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Math Help - Mulligan Probabilities

  1. #1
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    Mulligan Probabilities

    Hi all,

    I'm a magic player and I've been trying to find out how to calculate the probability of drawing distinct cards a_1, a_2,...,a_i out of b copies of each a_j in my deck (suppose b=4 for all a_j) given d cards in my deck (say d=60) , 7 cards in my opening hand, and the opportunity to mulligan my opening hand of 7 for a new hand at a cost of one card per mulligan.

    I know that for my opening hand and a_1=1,b=4 this is given by something like P(a_1=1)=
    100%- (56/60)*(55/59)*(54/58)*(53/57)*(52/56)*(51/55)*(50/54) = 39.9%, and that if I take a mulligam, P(a=1)=100%- (56/60)*(55/59)*(54/58)*(53/57)*(52/56)*(51/55)= 35.14%

    a)Does the above mean that I have a 39.9+35.14% chance of getting my specific a_1=1 card after two mulligans? In general, can I keep adding these probabilities for successive mulligans?

    b)The above only works for one specific card, I want to know the probability of distinct cards a and b (in general, referred to as a_1,...,a_i above)! I know I cannot simply multiply the two probabilities (i.e., P(A and B)=/=P(A)P(B) because they are not independent, I need P(A and B)=P(A)P(B|A), I need help calculating P(B|A). Can I use Bayes' theorem of transposed conditionals?

    c) Bonus points to whoever can do all of the above in addition to adjusting for the probability of having such distinct cards after mulligans AND a certain number of draw steps.



    Any help is much appreciated!
    Last edited by charmedquark; November 18th 2012 at 09:55 AM.
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  2. #2
    MHF Contributor
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    Re: Mulligan Probabilities

    Hey charmedquark.

    What did you get for the probability of P(a1 and a2 and a3 .... and an) in terms of the symbolic and numerical output?
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