Quote:
Originally Posted by Aesah
This line lets us bluff the turn 10/14 = 5/7 as often as we have value, which turns out to be the exact same as the previous strategy. Perhaps there is a simpler way to show that it'd be the same; while the result was not surprising, it definitely wasn't obvious to me that the result would be equal.
I usually don't post theory, since so few people appreciate it and those that do are probably the ones I need to worry about most, but you can show that betting X on the turn and Y on the river let's you have the same turn ratio of bluff to value as betting Y on the turn and X on the river.
Let a = the fraction of the pot you bet on the turn
Let b = the fraction of the pot you bet on the river
On the turn we bet a*Pot, so our opponent needs to win a/(1+2a) percent of the time in order to call. Hence we need to bet the river 1 minus that percent of the time, so (1+a)/(1+2a). So we have V+C = (1+a)/(1+2a).
On the river we bet b*Pot. We are offering our opponent 1+b:b in terms of odds. To make him indifferent to calling, we need to have b/(1+2b) percent bluffs out of our entire river betting range.
In other words, C = [b/(1+2b)] * (V+C)
Simplifying, we get V = [(1+b)/b] * C
Using V+C = (1+a)/(1+2a), now we can solve for C. The actual math is left as an exercise for the reader. Hehehe
We get C = [(1+a)(b)] / [(1+2a)(1+2b)]
Now we know that V = [(1+a)(1+b)] / [(1+2a)(1+2b)]
Remember, we also have V + G + C = 1. Our turn bluffs are C+G, which is equal to 1-V. Our turn value bets are V. The ratio is (1-V)/V.
As we can see, if we swap the a's and b's in the equation for V, V is still the same. Thus the ratio (1-V)/V is still the same. Hence, swapping our turn and river bets (in terms of percentages of the pot) does not change our turn bluff to value ratio.