Probability of no overcards to a pocket pair on the flop

[ChicagoVince]

I was in a discussion about mid-pocket pairs and I am not sure if I correctly computed the probablility of no overcards coming on the flop with pocket 9s. Here is my method, please tell me if I am going about this correctly.

C(30,3)/C(50,3) = 4060/19600 = 20.71%

C(30,3) because there are 20 overcards to your pocket 9s in the deck of 50 that you are trying to dodge. C(50,3) are the total number of possible flops.

I think I'm right, but some confirmation would put my mind at ease.

Thanks in advance

[LetYouDown]

Fair enough.

[ChicagoVince]

Thanks. I guess the tricky part if one wanted to create a table of such figures would be to adjust for the fact that some of the overcards are likely not in the deck, they are in the villians' hands. I haven't come up with a sound strategy, but using worst-case scenarios (like the computation above) is one way.

Any further suggestions are welcomed. My initial thought is to take one over out of the deck for each face card because people are more likely to play those than to play non-faces.

[Foghatlive]

Overcards: Part I
Brian Alspach
Poker Digest Vol. 2, No. 9, April 23 - May 6, 1999

Intuition is a much discussed - though somewhat elusive - topic. Many people when queried will claim they know what intuition is, but find it difficult to give a clear explanation if pressed. Mathematicians frequently talk about people possessing good ``mathematical intuition'', or talk about a problem having an ``unintuitive'' answer. What do they mean by these expressions? In the first case, I believe we are referring to someone who, when confronted with a research problem, has a feeling about the form the answer should take when the answer is unknown, or has a feeling about the route to follow in order to prove a certain answer is correct when the answer is ``known''. In the second case, I believe we are referring to an answer which is surprisingly different from the answer we would have guessed after a cursory inspection of the problem. Let's look at an example of the latter from a recent poker session in which I participated.

I was chatting quietly with my neighbor on the left when my neighbor on the right had his pocket queens lose to a pair of kings - one of which had appeared on the flop. Upon seeing the loser's unhappiness, the player on my left quietly said to me, ``Doesn't he know that almost half the time an overcard to queens will come in the flop?''

The losing player overheard and responded, ``That can't be right! There are only two ranks bigger than a queen.''

There it was! The losing player's intuitive answer is the likelihood of an overcard to queens being small because there are 10 ranks smaller than queen and only two ranks larger than queen. Let's go beyond this cursory examination and determine the exact probability of an overcard.

We take the viewpoint of the player with pocket queens. There are 50 cards he cannot see. We are choosing three cards from 50 for the flop, so there are C(50,3) = 50!/3!47! = 19,600 flops. How many of the flops contain at least one ace or king? The easiest way to determine this is to count the number of flops containing none of them and subtract it from 19,600. Of the 50 unseen cards, 42 remain if we disallow aces and kings. Thus, there are C(42,3)=11,480 flops containing neither an ace nor a king. Hence, there are 8,120 flops containing at least one ace or king. Then the probability of an overcard coming in the flop is 8,120/19,600 = 0.414. This undoubtedly would surprise the losing player and he would interpret it as an unintuitive result.

To calculate the probability of an overcard in the board, we perform similar calculations. There are C(50,5)=2,118,760 boards and there are C(42,5)=850,668 boards without an overcard. The number of boards with at least one overcard is 1,268,092. This implies the probability of at least one overcard in the board is 0.599.

We have seen that 60 percentof the time an overcard to queens will appear in the board. We calculate in the same way the probabilities of overcards to any pocket pair. The probabilities are given in the table below. The column headed ``flop'' is the probability of at least one overcard appearing in the flop, and the column headed ``board'' is the probability of at least one overcard appearing in the board. Readers may find many of these values unintuitive.


http://www.math.sfu.ca/~alspach/mag20/

[AaronBrown]

If you take cards out based on what people are likely to play, you also have to consider the folded hands.

In a 10-handed game, there are 18 unseen pocket cards. If one person stays in the pot with you, it's true she's more likely to have high cards than an average hand. But the 8 people who folded are less likely to have high cards.

If no one stayed in with you then getting the overcards would be somewhat more likely, but then the flop would not be dealt. If lots of people stayed in with you, you might think there is less chance of getting the overcards. But even that is suspect, because the more people in the pot, the more attractive it is to go with drawing hands and the less attractive it is to go in with high card hands.

If you look at data of cards actually dealt, you'll find there's only a tiny bias against high cards, due to the fact that lots of high cards dealt as pocket cards makes it more likely that two or more players will stay in to see the flop.

[ChicagoVince]

Thanks, Aaron. Can you refer me to anywhere that I might be able to find the magnitude of the bias. My real question is whether or not the bias is so small as to be negligable, if there should be some correction factor included. I would think that the impact of the bias would decrease as the held PP decreases. That is, if a player holds kings and even one ace is out of the deck, the percentage difference of an ace landing on the flop is big. Something like 22.6% versus 11.8%.

Thanks

[ChicagoVince]

Thanks, Fog. It looks like he used the same method that I did where you assume that all of the overcards are in the deck and can appear on the board.