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Thread: Probability of getting at least a specific value (more or equal to). Advanced problem

  1. #1
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    Probability of getting at least a specific value (more or equal to). Advanced problem

    This is a big one so read carefully.
    You have a room with 24 players. For each player an enemy will spawn. For each enemy that is defeated, 8 randomly picked players will receive an item. So 8 items per enemy is given out.

    What is the probability for a specific player of receiving at least 8 items? (8 or more)

    I can calculate the probability for no items but not for "8 or more".
    No items = (1-1/24*8)^24.
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  2. #2
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    Re: Probability of getting at least a specific value (more or equal to). Advanced pro

    I presume that, since there are 24 players and "for each an enemy will spawn", that there are 24 enemies. Now, what is the probability that an enemy will be defeated or can we assume that all 24 enemies are defeated? If the latter is true, then we can calculate the probability that a given player is "chosen" at least 8 times in the 24 possible times. Since there are 24 players, and, for each defeated enemy, there are 8 players chosen, the probability of a specific player being chosen each time is \frac{8}{24}= \frac{1}{3} and the probability of not being chosen is 1- \frac{1}{3}= \frac{2}{3}. The probability of being chosen n times out of 24 is \frac{24!}{n!(24- n)!}(1/3)^n(2/3)^{24- n}. The probability of "no items" (n= 0) using that is \frac{24}{0!24!}(1/3)^0(2/3)^{24}= (2/3)^{24}= 0.000059. Is that what you got?

    The simplest way that I see for finding the probability of being chosen "8 or more times" out of the 24 is to calculate that for n= 0, 1, 2, 3, 4, 5, 6, and 7, add them to get the probability of being chosen "7 or less times" and subtract from 1.
    Last edited by HallsofIvy; Nov 28th 2016 at 08:45 AM.
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    Re: Probability of getting at least a specific value (more or equal to). Advanced pro

    Thank you! Yes this is exactly what I was looking for! God bless your talents. Also, yes indeed, I got 0.000059.
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    Re: Probability of getting at least a specific value (more or equal to). Advanced pro

    you've omitted the probability of defeating an enemy and the results of not defeating an enemy. Is the pool of players reduced because they've died?

    Can we assume all enemies are killed and the player pool remains the same? If this is the case then HallsOfIvy has answered the question.

    If we assume that only some fraction of the enemies are killed, but they run away at some point leaving all players alive then we solve as follows.

    $P[\text{k enemies defeated}] = {_{24}C}_k ~p^k (1-p)^{24-k}$

    Now given $k$ enemies were defeated we can find the distribution of how many items a player receives.

    For each enemy 8 items are randomly distributed. So for $k$ enemies defeated $8k$ will be randomly distributed.

    For each item the probability a given player receives it is $p_r=\dfrac 1{24}$

    The probability a player receives $n$ items is then

    $P[\text{player receives n items | k enemies defeated}] = {_{8k}C}_n~p_r^n (1-p_r)^{8k-n},~n \in \mathbb{N},~0\leq n \leq 8k$

    and thus the overall probability of a player receiving $n$ items is

    $P[\text{player receives n items | k enemies defeated}]\cdot P[\text{k enemies defeated}] = {_{8k}C}_n~p_r^n (1-p_r)^{8k-n} \cdot {_{24}C}_k ~p^k (1-p)^{24-k},~0\leq k \leq 24,~0\leq n\leq 192$

    this can be fully evaluated once $p$ is specified.

    (looks like all this was unnecessary )
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