Card Removal Theory in Texas Holdem Part I

Edjon

This poker theory is about how exposed cards (e.g. your hole cards) influence the chances that your opponents has been dealt certain starting hands or cards. If you are dealt an ace for instance, then the chance that your opponent has AA or AK decreases. Players holding AK often say: “Because I got AK, the chance that my opponent has AA or KK is small”. In this article I will calculate the alteration in chance of your opponent having certain starting hands, if you are dealt specific cards.

Basics on Card Combinations in Texas Holdem

A poker deck contains 52 cards. In Texas Holdem each player is dealt 2 hole cards. The number of hole card combinations is therefore 52*51= 2652. Because order is irrelevant we divide this number by 2 to obtain 1326 card combinations. What is the chance that a player is dealt pocket aces? Now we have to count the number of pocket aces combinations: AcAd, AcAh, AcAs, AdAh, AdAs, AhAs, which makes 6 combinations out of a total of 1326 combinations. Hence, the chance that a player is dealt pocket aces is 6/1326= 1/221.

I got an Ace in my hand

Now suppose I have an Ace in my hand, how does this decrease the chance that my opponent (assume that we are heads up in a pot) is dealt a value hand like AA, AK or AQ. To calculate this I assume I am dealt the Ad (it doesn’t matter which ace I am dealt, since each ace impacts the difference in chance equally).
In the following table I put all possible AA, AK, AQ combinations:

Table 1. All possible AA, AK and AQ combinations

As you can see there are 6 combinations of AA, 16 combinations of AK and also 16 combinations of AQ, which makes a total 38 combinations. Remember that there are 1326 possible combinations, so the chance that an opponent is dealt one of these combinations is 19/663= 2.9%

If we are dealt the Ad, our opponent cannot have card combinations containing the Ad. In the following table I have made these card combinations yellow, which means that our opponent cannot have these combinations.

Table 2. All possible AA, AK and AQ combinations, where combinations containing the Ad are yellow

This means that there are 27 combinations of AA, AK and AQ left. Our Ad decreases the number of total combinations to 1326-51=1275. Hence, the chance that our opponent has AA, AK or AQ, when we are dealt one ace is 9/425= 2.1%.

The drop from 2.9% to 2.1% implies that the chance that your opponent is dealt AA, AK or AQ decreases 26% when you are dealt an ace.

I have AK. What is the chance that my opponent has AA or KK?

AK is a strong holding, because it dominates many starting hands, it does well against underpairs and because an ace and a king is out, the chance that your opponent holding aces or kings is small. But how small precisely and how does your AK impact the chance your opponent having aces or kings? We assume again that we are heads up and we are dealt AdKd (again, for this calculation it does not matter which AK combinations we have).

In the following table I put all possible AA and KK combinations. There are 6 combinations each of AA and KK. This makes the chance being dealt AA or KK 12/1326, which is 2/221 or 0.9%.

Table 3. All possible AA and KK combinations

If we are dealt AdKd, the number of combinations of AA and KK our opponent can have drops from 12 to 6. Since the number of total possible combinations drops from 1326 to 1225 (there are 51 combinations each with Ad or Kd, which makes a total of 101 unique combinations containing Ad or Kd, since we count AdKd only once), the chance our opponent has AA or KK is 1/204 or 0.5%.
The drop of 0.9 to 0.5 implies that the chance your opponent is dealt AA or KK decreases 46% when you are dealt AK.

Table 4. All possible AA and KK combinations, where the combinations containing Ad or Kd are yellow

In the next article I will discuss what the implications are of the above calculations and how these affect your decision making. If you have any remarks or questions, feel free to respond.

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