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Poker Dice Probability Problem

  • Thread starter cepheid
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  • #1
cepheid
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The assigned problem:

Poker dice is played by simultaneously rolling 5 dice. Compute the probabilities of the following outcomes:

b) two pairs (aabbc, with a, b, c distinct; answer: 0.2315)

My solution attempt:

Let E be the event specified. The number of ways it can occur is given by:

[tex] {6 \choose 1} {5 \choose 1} {4 \choose 1} \cdot \frac{5!}{2!2!1!} [/tex]

= 6*5*4*(120/4) = 120*30 = 3600

[tex] P(E) = \frac{3600}{6^5} = 0.4630 [/tex]

Explanation of my solution: I considered aabbc to be a multiset with 3 distinct elements: a, b, and c. a and b just happen to have multiplicity 2. So, [itex] {6 \choose 1} [/itex] is the number of ways of choosing a from the original six possible numbers. [itex] {5 \choose 1} [/itex] is the number of ways of choosing b from the remaining five possible numbers. [itex] {4 \choose 1} [/itex] is the number of ways of choosing c from the remaining four possible numbers. Now that I have chosen the 3 distinct elements, how many ways are there to permute them? For a set of five distinct numbers, there would be 5! ways. But, two of the numbers are repeated once each, and permutation with repeated elements is given by the formula I used. Another way of expressing it (and one that I understand better) is that this is the multinomial coefficient: it expresses the number of ways of choosing a position out of five for each of our three elements (two of which have multiplicity 2) when you don't care in what order they occupy those positions.

Problems with my solution:

i) The obvious one is that my answer is twice the one given by the professor. In fact, 0.4630 is the answer he gave for part c), the probability of getting only one pair.

ii) The fault, I believe, lies with me. For when I attempted to use this same method for part c) (probability of getting only one pair), I ended up with more possible desired outcomes than there were outcomes in the sample space!

Where did I go wrong?
 

Answers and Replies

  • #2
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hint:
you double count something, let's say
case 1) a=1,b=2,c=3...
case 2) a=2,b=1,c=3...
what is the different(if any) between these two case??
 
  • #3
cepheid
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Thank you for the hint. However, it only tells me that I have made a mistake, not what specific error in my reasoning has led to that mistake. The problem is that my reasoning makes perfect sense to me, so it is impossible for me to deviate from it without outside help. In other words, I'm stuck. Could somebody please tell me what the specific error in my explanation paragraph was, and how it can be remedied? Thanks.
 
  • #4
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hint:
if a and b are interchangeable... we are doing combination of a and b, not permutation...
once you have the answer of permutation... how do you correct it into combination?
 
  • #5
cepheid
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I got it...a and b are interchangeable because both are pairs of numbers, so although outcomes with c in a different place are different, outcomes with a and b interchanged are indistinguishable. To answer your question, divide by 2! in this case because:

[tex] {6 \choose 1}{5 \choose 1} = \ \ _{6}P_{2} [/tex]

and we actually want

[tex] _{6}C_{2} [/tex]

Thanks for the help.
 

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