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Homework Help: Poker Probability - the probabilities of each hand in a 2 man game winning

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    You will assess the probability of a certain hand winning after the starting hands are dealt. You only need to consider two players at a time. (At no point are we considering betting or folding playing a part in winning)

    Starting Points
    -Try and watch some games of Texas hold 'em either on television or youtube. They show probabilities throughout each game
    -Look at different possible starting hands, eg pairs, same suit....etc.
    -Look at the following example with two players;

    If player 1 has two fives (a heart and a spade) as his starting hand and player 2 has an ace of spades and a king of hearts.

    -What is the probability that player 1 will win this deal after the 5 community cars are dealt?
    -Select some different starting hands, two at a time, and calculate the probability that one hand will win - this will form the majority of your assignment.

    2. Relevant equations

    n!/(n-r)! for permutations

    n!/(n-r)!r! for combinations

    3. The attempt at a solution

    So far I have done mere scratchwork on the problem as I am not 100% sure where to start or what i should do.

    I have selected two random hands which are as follows

    hand 1 consists of a 7 of spades and a 4 of hearts
    hand 2 consists of a 5 of diamonds and a Jack of diamonds

    this hand is good because there are no pairs, no two cards follow each other numerically and it incorporates two crads of the same suit.

    I know that combinations are needed to complete this problem as order does not matter when the community cards are turned up.
    The probability of each hand winning does not add up to 100% like it should, because of the small chance that the hands tie.
    I also know that i need to go through the chances of each of the hands winning with the different combinations as shown below...

    High Card (no special combinations)
    One Pair (eg two jacks
    Two Pair (eg two aces and two fours)
    Three of a Kind (eg three queens)
    Straight ( all 5 cards in order, eg 7, 8 ,9 ,10, J)
    Flush (all 5 cards the same suit)
    Full House (three of a kind and a pair)
    Four of a kind (eg four kings)
    Straight Flush (all 5 cards in order and of the same suit)

    I know I need to calculate this doing combinations, but I also know that there are over 1 million different combinations of cards that could turn up and instead of calculating each of them out, if there is a shorter way to do it?
    If not are there any shortcuts I can take?

    Any help is greatly appreciated as I missed the lessons on probability and can use any help that I can get.
  2. jcsd
  3. Sep 23, 2009 #2
    Alright so i did some additional work to get some probability values and this is what i have done so far

    cards to help hand 1 win are

    12/13 = 92%

    cards to help hand 2 win are
    all diamonds (5 diamonds turning up)

    13/13 = 100%

    13C5 = 154,440

    12C5 = 95040

    95040/14440 = 0.62

    4C1 x 44C4 + 4C1 x 44C4 + 4C1 x 44C4 + 4C1 x 44C4 + 4C1 x 44C4 = 1,712,304 (amount of combinations that different cards could turn up

    probability of having a heart turn up is

    12C5 = 95040

    probability of having a spade turn up is

    12C5 = 95040

    probability of having a diamond turn up is

    11C5 = 55440

    probability of having a club turn up is

    13C5 = 154440

    this is where I'm restricted in my knowledge of combinations

    probability of having a certain number of a certain suit turning up is

    1/12/48 for hearts = 0.001736 x 100 = 0.1736%

    1/12/48 for spades = 0.001736 x 100 = 0.1736%

    1/11/48 for diamonds = 0.001893 x 100 = 0.1893%

    1/13/48 for clubs = 0.001603 x 100 = 0.1603%

    does anyone have any advice on how to start working out the probability of having a straight flush, my knowledge is limited and I could greatly use any help.
  4. Sep 23, 2009 #3
    Alright I did some more work and this is what i have so far, but I'm not 100% sure that it is correct, can anyone help me out

    Probability of getting a straight flush

    for diamonds (or hand 2)

    11C5 = 55440

    because there are 1712304 possibilities of 5 community cards turning up

    554400/1712304 = 0.032 x 100 = 3.2%

    for hand 1 (or spades)

    12C5 = 95040

    so 95040/1712304 = 0.056 x 100 = 5.6%

    for hand 1 (or hearts this time)

    12C5 = 95040

    so 95040/1712304 = 0.056 x 100 = 5.6%

    therefore 5.6 - 3.2 = 2.4% chance of a straight flush turning up

    Four of Kind

    1/48 + 1/47 1/46 + 1/45 + 1/44 = 0.10065 x 100 = 10.65%

    for a full house which is three of a kind and two pairs it is

    3C3 X 45C2 + 3C2 X 45C3 + 3C1 X 45C4 + 3C3 X 45C2 + 3C2 X 45C3
  5. Mar 13, 2010 #4
    This http://www.barrygreenstein.com/extra_stuff/" [Broken] may help. Essentially, you want to figure out the odds of someone catching up on the next street, and then use branches to figure out who the winner is.
    Last edited by a moderator: May 4, 2017
  6. Mar 13, 2010 #5
    Or, you can just download http://www.pokerstove.com" [Broken].
    Last edited by a moderator: May 4, 2017
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