"Blockers" Mathematical Proof?

    • metza
      metza
      Bronze
      Joined: 28.01.2012 Posts: 2,220
      Ok,
      I understand the basic mathematical reasoning behind "blocking", but I was thinking about it before going to sleep last night and it seemed dubious to me.

      Before anyone jumps and says I don't understand the concept, I do.
      eg A :diamond: 5 :club: means opponent cannot have A :diamond: A :club: , A :diamond: A :heart: , A :diamond: A :spade:
      Therefore in theory, opponent is 50% less likely to have AA

      But is this really true?

      The way I see it there are 52 'slots' to be filled by the cards, any number of order of cards (52!) are possible.

      Let's say you are SB and opponent is BB. You will get the 1st and 7th card of this string of 52 and your opponent will get the 2nd and 8th card in this order.

      How is this combination

      A :diamond: A :heart: x x x x 5 :club: A :spade: x x x x x x x x x x .....

      any less likely than

      2 :heart: A :heart: x x x x 7 :club: A :spade: x x x x x x x x.....

      or even

      A :heart: A :diamond: x x x x A :club: A :spade: x x x x x x x x......

      Where the x is another random card.

      In one case we get A5o and the other we get 72o. Any combination is not any more or less likely to happen than another. AA gets dealt at a specified frequency. So does A5. These frequencies are independent since every card is an independent unit in the deck and has just as much chance of being 1st in the deck as 28th. And sometimes they will overlap to give A5 vs AA.

      Though us having the ace of diamonds makes three combinations of AA impossible, to me all this seems to prove is that we are not going to run our A :diamond: 5 :club: into A :diamond: A :heart: , not that we are any less likely to be up against aces in any given hand just because we have been dealt A5.

      I am sure there is a way to prove that blockers do actually make the difference but to me it seems rather suspect and I don't want to just accept things blindly. I imagine the proof is something like the Monty Hall Problem which doesn't seem like it should work but it does. :f_biggrin:
  • 10 replies
    • DannyJQ
      DannyJQ
      Bronze
      Joined: 13.07.2011 Posts: 1,507
      Think in ranges then it will make more sense ;) .
    • dannywratten
      dannywratten
      Gold
      Joined: 11.05.2010 Posts: 1,462
      Originally posted by DannyJQ
      Think in ranges then it will make more sense ;) .
      This.

      Imagine a situation where you squeeze KJo for example, villain has less combinations of AK, KQ,KK,JJ,AJ etc to defend with against the 3b.

      Maybe not the best example, but might help a bit :)
    • metza
      metza
      Bronze
      Joined: 28.01.2012 Posts: 2,220
      Originally posted by metza

      Before anyone jumps and says I don't understand the concept, I do.
      eg A :diamond: 5 :club: means opponent cannot have A :diamond: A :club: , A :diamond: A :heart: , A :diamond: A :spade:
      Therefore in theory, opponent is 50% less likely to have AA

      Did you guys not read this? You just repeated what I already said I know, which is frustrating and makes me think you only read the title :f_biggrin:

      It's not like I don't get the concept. I'm not asking or debating about what it theoretically does to the opponents range. I just presented some ideas about preflop deck combinations and probabilities of any cards being in any position in the deck as being fixed and even, which seems to be in contradiction to this theory of blockers. What I want to know is how to rationalize these two things which both appear to be mathematically true but contradict each other.
    • Optroot
      Optroot
      Bronze
      Joined: 11.05.2008 Posts: 250
      You are looking at the probability of two specific hands (with suits known), and not taking into account the number of ways to make these hands. The probability of you having any two specific cards is the same, which I think is confusing you.

      P(7 :club: 2 :heart:) = P(A :diamond: 5 :club:) = P(A :heart: A :spade:)

      This is still true with dead cards. As you mentioned, if villain has A :spade: A :heart: , we are actually equally likely to have A :diamond: 5 :club: or 7 :club: 2 :heart: . So you are right to notice that.

      But we are ignoring the number of ways we can make these hands. You are correct that the two card strings have equal probability. When we have A :diamond: 5 :club: though, there are fewer total number of card strings that give him AA (there are 3)*, as opposed to KK (there are 6)*.

      When we compute the probability of villain having AA, we sum the probability of each of the possible events (each card string is an event) that give him AA, and each of them are equally likely. This is where the blockers come in. We limit the total number of card strings, hence we lower the probability of them having AA.

      * Ignoring order and the other player's cards. If you look at it another way, it will still be a factor of two different.
    • metza
      metza
      Bronze
      Joined: 28.01.2012 Posts: 2,220
      Originally posted by Optroot
      You are looking at the probability of two specific hands (with suits known), and not taking into account the number of ways to make these hands. The probability of you having any two specific cards is the same, which I think is confusing you.

