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Thread: A puzzling problem

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    A puzzling problem




    Hello all,

    I am posting here looking for help on solving a probability problem I have been struggling with for 2 days. It has been several years since my last statistics course during my undergraduate and I feel like I am overlooking something fairly simple. If you have any ideas on what I did wrong please let me know.

    So the problem is this.

    I play a trading card game. The deck consists of 99 cards. There is equal probability of drawing any one card in a perfect world. In this deck there are 7 cards that do the exact same thing. The game starts with a 7 card hand and you draw one card every turn.

    I am looking to create an equation to model the probability of drawing a certain number of the 7 similar cards in a certain number of turns. In particular I am looking just to know what the chance of drawing the first of these seven at any turn of the game. I.e., what is that chances that I draw my first of these particular 7 cards on turn 5, 15, 80, etc.

    I used some calculations from old text books and other forums but to no avail. Here is what I have.

    K=number of successes wanted= 1
    D=deck size=99
    C=# of desired card=7
    S=# of cards you have drawn so far (Hand) = 7 on the opening hand, 8 on turn 1, 9 on turn 2 etc.

    W=permutations of k desired cards in s card hand = (S choose K)*(C!/(C-K)!)
    WO=ways to draw all undesired cards in same hand = (D-C)!/(D-C-(S-K))!
    T=total ways of drawing s cards in a d sized deck = D!/(D-S)!
    P=Probability to draw k successes by a given turn = W*WO/T

    The first issue I run into when I enter this into excel to start plotting out probability is that after so many turns, the chance of having drawn a single success started going down. Which makes sense because after 50 something turns you would actually expect to have drawn 2 or more of these 7 cards and this equation was for exactly K=1. To resolve this I redid the entire calculation for K=2 through 7 and added it up to get me at any turn between opening hand and turn 92 (when you draw the last card) the probability of having drawn at LEAST one of the desired 7 cards.

    The issue I have now is although those numbers were close to what I was hoping to see. They aren't correct I don't think. Because in reality you would not receive 100% chance of having drawn one of these 7 cards until there is only 6 cards left in the deck. My spreadsheet gives me a 100% chance of having drawn at least one by turn 60... not turn 85.

    The entire intent of this is to manipulate the concentration in the deck to give me a balance between when I see diminishing returns adding more of a similar card and when it would be more beneficial to add cards of a different type instead. I have used the "rule of 7" for years and that is why I am testing with 7 similar cards. But if 8 showed significant increases that's important, etc.

    Any help would be appreciated. Thanks in advance.

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    Re: A puzzling problem

    This gives the probability of getting the first success when you receive, say, the 10th card (including those you started with).
    The probability of getting a collection with 1 success out of 10 when there are 7 possibilities out of 99 is hyper-geometric. In Excel you would use =HYPGEOMDIST(1,10,7,99). This includes the single success in any of the 10 positions so the probability that it will be last is 1/10 of that.
    In general, the probability that the first success will occur as you receive your n th card (including your original 7) is =HYPGEOMDIST(1,n,7,99)/n
    Another equivalent way is to calculate the probability of 9 failures, followed by 1 success = 92/99x91/98x ... 84/91x7/90
    I think, I hope, that this is what you were looking for! kat

  3. The Following User Says Thank You to katxt For This Useful Post:

    Jheath9131 (04-18-2017)

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    Re: A puzzling problem

    That seems a lot more simple than what I was trying to do. I will try this when I get a chance and see if it works. I'm really just hoping to determine the amount adding a single card of a similar type aids the chances of succeeding. Thank you again.

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    Re: A puzzling problem


    It worked wonderfully. The results surprise me at early numbers. You approach 50% chance of having drawn one of the successes by turn 3 (10 cards drawn)

    At exactly 93 cards drawn is the first time I hit 100% success (when results were measured to 20 decimal places). When only to 2 decimals it's 100% on turn 66 which is basically what I was getting before. I was close but think I wasn't paying attention to rounding.

    Yours is much easier and accurate though.

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