Where are the three extra cards coming from for this Texas Hold Em hand?

Question

I've recently gotten into odds/probabilities, and even though I don't fully grasp the concept of binomial coefficients/Combinatorics, I'm still trying to figure out some probabilities using good old fashioned.. probabilities. The scenario that I'm currently dealing with is a standard Texas Hold Em game, right before the River. The following cards are in hand and in the communal:

Hand: Ace of Clubs; 2 of Clubs

Community: 5 of Clubs; 4 of Spades; 8 of Diamonds; 6 of Clubs

Currently, I am trying to calculate the odds of the next community card creating a one pair with the current cards.

I created and ran a simulation 100,000 times, tracking the percentage of one pairs encountered, which was 46.238%

This is pretty close to the original probability that I generated prior, which was 6/13, or ~46.153%. The way that I came to this number, was that I had 13 ranks, and 6 of those ranks had a chance of producing a card that would actually complete my one pair.

Thinking about this, I realized I should be able to expand that information to the entirety of the deck.

There are 6 cards currently in play, which leaves us with 52-6=46 cards to be able to draw from. There are 6 ranks that can produce a one pair, with 3 cards that satisfy each rank. This should equate to 18/46 ~= 39.13%.

Unfortunately, that number is off. I took 6*(46/13) ~= 21.23. So the proper ratio is about 21/46.

So, my ultimate question is:

Where are these extra 3 possibilities coming from?

Answer 1

Elaborating on my comment, there is every reason to believe that your "simulation" was written incorrectly, most likely by not taking suits into account or drawing the river card from a full deck or drawing from a deck of only thirteen cards.

Your initial solution of $\frac{6}{13}$ is incorrect. Although there are indeed six ranks which will produce a pair out of the 13 available ranks, these ranks are not equally likely to be drawn in this scenario. There are fewer aces available in the deck than there are kings for example.

For extreme example, I have a sock drawer filled with two thousand white socks and only two black socks. I draw one sock uniformly at random. The probability that I draw a black sock is not going to be $\frac{1}{2}$ even though there are only two colors available and one of them is what I wanted, the probability of drawing a black sock has to do with the ratio of how many black socks there are over the total number of socks (not the total number of types of socks). The probability here will in fact be $\frac{2}{2002}$.

Remember that the common practice of $Pr(E)=\frac{|E|}{|S|}$, the probability being the ratio of number of desired outcomes over total number of outcomes, is only valid when the sample space is equiprobable. In common discrete probability questions like this, we must take extra care to ensure that the way in which we describe the sample space is in fact going to result in each outcome being equally likely. To ensure this is the case, we can often arbitrarily assign labels or colors to various objects (such as when rolling two dice, imagine one of the dice is red and the other is blue, or when selecting socks imagine a tag on each with a unique identification number) so that we may tell them apart. The existence (or nonexistence) of such a method of labeling won't affect the probabilities in any way, so we are free to have done so.

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Your second solutions is in fact the correct one. As there are 46 cards left in the deck, each of which are equally likely to be drawn for the river card, and 18 of those will result in a pair (6 ranks which result in a pair, 3 unused suits for the selected rank), the probability is going to be $\frac{18}{46}$.

There are no "extra three possibilities" that are missing.