Quote:
Originally Posted by flysohightosky
Also, I have no idea how they came to the conclusion on page 148 (paragraph right below the graph) that X should fold 100% even if Y bets only 1% of the pot. Makes no sense to me whatsoever.
This answer probably comes late, but I got intrigued by the same passage.
The Half-Street No Limit Clairvoyance game is solved for y = 1/2 (fraction of winning hands for Y). In the general case, solving for b the fraction of dead hands that Y bets as a bluff, we get (making X indifferent between call/fold):
b = s/(s+1) * y/(1-y). This respects the general result that Y bluffes s/(s+1) as many hands as he bets for value.
If s = 5 and y = 3/5, this yields
b = 15/12 > 1 which shows that Y cannot make X indifferent between calling and folding. The best Y can do is bluff all of his dead-hands (b=1), but since Y is still "under-bluffing", the best action for X is simply to fold all the time.
Now if we set s = 1 both for bluffing and value betting, then we would have a valid Nash Equilibrium if Y wasnt allowed to vary his bet size. The authors are saying that if Y can vary his bet size, then the bluffing frequency b and calling frequency c
alone do not guaranty a Nash Equilibrium, because Y can unilateraly increase his expectation by increasing the bet size without changing his bluffing frequency:
Indeed we have y = 3/5, b = 3/4 and c = 1/2. By locking these values and letting s "float", the expectation of Y is:
Ey = y*c*(+s) + (1-y)*b*(c(-s) + (1-c)*1) = 3(s+1)/20. Same goes for s=1%.
So Y can unilateraly increase his expectation by simply betting more chips. This new bet size will lead to new values for c and b, and so on until we reach b = 1 (starting from s = 2), at which point X does better by simply folding all hands.
Summary:
I think the authors are simply saying that there is no Nash Equilibrium at s = 1% if Y is allowed to vary both b and s. On the other hand, if s is
set in stone at 1% then we have the usual optimum, and X would be correct to call with frequency 1/(1+s).