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Question about cards (challenging problem)

  1. Jun 16, 2010 #1
    Hello!

    I have been doing a research about probability of facing a larger pair when holding let's say a KK. So the possible pairs larger than KK are AA (total 2_C_4=6).

    Now if I play with 2 opponents the probability of getting larger pair is:

    [tex]\frac{6*{48 \choose 2}}{{50 \choose 2}{48 \choose 2} \div 2!}}=[/tex]

    [tex]=0,0097959183673469387755102040816327[/tex]

    But now there is similar approach which is much efficient and good approximation:

    Here is the idea:

    You suppose that you play individually by two of the players.

    [tex]P(A_1 \cup A_2)= P(A_1) + P(A_2) - P(A_1 \cap A_2)[/tex]

    where [tex]P(A_1)[/tex] is the probability of getting larger pair of the first opponent and [tex]P(A_2)[/tex] of the second opponent.

    Now because [tex]P(A_1 \cap A_2)[/tex] is very small you can just compute [tex]P(A_1) + P(A_2)[/tex] and you will get:

    [tex]0.009795918367346938775510204081633[/tex] which is a almost perfect approx.

    But now the problem is that I want to subtract [tex]P(A_1 \cap A_2)[/tex] so that I can get the correct answer.

    Now because [tex]P(A_1)=P(A_2)=\frac{6}{{50 \choose 2}}=0.0048979591836734693877551020408163[/tex]

    [tex]2*P(A_1)=0.009795918367346938775510204081633[/tex]

    Now the error is about [tex]10^{-34}[/tex], so [tex]P(A_1 \cap A_2) \approx 10^{-34}[/tex]

    Now I only need to find [tex]P(A_1 \cap A_2)[/tex], that is "what is the probability that same pair of AA will come two the opponents"?

    For ex. what is the probability that A(spades) A (diamond) will come to both opponents?

    Here is my reasoning:

    [tex]P(A_1 \cap A_2) \approx (\frac{4}{50} * \frac{3}{49})^2 \approx 0,00002399[/tex]

    or the correct one [tex]\frac{(120}{(50*49)^2}=0,00001999[/tex]

    which is not even close to [tex]10^{-34}[/tex].

    Where is my error?

    Thanks in advance.
     
    Last edited: Jun 16, 2010
  2. jcsd
  3. Jun 17, 2010 #2
    bump!

    Anybody, what I am doing wrong?
     
  4. Jun 25, 2010 #3
    Hi, njama

    Your first solution,

    [tex]\frac{6*{48 \choose 2}}{{50 \choose 2}{48 \choose 2} \div 2!}}=[/tex]

    is exactly twice the probability P(A_1):

    [tex]P(A_1)=P(A_2)=\frac{6}{{50 \choose 2}}[/tex]

    So I don't think your first solution gives the probability you are looking for, that at least one opponent has a higher pair. The very small difference of 10^-34 between your first answer and P(A_1) + P(A_2) is probably due to rounding error. That could happen when you multiply by and then divide by 48 choose 2. Your second approach, using P(A_1 or A_2) = P(A_1) + P(A_2) - P(A_1 and A_2) is the right way to go about it. I don't understand how you are calculating P(A_1 and A_2), though.

    Is this question about hold'em poker, where each player gets two cards and there is a round of betting before the "flop" (first three community cards) are dealt?
     
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