Results 1 to 6 of 6

Math Help - hold em math

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    49

    Cool hold em math

    your playing a game of holdem you pick up ten ten ,
    your opponent goes all in pre flop
    you decide to call his all in
    what are the chances of hitting another ten with five cards to come ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member garymarkhov's Avatar
    Joined
    Aug 2009
    Posts
    149
    Awards
    1
    Quote Originally Posted by jickjoker View Post
    your playing a game of holdem you pick up ten ten ,
    your opponent goes all in pre flop
    you decide to call his all in
    what are the chances of hitting another ten with five cards to come ?
    In my view, you shouldn't care much about the answer to this question. The only time you should care a lot about the probabilities involved is when you are choosing a betting strategy - in this instance, you should be interested in the probability that your opponent has a higher pair before calling his all in.

    But maybe you're just curious and want to know the probability before seeing all the cards on the table. In that case, the answer seems to me to depend on how many players there are (how many people were dealt two cards before it was just you vs. your opponent), and whether you are "burning" cards (laying one card aside before the flop, another before the turn, and one more before the river). I'll think about that. And maybe someone else will chime in.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2006
    Posts
    42
    the chances of hitting your 3rd ten are 1 - (the chance of not hitting your 3rd ten).

    which is

    = 1 - (50/52)*(49/52)*(48/52)*(47/52)*46/52)
    = 33.13%
    (check the maths, i had to use a dodgy calculator)

    So you should hit that 3rd ten about 33% of the time. Bare in mind that person with the Pocket aces will also hit their 3rd ace about 33% of the time.

    with all this taken into account, the odds of your pocket 10's beating pocket aces (taking into account four of a kinds and straights) is about 20%.

    Hope that helps.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2006
    Posts
    42
    Quote Originally Posted by garymarkhov View Post
    But maybe you're just curious and want to know the probability before seeing all the cards on the table. In that case, the answer seems to me to depend on how many players there are (how many people were dealt two cards before it was just you vs. your opponent), and whether you are "burning" cards (laying one card aside before the flop, another before the turn, and one more before the river). I'll think about that. And maybe someone else will chime in.
    Gary, From this post and the roulette post I'm going to suggest that you lay off the gambling for a little bit

    The number of players playing and whether or not you are burning cards has no effect at all on the probability of making a set.

    Perhaps this somewhat strange example will give you an idea of why your logic is flawed.

    Bruxism and Gary are playing heads up texas hold-em. Bruxism holds pocket tens, while Gary holds pocket aces and they both go all in. Just as they put all their money in the pot, a goat walks into the room and starts to eat the pack of cards! Thinking quickly, Bruxism manages to wrestle the goat to the ground and save the last 5 cards before he eats them. Just enough cards for a flop, turn and river.

    Have the odds of hitting my ten been affected by the appearance of the goat?

    You can think of all cards that you haven't seen as being eaten by the goat. Players who folded preflop, burnt cards and even people who are still in the hand whos cards you haven't seen yet.

    I hope this makes things a little clearer.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member garymarkhov's Avatar
    Joined
    Aug 2009
    Posts
    149
    Awards
    1
    Quote Originally Posted by bruxism View Post
    Gary, From this post and the roulette post I'm going to suggest that you lay off the gambling for a little bit
    Haha. You know I wasn't suggesting the Martingale strategy was a good one, right? I thought it was clear that a near 1% chance ("the improbable") of losing your bankroll on any given fresh betting round is nothing to be cheery about if you play long enough.

    Quote Originally Posted by bruxism View Post
    Bruxism and Gary are playing heads up texas hold-em. Bruxism holds pocket tens, while Gary holds pocket aces and they both go all in. Just as they put all their money in the pot, a goat walks into the room and starts to eat the pack of cards! Thinking quickly, Bruxism manages to wrestle the goat to the ground and save the last 5 cards before he eats them. Just enough cards for a flop, turn and river.
    That's a wonderful analogy! The intuition it gives is fascinating and, I think, correct. From my perspective, any card I have yet to see is akin to a card that the goat eats. Similarly, if you were to deal from the middle of the deck, two thirds into the deck, etc., the chances of pulling up a ten are the same as if you were to deal from the top of the deck. So, I agree, burned cards and cards other players hold or held are not relevant to our calculation.

    However, let me humbly suggest that 1 - (50/52)*(49/52)*(48/52)*(47/52)*(46/52) may not be the right calculation. Because of the two tens I'm holding in my hand, I know that there are only 50 cards (the cards I've yet to see) that could possibly contain the other two tens. The calculation should continue to change as new cards appear face up on the table. There's a reasonably good chance that you will once again show me to be wrong, but I'll submit that the correct probability is 1-(48/50)*(47/49)*(46/48)*(45/47)*(44/46) = 19.2%.

    Curious to know what you think.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2006
    Posts
    42
    Quote Originally Posted by garymarkhov View Post
    However, let me humbly suggest that 1 - (50/52)*(49/52)*(48/52)*(47/52)*(46/52) may not be the right calculation. Because of the two tens I'm holding in my hand, I know that there are only 50 cards (the cards I've yet to see) that could possibly contain the other two tens. The calculation should continue to change as new cards appear face up on the table. There's a reasonably good chance that you will once again show me to be wrong, but I'll submit that the correct probability is 1-(48/50)*(47/49)*(46/48)*(45/47)*(44/46) = 19.2%.

    Curious to know what you think.
    Thanks, of course you are absolutely correct that is the calculation, can't believe i stuffed it up. Serves me right for coming on here instead of doing my maths assignment like i should be.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Why does this congruence hold?
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: April 3rd 2011, 04:16 PM
  2. Replies: 1
    Last Post: December 14th 2010, 08:53 AM
  3. Does the converse hold?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 26th 2010, 08:39 AM
  4. Force to hold a spring
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 10th 2009, 11:57 AM
  5. Parallelogram law does not hold in C([0,1])
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 14th 2008, 08:50 AM

Search Tags


/mathhelpforum @mathhelpforum