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Math Help - Poker Problem - Probability of Getting a Pair After the Flop

  1. #1
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    Question Poker Problem - Probability of Getting a Pair After the Flop

    I'm having trouble with a very specific problem.

    Given the following assumptions:

    1. Start with a standard 52-card deck of playing cards.
    2. Deal five cards from the deck.
    3. These cards are "visible", i.e., known to the solver of the problem.
    4. None of the five visible cards are of the same rank, i.e., there are no pairs of fives, sixes, jacks, queens, etc.

    Given these assumptions, what is the probability that, if two additional cards are dealt from the remaining 47-card deck, that the combined seven dealt cards will contain a five card hand that contains a pair (e.g., two fives or two sixes).

    Here's how I've analyzed the problem so far.

    Let C(n, k) be the binomial coefficient "n choose k". After the initial five cards have been dealt, there are 47 cards left in the deck, and there are C(47, 2) = 1081 ways to deal the two additional cards.

    However, only 15 of the remaining 47 cards would "pair up" with one of the five cards that have already been dealt. (Three remaining cards of the same rank for each of the five originally dealt cards.) In order for the combined seven dealt cards to contain a five card hand with a pair, then the last two dealt cards would have to contain one or more of these 15 cards.

    The probability of this happening is computed by choosing each of the 15 cards, pairing them with each of the remaining 46 cards, then dividing that product by the total number of outcomes for the final two cards, or 15*46/1081 = 63.83%.

    My problem is that I've also written some code to simulate this problem and produce a monte carlo result, and the numbers don't match! The results of my monte carlo simulation are approximately 58.7% and are fairly consistent.

    If any of you sees a flaw in the mathematical anlaysis I've presented here I would really appreciate a corrective post!

    Thanks!
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  2. #2
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    Quote Originally Posted by regenerator View Post
    I'm having trouble with a very specific problem.

    Given the following assumptions:

    1. Start with a standard 52-card deck of playing cards.
    2. Deal five cards from the deck.
    3. These cards are "visible", i.e., known to the solver of the problem.
    4. None of the five visible cards are of the same rank, i.e., there are no pairs of fives, sixes, jacks, queens, etc.
    Given these assumptions, what is the probability that, if two additional cards are dealt from the remaining 47-card deck, that the combined seven dealt cards will contain a five card hand that contains a pair (e.g., two fives or two sixes).

    Here's how I've analyzed the problem so far.

    Let C(n, k) be the binomial coefficient "n choose k". After the initial five cards have been dealt, there are 47 cards left in the deck, and there are C(47, 2) = 1081 ways to deal the two additional cards.

    However, only 15 of the remaining 47 cards would "pair up" with one of the five cards that have already been dealt. (Three remaining cards of the same rank for each of the five originally dealt cards.) In order for the combined seven dealt cards to contain a five card hand with a pair, then the last two dealt cards would have to contain one or more of these 15 cards.

    The probability of this happening is computed by choosing each of the 15 cards, pairing them with each of the remaining 46 cards, then dividing that product by the total number of outcomes for the final two cards, or 15*46/1081 = 63.83%.

    My problem is that I've also written some code to simulate this problem and produce a monte carlo result, and the numbers don't match! The results of my monte carlo simulation are approximately 58.7% and are fairly consistent.

    If any of you sees a flaw in the mathematical anlaysis I've presented here I would really appreciate a corrective post!

    Thanks!
    Suppose the five dealt cards are 2,3,4,5,6 and the two additional cards are 7,7.

    Also suppose the two additional cards are 6,6 how do you count that?

    CB
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  3. #3
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    Re: Poker Problem - Probability of Getting a Pair After the Flop

    Suppose the five dealt cards are 2,3,4,5,6 and the two additional cards are 7,7.

    Also suppose the two additional cards are 6,6 how do you count that?
    I think my analysis covers both these cases. In the former case, the 7,7 deal of the final two cards isn't included in the 15*46 potential "hits" that would result in a successful outcome. In the latter case, the 6,6 deal is considered in the potential hits since, once I pick a 6 as the first card in the pair of additional cards, I put the other two 6s back in the deck as potential second cards in the pair of additional cards. That's why I multiply 15 by 46 and not by, for example, 47-3 = 44...I need to account for the fact that, any of the fifteen ranked cards that could result in a five card hand with a pair could be paired with one of the other two cards with the same rank (remember the first card with that rank is already in the first five dealt cards).