      P(7 :club: 2 :heart:) = P(A :diamond: 5 :club:) = P(A :heart: A :spade:)

      This is still true with dead cards. As you mentioned, if villain has A :spade: A :heart: , we are actually equally likely to have A :diamond: 5 :club: or 7 :club: 2 :heart: . So you are right to notice that.

      But we are ignoring the number of ways we can make these hands. You are correct that the two card strings have equal probability. When we have A :diamond: 5 :club: though, there are fewer total number of card strings that give him AA (there are 3)*, as opposed to KK (there are 6)*.

      When we compute the probability of villain having AA, we sum the probability of each of the possible events (each card string is an event) that give him AA, and each of them are equally likely. This is where the blockers come in. We limit the total number of card strings, hence we lower the probability of them having AA.

      * Ignoring order and the other player's cards. If you look at it another way, it will still be a factor of two different.
      This is more the answer I was looking for, thanks. So it is a matter of conditional probability that negates the fact that all card strings are equally possible? Once we have this incomplete information we can then discern information about the card string after the fact?

      I wouldn't exactly say I was ignoring the different ways to make these hands, I have acknowledged this in the first post, but more the fact that yes we do "limit" the total number of card strings, but by the time we already have A5, we're already within in one specific fixed card string of 52! possible so it's an after-the-fact limitation.

      The thing then that I am dicey on is that its very hard to be certain that the overall strings will deal less often (again just an example hand) any variant of A5 vs any variant of AA less often than any variant of Q6 vs any variant of AA, simply because there are 52! (which is a shitload) iterations of card strings, therefore its very hard to know/solve if this does in fact occur less often, so I guess I will just have to trust that it does.
    • ilidek
      ilidek
      Bronze
      Joined: 06.07.2010 Posts: 2,952
      http://en.wikipedia.org/wiki/Bayes%27_theorem

      This is what you're looking for. When you understand Bayes theorem you will understand this as well. Remember that you are looking for event one: ace was dealt, event two: somebody got two aces, you want to count that possibility that someone have two aces when ace was dealt. Then you should compare your outcome with posibilitty of event two without event one.

      Btw there are difference betwen two random cards and two sepcific cards. There are different exemplar (?) of dealing with that.
    • w34z3l
      w34z3l
      Coach
      Coach
      Joined: 03.08.2009 Posts: 13,295
      I'm not gonna go into the maths even though I could. I will just use simple logic--->

      You have a bag of 5 red stones and 5 blue stones. You take 2 red stones out of the bag.

      What is the chance your opponent takes a red stone. Is it 1/2?

      Of course not, some of the reds are gone - the chance is now only 3/8.
    • Optroot
      Optroot
      Bronze
      Joined: 11.05.2008 Posts: 250
      The card strings are always equally possible. The blockers only change the number of possible strings there are.

      Be careful with saying that we are already in a specific card string. We actually are technically not when we try to calculate the probability of an unknown hand, because according to us, we don't know which one we are in. Given that we have A :diamond: 5 :club: , there are a certain number of card strings possible that give the opponent AA, and a certain number that give them KK. The probability of each of these is proportional to the number of possible card strings we could be in.

      I'll try to make this more clear with an example. Let's say we are playing a two player game with a simplified deck. There are only three cards with only ranks A, K, Q all of spades.

      I think you would agree that there are 3! ways to order this deck, each of them equally likely. Suppose I get the first card, and you get the second.

      All possible orderings are

      P(AKQ) = 1/6
      P(AQK) = 1/6
      P(KAQ) = 1/6
      P(KQA) = 1/6
      P(QAK) = 1/6
      P(QKA) = 1/6

      Now suppose I am dealt an ace. Which of these strings could we still be in? There is only AKQ and AQK. So how many of these result in you getting a Q? We know that each of these are equally likely, so that means that each has probability 1/N = 1/2, where N is the number of possible card strings. Since only one of them results in you getting a queen (given I have an ace), then the probability of you having a queen given I have an ace is 1/2, as expected.

      If we do this for an actual deck, the way that we calculate the probability is by 'counting' the number card strings that give us the result we want, and dividing it by the total number of card strings (52!).
    • fl4m3r
      fl4m3r
      Bronze
      Joined: 26.11.2009 Posts: 56
      Go read about conditional probability.. I think this is a case of : Given you have Probability A, what is the probability of B happening?

      w34z3l gave a very good example..
    • NateMilla
      NateMilla
      Basic
      Joined: 10.06.2012 Posts: 38
      Yeah in overall play. But sometimes luck plays a factor.

      Then again I've shove A9
      On Ax99

      Called by AA.

      Say you have A :diamond: 5 :diamond: Now we know there's eleven :diamond: s left in the deck. Of which it is impossible for Villian to have A :diamond: . Your post reads, it isn't possible for A :diamond: A :heart: to exist. This is true. You can always assume you have the nut flush, except in those rare straight flush draw boards. Villian simply cannot have a card that you have, is all you said.