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Homework Help: Probability question (throwing dice)

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data

    An unbiased six-sided dice is thrown 10 times. What is the probability that exactly 4 of any one number alone will occur?

    2. Relevant equations

    Binomial equation
    Combinatorial equations

    3. The attempt at a solution

    You would be interested in the probability of choosing any number 4 times out of the 10 throws:

    6*(10 C 4) (1/6)^4 * (5/6)^6

    You would also be interested in the chance of throwing that same number that has already been thrown 4 times a 5th, 6th, 7th, 8th, 9th and 10th time (since that side of the dice isn't removed, it could still come up in the remaining 6 throws).

    ...(6 C 1) (1/6)^1 * (5/6)^5
    + (6 C 2) (1/6)^2 * (5/6)^4
    + (6 C 3) (1/6)^3 * (5/6)^3
    + (6 C 4) (1/6)^4 * (5/6)^2
    + (6 C 5) (1/6)^5 * (5/6)^1
    + (6 C 6) (1/6)^6

    And also the chance that one of the remaining 5 numbers could be thrown 4 times...

    5 * (6 C 4)*(1/6)^4 * (5/6)^2

    However, these last two sets of equations aren't entirely independent of one another...you could, for example, start by throwing four 3's...and then in your remaining 6 tosses you get another 3 and then four 5's and then a 6.

    I can't quite figure out how to separate out these values into truly independent events in order to combine them and get to the answer.

    Am I on the right path here or is there a simpler way to think about this that I have overlooked? I'm quite stuck here so any help would be greatly appreciated.
  2. jcsd
  3. Nov 29, 2008 #2
    The notation (10 C 4) is weird. I would write C(10, 4). What you wrote is not ''the probability of choosing any number 4 times out of the 10 throws'' as there is no probability associated with choosing 4 out of the 10 throws.

    Why would I be interested in that? We're only interested if a number shows up exactly four times.

    I think I understand your problem, which I will state in more precise terms: Let S be the number ways of throwing a regular die 10 times. Let S(x) be the number of ways of throwing a regular die 10 times such that x appears exactly 4 times. One may think that the probability of getting 4 of exactly one number in 10 throws is [S(1) + S(2) + ... + S(10)] / S. The problem here is double counting, e.g. throwing four 1's followed by four 2's and then a 5 is counted in both S(1) and S(2). But this is easily fixed: let S(x, y) be the number of ways of throwing a regular die 10 times such that x and y appear exactly 4 times and subtract this from S(1) + S(2) + ... + S(10) accordingly.
  4. Nov 30, 2008 #3
    Thanks for the reply.

    When I wrote (10 C 4) I was intending that to be read as "from 10 choose 4", or the number of ways of arranging 4 things from 10 (i.e. 4 successful dice throws from 10 total throws). What I wrote IS a probability, it is the number of sides on the dice that could be rolled 4 times multiplied by the number of ways of rolling a certain number 4 times from 10 throws, times the probability of rolling a certain number on a dice 4 times, times the probability of rolling a different number on a dice 6 times.

    Sorry if the notation seems strange, it's the first time I ever wrote "from this number choose that number" on a message board, so I wasn't sure of the proper notation.

    And you're right about the second point, my reasoning was incorrect.

    Your third bit of advice did help me though. I ended up with the probability:

    C(6,1)*C(10,4)*(1/6)^4*(5/6)^6 - C(6,2)*C(10,8)*(1/6)^8*(5/6)^2

    0.3253 (32.53%)

    That's the result I was looking for.

    Thanks again.
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