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Calculating the conditional probability of an event

  1. Feb 2, 2016 #1

    Zondrina

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    Hi,

    I found this screenshot on a website and I thought it was crazy. I want to calculate the conditional probability of this event occurring because it seems so impossible.

    Assume NLTH is being played. I want to calculate the conditional probability of this hand being dealt. Here is a screenshot of the hand and my work:

    Screen Shot 2016-02-02 at 11.26.25 PM.png
    Please note I'm computing this because I'm curious and I like to practice.

    I made a few assumptions along the way:

    First, I'm assuming the players who didn't turn over a hand are getting dealt cards that are not ranked 9, 10, J, Q, or A. This is a reasonable assumption because most of those cards are in the hands of the players who turned them over, and are on the board.

    Also, when I calculate the probability of the cards on the board, I assume there are 4 cards of rank 8 left in the deck. I think this is reasonable because there was originally a 32/50 chance that the next card was not a 9, 10, J, Q, or A after the first two cards are dealt. Only 4/32 of those cards were of rank 8, and only 8/32 of those cards are dealt to players pre-flop. So assuming you do get dealt one of those 32 cards, the chance of an 8 would be unlikely. I could go as far as to compute the conditional probability of an 8 being dealt along the way, but assuming all 8s are still in the deck seems reasonable on average.

    I also assume no cards of rank 8 are burnt.

    Finally, the person next to the right of the dealer button gets dealt to first.

    Assuming these things, I find myself looking at a number on the order of ##10^{-16}## (after multiplying by ##100 \%##). So according to that, this should happen roughly 1/10 quadrillion times? I don't think people have collectively played enough cards in a lifetime to see this one.
     
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  3. Feb 3, 2016 #2

    Simon Bridge

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    Any particular hand is equally likely.
     
  4. Feb 3, 2016 #3

    Zondrina

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    Considering randomness, truesay. Although 52 choose 5 is a lot of combinations, and I think some hands would be more likely than others.

    So I assume the calculation in the OP is reasonable.
     
  5. Feb 3, 2016 #4

    jbriggs444

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    This computation is for the conditional probability of what, exactly? Conditioned on what, exactly?

    The conditional probability that you have seen a hand unusual enough to be publishable on PF given that you have published that hand on PF is roughly 1.
     
  6. Feb 3, 2016 #5

    PeroK

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    Which hand would you say is the most likely and which hand the least likely?
     
  7. Feb 3, 2016 #6
    Harsh guys, I think the OP could do with a little help - although given the HW Helper status I'm going to be blunt too:
    1. If you have a question about probability it is a good idea to post it in the probability forum.
    2. You (and I) know that NLTH stands for No Limit Texas Hold-em but not everyone that can help you does: you should put the effort in to type that once rather than expect many people to put the effort in to look it up.
    3. The term "conditional probability" does not make sense here, this is just plain probability.
    4. You have fallen into the common trap of assuming that an event has to look almost exactly the same as the one you are looking at to be as "special". A more reasonable analysis is below.
    We want to have dealt 5 different hands of pairs of nines or better. So one hand must be dealt a 9-A (probability 24/52) and a matching card (3/51). Another hand must then be dealt a different card 9-A (20/50) and a match (3/49). This follows for a total of 5 hands to give ## \frac{24}{52} \times \frac{3}{51} \times \frac{20}{50} \times \frac{3}{49} \times \frac{16}{48} \times \frac{3}{47} \times \frac{12}{46} \times \frac{3}{45} \times \frac{8}{44} \times \frac{3}{43} \approx 3.1 \times 10^{-9} ##

    But there are a total of 9 hands so this can happen ## _9{C}_5 = 126 ## ways, so we have a probability of approximately ## 3.1 \times 10^{-9} \times 126 \approx 3.9 \times 10^{-7} ##

    We now need 4 cards on the table to match 4 of the pairs, so this is ## \frac{10}{42} \times \frac{8}{41} \times \frac{6}{40} \times \frac{4}{39} \approx 7.1 \times 10^{-4} ## - this can happen ## _5{C}_4 = 5 ## ways so the overall probability is approximately ## 3.9 \times 10^{-7} \times 7.1 \times 10^{-4} \times 5 \approx 1.4 \times 10^{-9} ##

    Probably 10-100 million hands of poker played every day so you would expect something similar with a frequency of somewhere between every week and every couple of months.

