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starting pineapple hands

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  1. #1
    Eric's Avatar
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    Default starting pineapple hands

    How many are there where we have no more than 2 of any suit?

    What is the math behind the answer?
  2. #2
    MadMojoMonkey's Avatar
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    I worked this out for the 14 card draws a while ago, so this is much simpler, but the math is the same.

    The probability of drawing 1 of each suit with 1 of the suits having a 2nd, to make a 5-card draw:
    4 * C(13,2)*C(13,1)*C(13,1)*C(13,1)/C(52,5) ~= 26.4%

    The probability of drawing 2 of 1 suit, 2 of another suit, and 1 of a third suit:
    12 * C(13,2)*C(13,2)*C(13,1)*C(13,0)/C(52,5) ~= 36.5%

    The 4 is because that's the number of ways you can permute the set (2,1,1,1)
    The 12 is because that's the number of ways you can permute the set (2,2,1,0)

    ***
    Here's a summary of all possible suit distributions in a 5 card draw:

    0.20% - Chance of drawing 5 cards of the same suit
    4.29% - Chance of drawing 4 of 1 suit and 1 of another suit
    10.30% - Chance of drawing 3 of 1 suit and 2 of another
    22.32% - Chance of drawing 3 of 1 suit, 1 of another suit, and 1 of a third suit
    36.52% - Chance of drawing 2 of 1 suit, 2 of another suit, and 1 of a third suit
    26.37% - Chance of drawing at least 1 of each suit

    ***
    What is the probability that a 5-card draw will contain no suit with 3 or more cards?
    ~63%

    ***
    Your actual question was "How many are there?"
    That result is the same as above, except you DO NOT divide by C(52,5), which is

    949,104 + 685,464 = 1,634,568
  3. #3
    MadMojoMonkey's Avatar
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    Summary of flushes on a 14 card FL draw in pineapple:

    Oh hey. I just noticed that last time I did this, I was counting a draw of 10 or more cards in 1 suit as 1 flush, instead of 2.

    These numbers are revised from wherever my previous result landed:

    13.2% - Chance of drawing 0 flushes
    66.2% - Chance of drawing 1 flush
    20.6% - Chance of drawing 2 flushes

    Note that some of those are straight-flushes.
  4. #4
    Eric's Avatar
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    Quote Originally Posted by MadMojoMonkey View Post
    That result is the same as above, except you DO NOT divide by C(52,5), which is

    949,104 + 685,464 = 1,634,568
    Hmm, that's a lot of hands. It's even more when we look at all the 3, 4 and 5 suited hands. I was asking because we have over 1 million pineapple hands in our db and we thought it might be cool to see which 5 card sets led to the most royalties. Maybe we don't have enough data for this type of thing yet?
  5. #5
    MadMojoMonkey's Avatar
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    Quote Originally Posted by Eric View Post
    Hmm, that's a lot of hands. It's even more when we look at all the 3, 4 and 5 suited hands.
    You just included every possible hand in C(52,5) ~= 2.6 million

    Quote Originally Posted by Eric View Post
    I was asking because we have over 1 million pineapple hands in our db and we thought it might be cool to see which 5 card sets led to the most royalties. Maybe we don't have enough data for this type of thing yet?
    There's never too little data to get started. If you know you will want the results at some point, then

    The algorithms you develop to analyze one sample can be used for another sample.
    The individual results can be combined to show population statistics.

    ***
    If your goal is to determine which 5-card sets led to the greatest royalties, then can you just sort the population by top royalties?

    Then, once sorted, pull out the 5-card set for some # of the top results of that sort?
  6. #6
    MadMojoMonkey's Avatar
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    Quote Originally Posted by MadMojoMonkey View Post
    949,104 + 685,464 = 1,634,568
    This counts a lot of similar hands, and you can probably ignore the 4 and the 12 in those equations, which is

    171,366 + 79092 = 250,458

    This smaller result counts
    2 , 2, 1
    as the same as
    2 , 2, 1

    etc.

