I'm wondering how exactly I would go about calculating the odds of getting exactly a pair on the flop (so two pairs and three of a kind do not count). I'm assuming that the player's hand is not a pocket pair.
So I tried calculating the odds of not getting a pair and subtracting it from one, but I can't figure out how to account for the flop containing a pair (so the flop would be something like queen of hearts, queen of spades, 7 of hearts, and the hand is 6 of clubs, 3 of clubs). To calculate the odds of a card in my hand pairing with a card from the flop I did
I used the formula
$$1 - \frac{\frac{44!}{3!(44-3)!}}{\frac{50!}{3!(50-3)!}}$$
- The top fraction is for the 44 cards which will not make a card with the hand
- The bottom fraction is pulling 3 cards from the 50 remaining
To calculate the odds of my hand not making a pair, I'm not sure how to approach the odds of the flop not containing a pair. And if one can explain how I'd alter this for different parts of the game(such as the turn and river) I'd really appreciate it. Of course, if you have an approach not involving subtracting the odds of not making a pair from 1 I would love to see it.
Just to clarify, assume that you have an Ace and a Deuce. Then, you want $1$ minus the probability that the flop will contain $3$ different ranks, none of which is an Ace or a Deuce.
The denominator remains $~\displaystyle \binom{50}{3},~$ so it is just a matter of re-calculating the numerator.
There are $11$ ranks left, so you have two choose $3$ of these ranks. Once this is done, you have to choose $1$ card from each of these ranks. So, the numerator changes to:
$$\binom{11}{3} \times 4^3.$$
So, the final computation is
$$1 - \left[\frac{\binom{11}{3} \times 4^3}{\binom{50}{3}}\right].$$
I am assuming that
straights, flushes, and straight flushes are to be ignored.
you intend one pair or better, looking at all $5$ cards : your two hole cards + the $3$ community cards.
Edit
As N.F. Taussig's comment indicates, based on your posting, it looks like you intend to focus on probability, rather than odds. To explain the difference:
Consider throwing $1$ die.
The probability of it coming up a $6$ is $\frac{1}{6}.$
In terms of odds, this is referred to as $5$ to $1~$ against.
In general, if the probability of an event is $~p~: 0 < p < (1/2)$
and $q = (1 - p)$
then you would typically refer to the odds against the event occurring as
$$q ~~\text{to}~~ p ~~\text{against}.$$