    Can't guarantee this is error free, I was out earlier and may have had some wine.
     
  8. Feb 4, 2016 #7

    Zondrina

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    For example: Treat each individual card rank as an event, so each event has 4 simple events. When the first card is dealt, we can calculate ##P(9) = \frac{4}{52}##. Now given that the first card is a nine, what is ##P(Q|9)##? Etc.

    Since all the cards are not visible, it's hard to calculate the probability of the overall event precisely. For example, I had to assume on the third card dealt that it was possible to be dealt an 8, even though I assume no 8s are dealt or burnt. I did this so I didn't influence the probability of all the other cards that could have been dealt as well.

    It's all highly approximate.

    Usually they have nothing :oldbiggrin:.
     
  9. Feb 4, 2016 #8

    jbriggs444

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    What is Q?
    The probability that the first card is a Queen knowing that the first card is a Nine is zero.
     
  10. Feb 4, 2016 #9
    No it isn't, the "invisible" cards are irrelevant - if a card is not in play then it doesn't matter whether it is still in the pack or it has been dealt and folded.
     
    Last edited: Feb 4, 2016
  11. Feb 4, 2016 #10

    PeroK

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    What would make sense is to calculate the probability that all 5 hands are different pairs. Regardless of what's in the middle.
     
  12. Feb 4, 2016 #11
    Well, different pairs of nine or better. That's what I did as the first step.
     
  13. Feb 4, 2016 #12

    PeroK

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    Yes, apologies, I had just looked at the thread from the reply to my previous post.

    I like your analysis, although it seems to me that any set of five pairs would be equally remarkable.
     
  14. Feb 4, 2016 #13
    That's where the game specifics become relevant - lower ranking pairs still in the game would be even more unlikely: with nine players at the table you are probably going to fold anything worse than a pair of nines.

    I did miss something though - this deal is won by the player holding the pair of tens as he makes a straight with the cards on the board. There's actually quite a few ways this can happen though, so it doesn't make the deal vastly more unlikely.
     
  15. Feb 5, 2016 #14

    Zondrina

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    I feel as if I may have been unclear at some points in this thread. Particularly the title. I'm trying to compute the probability of a large intersection of events using the rule:

    $$P(A \cap B) = P(A|B){P(B)}$$

    If we did know all of the cards, we could compute the probability precisely with repeated applications (or just use a probability tree).

    I thought it was interesting we could compute the probability of the entire hand using conditional probability. I mistakingly used the term conditional probability throughout the thread though when I intended to mean intersection; sorry about that.

    So to make it clear: My concern is the probability of the whole hand, which I can only approximate sadly. The technique is what I'm interested in.

    @jbriggs444 : ##P(9) = \frac{4}{52}, P(Q \cap 9) = P(Q|9)P(9) = \frac{4}{51} \frac{4}{52}##

    $$P( (A' \cap Q' \cap J' \cap 10' \cap 9') \cap (Q \cap 9) ) = P((A' \cap Q' \cap J' \cap 10' \cap 9') | (Q \cap 9)) P(Q \cap 9) = \frac{32}{50} \frac{4}{51} \frac{4}{52}$$
     
    Last edited: Feb 5, 2016
  16. Feb 5, 2016 #15

    jbriggs444

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    Thank you for the clarification. However, the notation above could use some work. You are speaking about a probability space in which two cards are drawn in sequence. By P(9), you mean "the probability measure of the set of outcomes where the first card drawn is a nine". By P(Q) you mean "the probability measure of the set of outcomes where the second card drawn is a queen". By ##P(Q \cap 9)## you mean "the probability measure of the set of outcomes where the first card is a nine intersect the set of outcomes where the second card is a queen".

