poker hand odds

I'm trying to come up with odds of events happening in Texas Hold Em poker.?

I would like to know about Texas Hold'em, which is a small probability Four of a kind had been beaten by more four of a kind or straight flush. , please assume that there are only 2 players in each game, and that each player has to repeat any two cards. In other words, it is a heads up, where one player is all-in lot. Please provide some basic statement of how the arrived at the answer. Thank you.

Here's how I would have this problem: First, you have four of a kind. One of three things happened: 1 The community cards to 2 quads Common cards are regarded as three of a kind you like the fourth and Card 3 Common to keep a few cards and keep a couple of other cases # 1, nothing interesting can happen. It is impossible for another person may have quads in government, so this is uninteresting. case # 2, there are four remaining cards – the other two community cards and your opponent two cards. So the following is possible: 2A. These four cards to create a larger four of a kind you. 2B. These remaining four cards can be combined to create a three of a kind straight flush. If # 3 is the remaining five cards – the other three community cards and your opponent two cards. This results in the following options: 3A. These five cards containing quadruplets, who wins his quads. 3B. These five cards form a straight flush. 3C. Four of these five cards combine a couple of the shows cards to create a straight flush. Now these are the only ways you can get. This is a complex issue, and so I'm assuming you have the ability to calculate probabilities already. The most important thing is to calculate the probability of each event takes place above and combine them together: P (2) * [P (2A) + P (2B)] + P (3) * [P (3A) + P (3B) + P (3C)] Note: The probability of this event happening. You may also be interested in the conditional probability Can you maybe give that you alreay have your quads. Then you can just leave the P (2) and P (3), to obtain P (2A) + P (2B) + P (3A) + P (3B) + P (3C) I do not really find an answer, but I am sure that it is very small!

Poker Counting odds and outs