# Issue with Platinum article on Nash Equilibrium and Bluffing/Calling Frequencies

• Bronze
Joined: 12.04.2010
For example, let's imagine a board of JsTs4 2, that you are obviously on a draw (straight or flush), and that player 2 has a made hand. If the river is a 6s, completing possible flush draws, and you hold KQ, then, to determine how often to bluff, you should consider how often you would have a flush.

(As9s-As5s, As3s, As2s, 9s7s, 8s7s, 8s6s, 7s6s, 7s5s, 6s5s, 5s3s) is an example for a realistic range, amounting to 14 different hands. You will hold a busted straight draw just as often (7 combinations each of KQ and 98, assuming that you would have played OESD+FD differently earlier on).

I thought there'll be 15 combinations each of KQ and 98?
e.g K Q , K Q , K Q
K Q , K Q , K Q , K Q
K Q , K Q , K Q , K Q
K Q , K Q , K Q , K Q

If my combinations are indeed correct, does that mean that in this case, my probability of having a busted straight draw = 30/44 = 0.681?
Hence a flush board is actually not as scary as a straight board? (Assuming the hand range for a straight draw is only connectors, KQ/ 89)
• 4 replies
• Bronze
Joined: 14.09.2009
there are 16 combos of KQ or 98. To find the number of combos remaining is:

number of card 1 x number of card 2

4 Kings x 4 Queens = 16.

If there was a K on the board you got 3 x 4 = 12 combo remaining.

Apologies if I am not clear.
• Bronze
Joined: 12.04.2010
K Q is out because that'll make a flush. So 15 combinations right?
• Bronze
Joined: 14.09.2009
Originally posted by livelydolphin
K Q is out because that'll make a flush. So 15 combinations right?
right sorry
• Bronze
Joined: 12.04.2010
np. thanks for the clarification