    But that notation leaves you with no way to express "the probability the the first card drawn is a queen" or "the probability that the second card drawn is a nine". In addition, it is not clear whether you really want to distinguish between a draw of a nine followed by a draw of a queen and a draw of a queen followed by a draw of a nine.
     
  17. Feb 6, 2016 #16
    Don't get hung up on terminology, it doesn't change the nature of the problem or its solution. Having said that, most mathematicians would call this a problem in combinatorics.

    But we don't need to know all the cards, we only care about the cards that are in play. It doesn't matter if the AH is still in the pack or was dealt to DarlusDriver. So we can use a probability tree, and in fact that is exactly what you do when you calculate ## P(A \cap B) = P(A|B){P(B)} ##.

    No, you don't want the probability of the whole hand, you want the probability of a hand that is at least as remarkable as this one for a similar reason - it doesn't make any difference whether PotHanger was dealt pocket aces or DarlusDriver was so we have to count both.

    You are using the right technique, you have just made a few errors. In my post #6 I have used the same technique (hopefully without error!) and it does give the probability of a hand which is equally remarkable as the one shown.

    Let's just make sure we agree on why this hand is remarkable: I have assumed it is because at the end of the hand there are 5 players holding pairs of nines or better pairs and exactly four of the pairs are matched by community cards. I have ignored the sequence of the community cards (because I have assumed that different sequences are equally remarkable), and I have also ignored the fact that the unmatched pair makes a straight (mainly because I didn't notice; if you think that makes the hand more remarkable then add it into the calculation - it probably adds a factor of about 10). I have also ignored any information provided by the fact that 4 hands have folded and by the betting.

    So what are the differences between your calculation and mine?
    1. By taking ## \frac{4}{52} \frac{4}{51} ## you have only counted hands where the player to the left of the dealer holds a pair of nines and the next player holds a pair of tens. I have corrected this by (i) counting hands where the first player left in the deal holds any pair better than eights (ii) multiplying by the number of similar combinations so that it doesn't matter which of the 9 hands dealt holds which pair.
    2. By taking ## \frac{32}{50} ## you have assumed that a hand where the 9S is dealt to the third player is more remarkable than one where it is still in the pack. You need to ignore the unseen cards.
    3. You have made the same mistakes with the community cards. By taking ## \frac{2}{33} \frac{2}{32} \frac{2}{31} ## you have only counted hands where the flop is 9, Q, J in that order - it doesn't make any difference. And again by taking 33 as the denominator you have distinguished between hands where the 9S is dealt to the third player and where it is still in the pack.
    So you have calculated the probability of exactly the hand that is partially shown (including cards we cannot see) being dealt and come to an miniscule number. As others have pointed out, all hands that are actually dealt have equal, miniscule, probability. But when we say a hand is unusual or remarkable, we mean that the hand has certain features which are worthy of note. When calculating the probability of such a hand we must ignore features that we cannot see (such as which cards were folded) or are not significant (such as whether the dealer holds a pair of queens and the blind a pair of nines or vice versa).
     
  18. Feb 6, 2016 #17

    Zondrina

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    ##Q## is the event a Q is dealt, ##3## is the event a 3 is dealt, etc. I assume the leftmost event in an intersection is the latest event to occur because I want to keep things consistent. So when computing ##P(Q \cap 9)##, I assume the queen is the last card dealt during the event, and I read it as "the probability a 9 is dealt and a queen is dealt".

    Assume the cards were just shuffled, what is ##P(Q)##? (The probability a queen is dealt)

    Assume the first card was actually a queen for simplicity, what is ##P(9|Q)##? (The probability a 9 is dealt, given a Q is dealt)

    This can be ambiguous because of commutativity of the intersection. It's good to adopt a convention and stick with it as a result, but it's also good to be adaptive to different problems.
     
  19. Feb 6, 2016 #18

    jbriggs444

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    A notation without a defined meaning is a waste of time.
     
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