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Thread: Card draw and joint probability

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    Card draw and joint probability




    I wonder if I'm overthinking this and want confirmation...

    I'm in thought about how to apply (1) to find the probability of 2 not independent events A & B=P(A) * P(B/A) & (2) the probability of 2 independent events is P(A) * P(B).

    I built an Excel spreadsheet using the, "HYPGEOMDIST" function, which is extremely useful when attempting to calculate the probability of drawing specific cards out of a deck (population).

    For example, if I draw 5 cards, out of a standard deck of 52 cards, the probability of drawing at least 1 Ace is 34.12%. (By using HYPGEOMDIST(0,5,4,52) you will get the chance for not drawing the card. Therefore, if you want to check for the chances of drawing an Ace, you would subtract the result from 1)

    Or, if I draw 5 cards, out of a standard deck of 52 cards, the probability of drawing at least one 3 OR 9 is 58.21%.

    Now, if I wanted to calculate the probability of drawing BOTH an "Ace" AND a "3 OR 9" in the first 5 cards, I should just multiply them together, pursuant to the rule above, yes? (and calculate ~19.86%)

    HOWEVER, I'm getting stuck on the idea of those being, "Independent" events or not. Why? Because the conditional and unconditional probabilities wouldn't equal if I knew I already drew an Ace, for example.

    Am I overthinking this?

    Since I'm not revealing the cards 1 at a time, I'm thinking that it is an independent event, but I'm not 100% with it. If it is a situation where they would be, "not independent," how would I adjust the calculation?

    Does that make any sense?

    Or, flip it this way: if I wanted to calculate the probability of drawing an "Ace" OR a "3 OR 9" in the first 5 cards, the calculation should be= P(A or B)=P(A) + P(B) - P(both A and B). Am I correct when I calculate it as follows: 34.12%+58.21%-19.86%=72.47%? This doesn't make any sense at all!

    Any help would be appreciated.

    Thanks
    Last edited by Lunaticfringe; 08-06-2013 at 08:11 PM.

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    Re: Card draw and joint probability

    The events are not independent; you need to extend the idea a little bit to multi-hypergeometric distribution - essentially you extend from 2 groups (ace/non-ace) to 3 groups now, and the idea is similar.

    For the or part it is correct to apply the inclusion-exclusion principle. But as usual you need to get the intersection part correct first.

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    Re: Card draw and joint probability

    Quote Originally Posted by BGM View Post
    The events are not independent; you need to extend the idea a little bit to multi-hypergeometric distribution - essentially you extend from 2 groups (ace/non-ace) to 3 groups now, and the idea is similar.

    For the or part it is correct to apply the inclusion-exclusion principle. But as usual you need to get the intersection part correct first.
    Thanks so much.

    Can you give me a push in the right direction in terms of solving this?

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    Re: Card draw and joint probability

    Over 100 views and no additional help?

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    Re: Card draw and joint probability

    The following should be the answer:

    ([4!/(1!*3!)]*[8!/(1!*7!)]*[50!/(3!*47!)])/[52!/(5!*47!)]

    ([number of ways of drawing one ace]*([number of ways of drawing either a 3 or 8])*([number of ways of drawing the three remaining cards]))/([number of ways of drawing 5 cards out oof 52])

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    Re: Card draw and joint probability

    Quote Originally Posted by Englund View Post
    The following should be the answer:

    ([4!/(1!*3!)]*[8!/(1!*7!)]*[50!/(3!*47!)])/[52!/(5!*47!)]

    ([number of ways of drawing one ace]*([number of ways of drawing either a 3 or 8])*([number of ways of drawing the three remaining cards]))/([number of ways of drawing 5 cards out oof 52])
    I don't think this is correct. The main issue is that you're double counting some events. For instance any case where you have multiple aces (or multiple cards that are either 3 or 8) gets counted more than once using this method.

    Through simulation and a combinatorial argument that I'm too lazy to write up at the moment I get a solution of

    1-(choose(44,5)+choose(48,5)-choose(40,5))/choose(52,5) = 0.1764767
    I don't have emotions and sometimes that makes me very sad.

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    Re: Card draw and joint probability

    Quote Originally Posted by Dason View Post
    I don't think this is correct. The main issue is that you're double counting some events.
    Yes, I did that on purpose. If that is the case, that you don't count the events that render in more than one ace et cetera, then the answer should be

    (4 nCr 1 * 8 nCr 1 * 40 nCr 3) / 52 nCr 5 = 0.1216486594637855

    Or shouldn't it?

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    Re: Card draw and joint probability

    Quote Originally Posted by Englund View Post
    Yes, I did that on purpose. If that is the case, that you don't count the events that render in more than one ace et cetera, then the answer should be

    (4 nCr 1 * 8 nCr 1 * 40 nCr 3) / 52 nCr 5 = 0.1216486594637855

    Or shouldn't it?
    Well you want to count the cases where there are more than one ace and at least on card that is 3 or 8. The issue was that your method counted those cases multiple times instead of just once. I still say the answer is .176 answer I gave above. When I get back to an actual computer I'll post my logic. Although I could be wrong in which case I'll feel a little silly.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Card draw and joint probability

    What probability are you two calculating? The "Ace" AND a "3 OR 9" in the first 5 cards? Are you calculating this as exactly 1 of each or at least 1 of each?

