simple game theory question

[Alan McIntire]

With 2 players, there are 5 cards exposed at the river,
you see 2 additional cards in your hand, that leaves 52-7
=45 cards remaining. Your opponent could have one of
45*44/2=990 hands. If only 1 hand is the nuts, You can NEVER bet more than 989 times the pot, else you're guaranteed to be making a negative expectancy bet. 989 is an upper limit.

If you have Kxs with a flush, 3 of your suit on board and none is an Ace, there are 7 ways your opponent could have the nut flush- A and any of the 7 remaining cards. In that case, he'll have the nuts 7/990 of the time so you can't bet
more than 990/7 = 141 3/7 of the pot or you're guaranteed to be making a negative expectancy bet.

If you bet 9 times the pot, and your opponent calls you
less than 1/(9+1) of the time, you'll win money by betting anything.
If you bet 989 times the pot and your opponent calls you less than 1(989 +1) of the time you'll win money, but he'll have the nuts at least 1/990 of the time if you don't have something like A of the suit to prevent this.

If you bet over 1/(141 3/7) of the pot with Kxsuited and 3 of your suit on board, your opponent will call you
1(141 3/7) of the time and you'll lose money. - A. McIntire