Odds of getting a pair in texas holdem

Hi, thank you to everyone who contributes to these forums. I have been searching everywhere for the odds of, if I am all in preflop in texas holdem with any 2 cards that is not a pair in my pocket cards, what are the odds that I will hit a pair by the end of the hand. I am trying to work out the odds that at least one of my 2 pockets cards will pair as a whole hand rather than on the flop or on the turn or on the river but by the end of the hand. Also what is the difference in the odds by the end of the hand if I was to include the community cards. Thank in advance for your help

One useful fact to know when figuring out probabilities is, the probability of something happening = 1 minus the probability of it not happening.

You have two unpaired cards - say, an Ace and a 2.
You want to determine the probability of getting a pair. It is probably easier to determine the probability of not getting a pair, and then subtracting that from 1.

If you are including pairs within the flop:
For the first flop card, there are 50 possible cards, 44 of which are neither A nor 2. Let's say it's an 3.
For the second flop card, there are 49 possible cards, 40 of which are not A, 2, or 3 (otherwise you pair the two cards in the flop). Let's say it's a 4.
For the third flop card, there are 48 possible cards, 36 of which are not A, 2, 3, or 4. Let's say it's a 5.
For the turn, there are 47 possible cards, 32 of which are not A, 2, 3, 4, or 5. Let's say it's a 6.
For the river, there are 46 possible cards, 28 of which are not A, 2, 3, 4, 5, or 6.
The probability of having no pairs = 44/50 x 40/49 x 36/48 x 32/47 x 28/46 = 0.2232853
The probability of having at least one pair = 1 - 0.2232853 = 0.7767157.

If you are not including pairs within the flop:
Again, assume you have an Ace and a Deuce.
In every case, there are 6 cards (the three remaining Aces, and the three remaining Deuces) that will pair one of your hole cards.
Not pairing with the first flop card = (50 - 6) / 50
Not pairing with the second flop card = (49 - 6) / 49
Not pairing with the third flop card = (48 - 6) / 48
Not pairing with the turn = (47 - 6) / 47
Not pairing with the river = (46 - 6) / 46
The probability of not pairing either hole card = 44/50 x 43/49 x 42/48 x 41/47 x 40/46 = 0.512568
The probability of pairing at least one of the hole cards = 1 - 0.512568 = 0.487432

Part of your question reminds me of something I often state at my poker league games.

Often, at the showdown, someone will announce that they have Two Pair. Sometimes I'll tell them, "Two pair ain't nothing to be that proud about, when one of the pairs is on the board...."

Quote: DJTeddyBear

Part of your question reminds me of something I often state at my poker league games.

Often, at the showdown, someone will announce that they have Two Pair. Sometimes I'll tell them, "Two pair ain't nothing to be that proud about, when one of the pairs is on the board...."

If someone says "I have two pair - Sixes and Sixes" Are you allowed to kick his ass?

Quote: FinsRule

If someone says "I have two pair - Sixes and Sixes" Are you allowed to kick his ass?

LOL. Only of they said, "Red Sixes, and Black Sixes."

deleted

Quote: Ibeatyouraces

my opponent asked me " How big is your kicker?" I replied "Jack high." He call so I revieled my hole cards as pockect jacks and me saying "I have a pair of jacks with a jack kicker." He was pissed.

That is awesome!

Quote: FinsRule

If someone says "I have two pair - Sixes and Sixes" Are you allowed to kick his ass?

There should be rules about this, as it can players who otherwise wouldn't reveal their hand to show what they were holding. Same with the jack high comment as well. The only time I say Jack high is when referring to a flush or straight, but in my mind, I assume that the player knows I have a flush or straight.