I have a question regarding an example from the book Mathematics of Poker by Chen and Ankenman. There on p. 48 he gives
Example 4.1 Two players play headsup limit poker on the river. Player A has either the nuts (20% of the time) or a valueless (or dead) hand (80% of the time), and Player B has some hand of mediocre value - enough to beat dead hands, but which loses to the nuts. The pot is four big bets, and A is first. Lets us first consider what will happen if A checks. B could bet, but A knows exactly when he has B beaten or not; hence he will raise B with nut hands and fold at least most of his bluffing hands. B cannot gain value by betting; so he will check. As a result, A will bet all of his nut hands. A might also bet some of his dead hands as a bluff; if B folds, A can gain the whole pot.
We'll call the % of total hands that A bluff with x. A's selection of x is his strategy selection. B loses one bet for calling when A has a nut hand, and wins five bets (the four in the pot plus the one A bluffed) when A has a bluff. B's calling strategy only applies when A bets, so the probability values below are conditional on A betting. The expection of B'S hand if he calls is:
[B, call] = P(A has nut)(-1) + P(A has bluff)(+5)
[B, call] = (0.2)(-1) + 5x
[B, call] = 5x - 0.2
If B folds, his expection is simply zero.
[B, fold] = 0
Remark: [B,call] denotes the expection if B chooses to call, likewise [B,fold] if he chooses to fold.
I guess this analysis is not correct, according to the sentence "so the probability values below are conditional on A betting" we should use conditional probabilities instead. Assume A has 20% of the time the nuts, x*100 % of the time he bluffs and the rest he simply checks. Then the probabilty that he has the nuts if he bets is
P(A has nuts | A bets) = P(A has nuts and A bets) / P(A bets) = ( 0.2 * 1 ) / (0.2 + x)
and accordingly the probability that he bluffs if he bets is
P(A bluffs | A bets) = P(A has dead hand and A bets) / P(A bets) = x / (0.2+x)
- P(A has nuts and A bets) = 0.2
- P(A has dead hand and A bets) = x
- P(A has dead hand and A checks) = 1 - (0.2+x)
- P(A bets) = 0.2 + x
so that the correct expression for B calling should be
[B, call] = P(A has nuts | A bets) * (-1) + P(A bluffs | A bets) * 5
[B, call] = - 0.2/(0.2+x) + 5x/(0.2+x)
[B, call] = (5x-0.2)/(0.2+x)
Am I right or is the book right?