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Is it possible to work this out? Even roughly.

ShaunyTShaunyT Member Posts: 619
So, I'm trying to work out the simplest way of asking this.

Assuming there are only 2 people left in the hand (and obviously you cant know what folded cards were). If you are dealt AK suited or unsuited, how much does this alter the probability of your opponent having AA or KK vs when you are dealt QQ?

I'm talking about those pre flop raising situations where its obvious both parties have a monster, I was just wondering how much it factors in that you have blockers to AA KK when you have AK in your hand rather than QQ.

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    chicknMeltchicknMelt Member Posts: 1,159
    edited April 2020
    what your asking about is combinatorics.

    There are 6 possible combinations of each pocket pair, and 16 combinations of every other hand (4 suited and 12 offsuit)

    I'll start you off with the example of holding AK yourself

    If you hold an Ace and a King, it reduces the possible combinations of AA and KK to 4 each and AK to 9 combos.

    so there there are 6 combos of QQ, 4 of KK, 4 of AA and 9 of AK. so 23 in total, 6 of which are QQ.

    6/23 = 26%.
    8/23 or 35% that the have Kings or aces
    12/23 or 52% that they also have AK


    In the scenario that you have QQ
    2 combos of QQ left
    6 each of Aces and Kings
    16 of AK
    30 in total.

    12/30 or 40% they have AA or KK
    16/30 or 53% they have AK
    2/30 or 7% they have QQ

    ---edit--
    3 combos of KK and AA in the Ak example
    and its one combo left of QQ, not 2 in the holding QQ example. doh.

    in my defence I wrote this while playing..


    just as well Pingu is more thorough!

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    EvilPinguEvilPingu Member Posts: 3,462
    ShaunyT said:

    Assuming there are only 2 people left in the hand (and obviously you cant know what folded cards were). If you are dealt AK suited or unsuited, how much does this alter the probability of your opponent having AA or KK vs when you are dealt QQ?

    I'm talking about those pre flop raising situations where its obvious both parties have a monster, I was just wondering how much it factors in that you have blockers to AA KK when you have AK in your hand rather than QQ.

    There are six ways to be dealt a pocket pair - Taking Aces for example, you have
    AhAd
    AhAc
    AhAs
    AdAc
    AdAs
    AcAs
    Having one blocker (as is the case with AK) reduces this to 3 combinations.

    Meanwhile, there are four ways to have AKs and a further 12 ways to have AKo, for a total of 16 AK combos. Suppose we have AK ourselves - suited or not makes no difference, so let's say we have AdKs.

    That removes all the combinations of AK with Ad (AdKd, AdKc, AdKs, AdKh) and the combinations of AK with Ks (AdKs, AcKs, AhKs, AsKs), leaving 9 combinations of AK remaining (not 8, since the above has removed AdKs twice).

    Thus, having AK when we get stacks in means there are the following combos:
    9 x AK
    3 x AA
    3 x KK (by same reasoning as AA)
    6 x QQ

    As opposed to
    16 x AK
    6 x AA
    6 x KK
    6 x QQ
    if we have no A, K or Q in our hand

    Or
    16 x AK
    6 x AA
    6 x KK
    1 x QQ
    when we have QQ ourselves.

    NB: AhAd and AdAh, for instance, are counted as the same hand and one combination, since the order the cards are dealt to you has no effect on anything.
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    ShaunyTShaunyT Member Posts: 619
    Great answer. Thank you.

    So if I've understood that correctly. You holding AK reduces the chance of them having AA or KK by 18% as opposed to holding QQ.

    That's quite a big decrease. Also I learned a new word "combinatorics". Brilliant.
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    EvilPinguEvilPingu Member Posts: 3,462
    edited April 2020
    EvilPingu said:


    Thus, having AK when we get stacks in means there are the following combos:
    9 x AK
    3 x AA
    3 x KK (by same reasoning as AA)
    6 x QQ

    Or
    16 x AK
    6 x AA
    6 x KK
    1 x QQ
    when we have QQ ourselves.

    Hence:
    Having AK means:
    P(AA) = 1/7
    P(KK) = 1/7
    P(AK) = 3/7
    P(QQ) = 2/7

    Having QQ means
    P(AA) = 6/29
    P(KK) = 6/29
    P(AK) = 16/29
    P(QQ) = 1/29

    *NB: Notation P(__) = Probability of (opponent holding). In reality, the villain won't be limited to those four hands and they won't be playing them the same way every time, either, so take the numbers with a big pinch of salt.

    Additionally, equities are important to consider when using combinatorics to make decisions - If you need 40% to call correctly, it's no good having 50% equity against half of the villain's range if we have 20% equity against the other half, because you can potentially find yourself using the "I'm right more than half the time" to justify a bad decision.

    I recommend playing about with https://www.pokerstrategy.com/poker-software-tools/equilab-holdem/ - Put individual hands in, toggle the # of combos display on and you'll be able to see what I've explained above in action when you attempt to input a range for the opponent. It does a lot more than just equities and combos, too. It's a pretty decent piece of software for a freebie.
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    EvilPinguEvilPingu Member Posts: 3,462
    Take this situation - You want to know if you should jam 98s and A2o . You have a range that you think the villain is going to call with if you shove on him.

    The equities are almost identical, so you should probably be shoving them both or folding them both, right?

    But actually, the shove with A2o will get called less than a shove with 98s in an identical situation because of the Ace blocker meaning there are less combos of Aces which our opponent can have, which is obviously good for us since we're behind against a calling range.





    Once you start to get into more extreme situations where you have to consider ICM (such as in a DYM where getting called is generally terrible for us, also calling ranges are super tight and consist of a ton of strong Ax and big pairs), it's entirely possible that you'll find situations where A2o is a jam and 98s is a fold, due to the Ace blocker making it harder for our opponent to have a hand to call with.
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    WUBBZY1401WUBBZY1401 Member Posts: 109
    G00D T0 S33 Y0U 1NV35T3D 1N 4 N3W K3YB04RD 5H4UNY
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    ShaunyTShaunyT Member Posts: 619

    G00D T0 S33 Y0U 1NV35T3D 1N 4 N3W K3YB04RD 5H4UNY

    Haha , not quite. Typing from my phone. If you chat at the tables 1T W1LL Be th1s rubb1sh
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