assumed part of the game that everyone knows.

Other than that, this is an excellent section. X goes first, but we're assuming he always checks, so effectively Y is going first. The payoff matrix is correct. The bluff EV when X calls correctly is

P*s/(P+s) - s*P/(P+s) = 0.

Remember, s is the amount Y bets, and in a limit game, s=1 (as in the payoff matrix which is for limit). Y wins the pot P when X folds which he does with probability s/(P+s), and he loses his bet s when X calls which he does with probability P/(P+s).

ok,I get it. X checks first beacuse Y can see his hand and thus play really accurately against him (when X bets first). Now Y can bluff some of his weak hands, so here we're trying to find optimal calling strategy for X right? And it is 1-a.
And for example for P=1, Y bets 75% of the time (50% with nuts and a/2=25% bluffs)??

We're trying to find the optimal calling strategy for X so that it doesn't matter how often Y bluffs, and the optimal bluffing strategy for Y so that it doesn't matter how often X calls. That is, neither can exploit the other by changing their strategy. If X calls more than the optimal amount, then Y would do better if he bet every time, and if X called less than the optimal amount, Y would do better if he never bluffed. Similarly, if Y bluffed more than the optimal amount, X would do better if he called every time, and if Y bluffed less than the optimal amount, X would do better if he always folded to a bet.


Yes.

ty! you are unquestionably cool guy

oh and also one more thing; why in the first table on page 120 we win only +1 when x's hand is [y1,x1*]? Shouldn't it be P+1? the same thing in the same table with [x1*,1] - I think here we win pot when x folds. And also the same thing in the next table with [y0,1]. Here our <Y,bet> is equal to P is it?

No, because we are considering ex-showdown value as explained at the beginning of chapter 11. If x's hand is [y1,x1*] or [x1*,1] we win the pot no matter what we do. We are only concerned with the value of betting which gains 1 bet if we bet in the first case when x calls, and gains 0 in the second case when x folds.

It was the same issue in the clairvoyance game. If Y has the nuts, he can only win 1 bet since the pot is already his no matter what. If he bluffs, he can only lose 1 bet since he loses the pot anyway. He only wins the pot when he bluffs and X folds a better hand.

There are some obvious typos in the second table. The product of x1* and -1 is -x1*, not x1. Also, if X's hand is [y0,1], P(X's hand) is 1-y0, not 1-x1*. That was a copy and paste error from the first table. The final result is correct.

yeah so when Y bets when X hand is [y0,1] and we know he always fold this hand, why <Y,bet> here is 0 not P? P is here our ex-showdown value right?

No it's 0 because we would have won the pot even if we had checked since we have a better hand y1.

ok ,ok now i get it I don't know why I ignored this when I was trying to understand the table ty

They didn't make it very clear when they were explaining ex-showdown value. They said that we were ignoring bets that had been made before, but that in some cases we would be awarded the pot. They didn't explain that those cases were characterized by the fact that we wouldn't have won the pot had we done something else. You really need BruceZ's Companion Guide to The Mathematics of Poker by Bat Crazy Publishing.

wow, is there something like that? could you link me up this?
can't find this guide anywhere

I'm really having a hard time understanding what their actual definition of ex-showdown is in a way which would let you calculate things without relying on the verbal arguments that they try to cook up. Is it something like:

If we assume that our opponent plays optimally, then we have

Value of holding Hand Z = Sum (Value of each available strategic option with Z)


We sum over all strategic options and get (for the situation where we may bet or check)

Value of holding hand Z = Value of betting Z + Value of checking Z

This gives us that the value of betting Z is equal to the difference of holding Z (if everybody plays optimally) minus the value of checking. This doesn't seem to be quite what they're doing, though, because frequently the "ex-showdown value" is just the value of betting minus the value of checking.

For example, if we know our opponent always calls, then intuitively, we gain 1 unit by betting, and the "ex-showdown" value of betting the nuts for 1 unit into a pot of P is represented by the difference between betting and checking:

(100%)(P+1) - (P) = P + 1 - P = 1


Is the idea here that we just take the ex-showdown value of one player and to determine the other player's, we just take the negative of their opponent's?

I really wish this book would actually explain these things in some kind of reasonable way.