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If you hold KQs, the odds that someone has Ax of your suit is (Chance they have the ace of your suit)*(chance they have one of the other 10 of that suit) = 1/50*10/49 = approx 1 in 250, or .4% of the time.
Assuming that the hand in question is one of these 250 times, the chance you both make a flush is (chance you make it on the flop)+(chance you make it on the turn)+(chance you make it on the river).
The following math assumes that THREE and ONLY three of the suit in question appear on the board by the river...If 4 of the suit come up, the chance you're beat with KQs increases, because now the opp only needs the ace of that suit, not the x of that suit to go with it.
Both make your flush on the flop = 9/48*8/47*7/46 = 504/103,776 =.48%
Both make your flush on the turn = 3*(9/48*8/47*39/46) * 7/45 = 55,944/4,669,920 = 1.26%
Both make your flush on the river = [6*(9/48*8/47*39/46)*38/45*7/44] = (4,481,568/205,476,480) = 2.181%
So *IF* you have KQs and your opp has Ax of your suit (only 1 in 250 times that you have KQs...(which is 4/52*1/51 = .18% of hands)) ...so a big if...You'll both make the flush only .48+1.26+2.181 = 3.826% of the time.
You are dealth KQs .18% of the time, or 1/500 hands. 1/250 of those times, will you be up against Ax of your suit.
I think 3.826% of the time is good enough to call an all in every single time no matter what on an unpaired board if you have the flush, and only 3 to the flush is on the board.
Edit: This math all assumes you're up against only one opponent...the chance increases as the amount of players who see the flop increase, obviously. Someone else can take it from there if you wanna find the chance this happens to you in a full ring...still pretty slim.
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