    The 1.6 million counts the total number of hands that have less than 3 of any suit.
    The 250,458 number counts the same, but assumes all suits are of equal rank.

    This is the same as treating all AKo as the same pre-flop.
  7. #7
    Eric's Avatar
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    Does this count the hands below as two different hands?
    A K Q A K
    A K Q A K

    If so then what is the total combination of hands ignoring suit from the pool of hands where we have no more than 2 of any suit? In other words, these are both in our pool but they should only be counted as one hand.
  8. #8
    MadMojoMonkey's Avatar
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    Quote Originally Posted by Eric View Post
    Does this count the hands below as two different hands?
    A K Q A K
    A K Q A K

    If so then what is the total combination of hands ignoring suit from the pool of hands where we have no more than 2 of any suit? In other words, these are both in our pool but they should only be counted as one hand.
    The 250,458 result counts those as the same.
    The 1.6 million result counts them as different.

    I bolded a part that's a bit confusing. I think what you're saying is not to ignore suit, but to acknowledge that all 4 suits are equally-valued. If that's correct, then again, the 250,458 result is correct.

    The value of the cards is not relevant in any case, only the suits are considered.
    Your first example is considered as:
    Your second example is considered as:
  9. #9
    MadMojoMonkey's Avatar
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    OK, I see what you were saying. Here's the confusion. When I said this:

    This smaller result counts
    2 , 2, 1
    as the same as
    2 , 2, 1

    The numbers represent how many clubs, how many spades, etc.

    So I was saying:
    =

    In the result that is 250,458.
  10. #10
    Eric's Avatar
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    Ok, so when all aakkq hands with no more than 2 of a suit are counted as 1 then our total drops down to 250kish, right? In other words, we also count A K Q A K as the same hand, right?
  11. #11
    MadMojoMonkey's Avatar
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    Quote Originally Posted by Eric View Post
    Ok, so when all aakkq hands with no more than 2 of a suit are counted as 1 then our total drops down to 250kish, right? In other words, we also count A K Q A K as the same hand, right?
    The card value is not a factor.

    That suit distribution is counted.
  12. #12
    MadMojoMonkey's Avatar
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    Wow, I was tired last night. I think I got hung up on the card-values, but I wasn't seeing your point.
    In fact, your use of card values is spot-on, and I should have realized.

    If this helps:

    A starting hands chart for Hold Em shows the C(52,2) = 1,326 possible unique starting hands on a grid that is 13 x 13 = 169 squares.

    The 1.6 million result is counting every possible combo, analogous to the 1,326 unique starting hands in HE. This counts every hand as unique, and does not consider any hand to have "equal value" to another.

    The 250k result is analogous to counting the starting hands, acknowledging that all suits are equal. This counts hands with the same card-values as equally valued, regardless of suit.

    ***
    I was thinking, just take 2 from a set of 13; it doesn't matter what the "value" of the 2 is. However, if that were true, then there would be only 1 way to draw 2 from 13, since the set of 13 is identical, there is no way to distinguish them, so any one is equal to any other, so when I pick one, I can't tell which.

    So picking 13 from 2 in that case has only 1 result. In my calculation, I acknowledge that the 13 are different and if I want to pick 2 from 13 there are 78 ways I can do that.

    ***
    Sorry for the rustiness.

    The card values are definitely counted, and any permutations of suits which has the same distribution of suits and card values is equal.

    The 250k result is what you want.

    If the 250k result has accounted for a hand, say A A K K Q
    Then it has counted all hands which have the card-values AAKKQ, which also have the suit distribution 2-2-1-0 (In this hand, it's 2 clubs, 2 diamonds, 1 heart and 0 spades.)

    So all the hands that have AAKKQ and 2-2-1-0 count as 1 result.
    All the hands that have AAKKQ and 2-1-1-1 count as 1 separate result.

    When I write 2-2-1-0, I mean that 2 cards are of a first suit, 2 cards are of a second suit, 1 card is of a third suit, and no cards are from a fourth suit. The order of the suits does not make a difference. As long as the distribution matches, it's considered the same.

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