Symmetry of adding probabilities - poker example

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dapaulcz

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I'm having trouble getting my head around a probability question, and could use some help. The setup = Texas holdem, 8 players, after the turn (only one card left to be flipped), and no one has folded. The board has a pair - let's say two red jacks. So:

Total number of players' hole cards = 8 x 2 = 16
Total number of cards not held by players and not already known = 52 - 4 (on the board) - 16 (hole cards) = 32

Question: What is the probability of at least one of the remaining jacks being held by a player in his/her hole cards?

My thinking: Probability of jack of clubs being held = 16/(16+32) = 1/3, Probability of jack of spades being held = 16/(16+32) = 1/3, so odds of EITHER one being held is 1/3 + 1/3 = 2/3

Where my thinking breaks down: If I reverse the question, i.e. What is the probability of at least one of the remaining jacks being in the unknown non-hole cards?

Applying similar thinking gets me 2/3 + 2/3 = 4/3! Obviously nonsense.

Where am I going wrong here? What is the correct answer?
 
dapaulcz said:
I'm having trouble getting my head around a probability question, and could use some help. The setup = Texas holdem, 8 players, after the turn (only one card left to be flipped), and no one has folded. The board has a pair - let's say two red jacks. So:

Total number of players' hole cards = 8 x 2 = 16
Total number of cards not held by players and not already known = 52 - 4 (on the board) - 16 (hole cards) = 32

Question: What is the probability of at least one of the remaining jacks being held by a player in his/her hole cards?

My thinking: Probability of jack of clubs being held = 16/(16+32) = 1/3, Probability of jack of spades being held = 16/(16+32) = 1/3, so odds of EITHER one being held is 1/3 + 1/3 = 2/3

Where my thinking breaks down: If I reverse the question, i.e. What is the probability of at least one of the remaining jacks being in the unknown non-hole cards?

Applying similar thinking gets me 2/3 + 2/3 = 4/3! Obviously nonsense.

Where am I going wrong here? What is the correct answer?
Click to expand...

You only know 6 cards leaving 46 cards. There are 46C2 [= 46!/(44! 2!) = 45*46/2] different possible 2 card hands. Of those 44C2 do not have a jack leaving, roughly, 9% which have both jacks possibly in two different hands or the same hand. There are 2 [spade or club jack]*44[any non jack]=88 with possibly 1 jack and 1 with possibly both jacks.

If you are going to approach it from the other direction, i.e. only 32 cards in the deck remaining, you will also have to assume something about the other 14 cards held by the other players. The accuracy of your 'guesstimation' will depend on the accuracy of what you assume is held by them. For example if you are heads up against a very tight player (KQ suited or better to see the flop), the player probably has both jacks or neither jack.
 
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