    Does that make sense?
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  4. #4
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    Captian Black,

    Thanks again for the reply, I obviously didn't read it carefully enough! I was missing the "7 7" case, and with that observation I've been able to solve the problem:

    There are three favorable combinations of the last two cards to consider:

    1. Some combination of the fifteen cards that would pair up with the cards already showing. There are C(15,2) = 105 such combinations.
    2. Some combination of one of those 15 cards with some other card. There are 15 * (47 - 15) = 480 such combinations.
    3. As you observed, a pair of cards, neither of which pair up with the first five cards. There are 13-5=8 remaining unseen ranks, and 4 cards for each rank, so the number of these combinations is 8 * C(4,2) = 8 * 6 = 48.

    So the total number of "hit" possibilities is 105 + 480 + 48 = 633. Therefore the hit probability is 633 / 1081 = 58.557%, which perfectly matches my new and improved monte carlo simulation.

    Thanks again for the help!
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  5. #5
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    Quote Originally Posted by regenerator View Post
    I'm having trouble with a very specific problem.

    Given the following assumptions:

    1. Start with a standard 52-card deck of playing cards.
    2. Deal five cards from the deck.
    3. These cards are "visible", i.e., known to the solver of the problem.
    4. None of the five visible cards are of the same rank, i.e., there are no pairs of fives, sixes, jacks, queens, etc.

    Given these assumptions, what is the probability that, if two additional cards are dealt from the remaining 47-card deck, that the combined seven dealt cards will contain a five card hand that contains a pair (e.g., two fives or two sixes).

    Here's how I've analyzed the problem so far.

    Let C(n, k) be the binomial coefficient "n choose k". After the initial five cards have been dealt, there are 47 cards left in the deck, and there are C(47, 2) = 1081 ways to deal the two additional cards.

    However, only 15 of the remaining 47 cards would "pair up" with one of the five cards that have already been dealt. (Three remaining cards of the same rank for each of the five originally dealt cards.) In order for the combined seven dealt cards to contain a five card hand with a pair, then the last two dealt cards would have to contain one or more of these 15 cards.

    The probability of this happening is computed by choosing each of the 15 cards, pairing them with each of the remaining 46 cards, then dividing that product by the total number of outcomes for the final two cards, or 15*46/1081 = 63.83%.

    My problem is that I've also written some code to simulate this problem and produce a monte carlo result, and the numbers don't match! The results of my monte carlo simulation are approximately 58.7% and are fairly consistent.

    If any of you sees a flaw in the mathematical anlaysis I've presented here I would really appreciate a corrective post!

    Thanks!
    Hi Regenerator,

    The problem with the computation 15 * 46 / 1081 is that the 15 and 46 have some cards in common. This results in over-counting.

    Let's see if we can avoid this problem, starting by defining more precisely what is meant by "a pair". A hand might include exactly one pair, two pairs, or three of a kind. Any of these combinations might be said to include a pair. Let's analyze each of the three possibilities separately.

    Three of a kind: There are 5 choices for the rank of the card, since it must match one of the first 5 cards dealt, and then there are \binom{3}{2} = 3 ways to choose the 2 cards. So
    P(\text{three of a kind}) = 5 * 3 / \binom{47}{2} = 0.01388.

    Two pairs: There are \binom{5}{2} = 10 ways to choose the two ranks from among the first 5 cards dealt, and then there are 3 * 3 = 9 ways to choose the cards, so
    P(\text{two pairs}) = 10 * 9 / \binom{47}{2} = 0.08326.

    Exactly one pair: This can happen in two ways. First, we could match exactly one of the first 5 cards dealt. There are 5 ways to choose the rank and 3 choices for the card, and then there are 47-15 = 32 choices for the non-matching card. Second, we could draw a pair which does not match any of the previous cards. There are 8 ways to choose the rank and \binom{4}{2} = 6 ways to choose the two cards. So
    P(\text{exactly one pair}) = \frac{5 * 3 * 32 + 8 * 6}{\binom{47}{2}} = 0.4884.

    If you are interested in any one of these mutually exclusive possibilities then the total probability is 0.01388 + 0.08326 + 0.4884 = 0.5855, which is close to your simulation result.
    Last edited by awkward; November 3rd 2008 at 05:09 PM. Reason: corrected typo
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  6. #6
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    Possible approach to matching piar problem

    Consider this as a NEARLY binomial problem. Nearly binomial because we are not using replacement. However if we define p(success)=15/47 because there are 15 cards that can pair the 5 known cards. Now use binomial probability with:

    number of trials = 2 (2 more cards drawn)
    number of successes =0 (we will use the complement of zero successes for at least one success)
    probability of success on one trial = 15/47 = 0.319

    The result P(at least one success, matched pair) = (1 - .463)= .537 or 53.7%

    This answer underestimates the Monte Carlo answer of 58.7% by about as much as your analysis overestimates the MC answer.
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