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How likely is it to flop the nuts and have your opponent suck out?
I was watching high stakes poker the other night, and Daniel Negreanu got me thinking. How likely is it to flop the stone nuts and still lose the hand? In mathematical terms? Is there any way to calculate the probability of flopping the nuts and then having the nuts change, and your opponent holds the necessary cards? I know it's not likely, but I once watched a guy flop quad kings, and lose when the guy he was playing against had AA and caught runner-runner aces. What are the odds of that happening in a general sense? Are there too many variables to calculate?
Public Comments
- Well, on most tv poker shows they do show the percentages during the hands. It's pretty easy to calculate the odds. After the flop figure out which cards you would need to suck out, count those cards, and divide by the number of cards left in the deck. There's your suck-out odds. In poker, people flop the nuts and lose all the time. It's not a common occurrence, but it does happen.
- you can rearrange the words in your question to say: and how the nuts flop out is likely to have your opponent suck it
- too many variables, it would depend on the exact flop and the number of drawing possibilities on it...for example, a flop of K-8-2 rainbow has no straight or flush draw possibilities, so the nuts can only change with an A, K, 8, or 2, so it's not very likely...on the other hand, a flop like Q-10-7 with two diamonds, the current nuts would be Q-Q for top set, but any card 6 or higher or any diamond would change the best possible hand, so it's a lot easier to be drawn out on generally speaking
- Just go here .... http://www.bluffmagazine.com/tools/index.asp . And use the Poker Calculator