    I'm now attempting to calculate this more from a long-hand perspectve and I'm using the "COMBIN" formula in Excel. At first, I calculated the probability of at least 1 "Ace" and at least 1 "3" (I'm not including the "9" in this initial calculation--just to keep it as simple as possible at first). I get a probability of 10.018%. If I changed it to exactly 1 of each, I get 8.153%.

    Can either of you tell me what you calculate in regards to the probability above?

    Any help would be great!
    Last edited by Lunaticfringe; 08-06-2013 at 08:31 PM.

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    Re: Card draw and joint probability

    Quote Originally Posted by Lunaticfringe View Post
    Can either of you tell me what you calculate in regards to the probability above? Why? Because if we agree that the numbers above are correct, why wouldn't I just double those percentages to calculate the probability of an "Ace" AND a "3 OR 9" in the first 5 cards?

    In other words, I'm now calculating my, "AND" question from the original post as being 20.036%, which doesn't agree to either of your calculations above! ARGH!!

    LOL

    Any help would be great!
    Just doubling the probability clearly doesn't work. I don't even understand your logic there.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Card draw and joint probability

    Quote Originally Posted by Dason View Post
    Just doubling the probability clearly doesn't work. I don't even understand your logic there.
    Sorry. I tried to edit my post before any replies, but failed. Yes, that clearly doesn't work. I'm still thinking about this though. What do you calculate as to (at least) 1 Ace and one 3? Do you get the same percentage as I do?

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    Re: Card draw and joint probability

    GOT IT!!

    DASON---I just got EXACTLY the percentage that you got!

    If you would be so kind as to explain your calculation to me, as I did this much more long hand and I'm very interested in how you calculated it.

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    Re: Card draw and joint probability

    Mine was a little more long-winded than necessary and I probably could condense it but how about you explain how you got your solution and I'll look over it and see if there is anything wrong with the logic.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Card draw and joint probability

    Not only am I calculating exactly what you got, but I also got exactly what Englund got as well. It seems that he calculated, "exactly 1," while you calculated, "at least" 1.

    The way I calculated it was as follows:

    As discussed above, I put together an Excel spreadsheet using the "combin" formula. This returns the number of combinations for a given number of items. Then, I literally typed out every single combination of items, then added them up. Ultimately, I'm looking for take the total number of successes divided by the total number of combinations.

    For instance...

    There are 4 Aces in a standard deck. Out of those 4 Aces, if I'm looking to draw exactly 1 Ace, there are 4 different ways of drawing such Ace. Similarily, if I drew 1 card from a pool of 8 cards (since it's the 3 or 9), the total number of combinations is 8. Extended further, if I was looking to draw 2 Aces, out of the total of 4 Aces, there are 6 different combinations of drawing 2 Aces. etc...

    To help me visualize it (as well as confirm my understanding of the Excel calculation), I literally wrote it out, something like this (confirming, say, the different number of combinations of 2 Aces from 4 cards):

    1: A, A, x, x
    2: A, x, A, x
    3: A, x, x, A
    4: x, A, A, x
    5: x, x, A, A
    6: x, A, x, A

    The next part of the calculation is slightly tricky, as I then was trying to calculate the remaining # of combinations. I took the total # of cards in the deck (52) and subtracted 12 from it (4 + 8). Further, I reduced the number of cards drawn from 5 to 3 (as I'm looking to calculate the probability of exactly 1 of each). Using the COMBIN formula, Excel told me that the total # of remaining combinations was 9,880.

    Lastly, I used the COMBIN formula to tell me what the total # of combinations can be made from drawing 5 cards out of a deck of 52 and it gave me 2,598,960.

    This then solves as (4*8*9,880)/2,598,960=12.165%, which is exactly what Englund calculated.

    But this is only for getting exactly 1 Ace and 1 "3" or "9." In order to get, "at least" 1 of each, I typed out each possible combination and summed the probabilities of each. Example: I did the exactly same calculation as above, except said, "what if I drew TWO Aces and one 3 or 9, then THREE Aces and one 3 or 9. I repeated this until I solved for every single combination. Ultimately, there are a total of 10 combinations, when drawing 5 cards, of the number of Aces that could be drawn (1 through 4), as well as one through four 3 or 9s. Adding them ALL up, I summed to 17.648%.

    Since I'm tying out to what both of you calculated above, I'm like 99% sure I'm correct, but I'd be VERY interested about exploring an option that's more of a direct calculation, as my method seems like more of a long-handed way of calculating this.

    Does that make any sense?!?!
    Last edited by Lunaticfringe; 08-07-2013 at 02:21 PM.

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    Re: Card draw and joint probability


    To sum up: when you are calculating exactly 1 ace and 1 "3 or 9", you invoke the multi-hypergeometric pmf:

    \frac {\displaystyle \binom {4} {1} \binom {8} {1} \binom {40} {3}}
{\displaystyle \binom {52} {5}}

    When you want to calculate at least 1 ace and at least 1 "3 or 9", you can sum the pmf over all the possibilities as you did, or you consider the complementary event as Dason did. Note that the complementary event is no ace or no "3 or 9", therefore by inclusion-exclusion principle, the required probability can be expressed as

    1 - \left(
\frac {\displaystyle \binom {4} {0} \binom {48} {5}} {\displaystyle \binom {52} {5}} 
+ 
\frac {\displaystyle \binom {8} {0} \binom {44} {5}} {\displaystyle \binom {52} {5}} 
- \frac {\displaystyle \binom {4} {0} \binom {8} {0} \binom {40} {5}}
{\displaystyle \binom {52} {5}}\right)

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