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Thread: Probability of 20 people drawing the same card 9 or more times in a deck of cards.

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    Probability of 20 people drawing the same card 9 or more times in a deck of cards.




    Hello! This is my first post so apologies if I'm out of line :-S! I've looked EVERYWHERE and haven't found a problem that I an use as a model... I've come close and I'll explain in a bit why it's different.

    I'm working on a problem for my job... let me give a simple toy problem that I believe is equivalent.

    We have a deck of 52 cards. let's draw 20 (replacing the last card every time). What is the probability that we'd draw the same card 9 or more times?

    I saw a similar problem involving dice here: http://math.stackexchange.com/questi...ility-question However, they're calculating the following: roll a 4 sided die 5 times. What is the probability that we'd roll the same number 3 or more times?

    This is different, in my opinion, because there can only be 1 set of 3 in the 5 tries. There can be 2 sets of 9 in our draw of 20! This opens a new can of worms and I have NO clue where to start.

    The real scenario with my job? I wrote a PHP script that will be used by 50 people at a time (20 draws in my reasoning?). They refresh the page every 90 seconds (52 cards in my reasoning?). The site will break in the event that 24 of these refreshes occur at the same time (draw the same card 9 or more times). LOL Don't judge my code, haha! I mean the API connection only allows 25 calls per second but I want to keep this problem simple. I figure that these 50 people log in at equally probable times throughout the 90 second interval.

    I really want to learn how this works so I'd appreciate as much as you're willing to teach!!! Thanks so much in advance. I can't express my gratitude enough for any help you can provide. I did take a intro probability/stats class in my Electrical Engineering degree at Georgia Tech but for some reason I have so much trouble with this subject .
    Last edited by imackenzie; 07-25-2015 at 12:50 PM.

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    Re: Probability of 20 people drawing the same card 9 or more times in a deck of cards

    For your toy model: Assume you mean the favorable outcome is "at least 1 set of 9 or more identical cards", then you can use inclusion-exclusion principle to help as you are dealing with a union of events.

    Let A_i be the event that the i-th card appearing 9 or more times in the 20 independent draws, i = 1, 2, \ldots, 52

    Note that A_i \cap A_j \cap A_k = \varnothing when i, j, k are distinct - i.e. you cannot have 3 (or more) set of 9's happen together. So when you applying the inclusion-exclusion principle, you can terminate at the second term.

    The required probability is

    \Pr\left(\bigcup_{i=1}^{52} A_i\right)

    = \sum_{i=1}^{52} \Pr(A_i) - \sum_{i=1}^{51}\sum_{j=i+1}^{52}\Pr(A_i \cap A_j) (by inclusion-exclusion principle)

    = 52\Pr(A_1) - \binom {52} {2} \Pr(A_1 \cap A_2) (since they are equally-likely)

    The difference with the dice example you provided, as you mentioned before, is that you need to deal with the second summation here as it is possible to have 2 sets. The first term is easy, you just need to use the Binomial CDF.


    Let (X_0, X_1, X_2) be the number of card 3-52, card 1 and card 2 drawn respectively. So they jointly follows a multinomial distribution. And the above probability become

    52\Pr\{X_1 \geq 9\} - \binom {52} {2} \Pr\{X_1 \geq 9, X_2 \geq 9\}

    For the second term, you need to count all the possible outcomes for (X_0, X_1, X_2) satisfying X_1 \geq 9, X_2 \geq 9:

    (2, 9, 9), (1, 10, 9), (0, 10, 10), (0, 11, 9), (1, 9, 10), (0, 9, 11)

    and you sum the corresponding multinomial pmfs to obtain the second term.


    Let see if you understand all these first. Your real application problem seems similar with the size of deck increase to 90 and the number of draws to 50, and require 24 or more appearances.

    Remark: The second term here only provide a minor adjustment if \Pr(A_1) is a very small number, as we know \Pr(A_1 \cap A_2) < \Pr(A_1) will be a much smaller number. So in that case even you only calculate the first term, not only it will give you an upper bound, it will be also a good approximation.

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    Re: Probability of 20 people drawing the same card 9 or more times in a deck of cards


    Wow, thanks so much for such a meaningful response. That does really help me break the problem up better! Wouod have taken me a cery long time to think of it that way but it makes sense. Let me consider this for a while...

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