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Math Help - Dice session :-)

  1. #1
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    Dice session :-)

    10 dice is rolled, what is the probability of:

    a) getting one & two once and all other numbers exactly twice
    b) getting one & two exactly 3 times
    c) getting four even and six odd numbers
    d) getting one & three two times, if we know that two appeared once and five 3 times
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  2. #2
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    The correct answer to part a) is \frac{10!}{(2^4)(6^{10})}.
    That is the number of ways to rearrange the string 1233445566 times the probability of any string.
    Last edited by Plato; May 6th 2010 at 02:26 PM.
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    I'm not sure I got it, which distribution is used in this solution?
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    Quote Originally Posted by losm1 View Post
    I'm not sure I got it, which distribution is used in this solution?
    I agree with Plato's answer.

    See this Wikipedia article on multinomial coefficient.

    Let E = the event that we get one 1, one 2, and two of every other face value.

    Let M = {1, 2, 3, 3, 4, 4, 5, 5, 6, 6}

    Say the dice are labeled from 1 to 10, and that after rolling them you put them in ascending order from left to right. This is the same as rolling a single die 10 times in a row and writing down the result from left to right.

    There are 6^10 possible combinations, as Plato says, strings.

    The number of ways to get E is the number of permutations of M, which is

    \frac{10!}{1!1!2!2!2!2!} = \frac{10!}{2^4}

    Then we simply write

    P(E)=\frac{\text{number of ways to get E}}{\text{number of ways to roll dice}}=\frac{10!}{2^4\cdot6^{10}}
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  5. #5
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    What about b)? I'm not sure how to apply multinomial distribution formula on this one?

    Formula is \frac{n!}{k_1! k_1! \cdots k_n!}\times p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}

    \frac{10!}{3!3!\cdot 6^{10}} doesn't feel right.
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  6. #6
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    Quote Originally Posted by losm1 View Post
    What about b)? I'm not sure how to apply multinomial distribution formula on this one?
    Formula is \frac{n!}{k_1! k_1! \cdots k_n!}\times p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}
    \frac{10!}{3!3!\cdot 6^{10}} doesn't feel right.
    You really need to learn to model the outcome space.
    For part (b) the desired outcomes are rearrangements of the string 111222XXXX.
    Where each X can be 3,4,5,\text{ or }6.
    The string can be arranged in \frac{10!}{(3!)^2(4!)} ways.
    But each X can be replaced in four ways that gives \frac{(10!)(4^4)}{(3!)^2(4!)}.
    Now each of those strings has probability \frac{1}{6^{10}}.
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    Thanks Plato!
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  8. #8
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    Quote Originally Posted by Plato View Post
    You really need to learn to model the outcome space.
    For part (b) the desired outcomes are rearrangements of the string 111222XXXX.
    Where each X can be 3,4,5,\text{ or }6.
    The string can be arranged in \frac{10!}{(3!)^2(4!)} ways.
    But each X can be replaced in four ways that gives \frac{(10!)(4^4)}{(3!)^2(4!)}.
    Now each of those strings has probability \frac{1}{6^{10}}.
    I don't know if this post will be of value to anyone, but here is another method which is much more complicated than Plato's and gives the same result. I outlined it originally because I was having trouble following Plato's reasoning, even though it seems really easy to follow now.

    Consider each possible type of string, by which I mean:

    111222XXXX {4}
    111222XXXY {1,3}
    111222XXYY {2,2}
    111222XYZZ {1,1,2}
    111222XYZW {1,1,1,1}

    where X, Y, Z, and W can be 3, 4, 5, or 6.

    This corresponds with the ways 4 can be partitioned as a sum of positive integers. (For partition function P, P(4) = 5.)

    Then you count the ways to get each one and multiply by the corresponding ways to arrange the string.

    4 ways to get {4} resulting in \frac{(4)(10!)}{(3!)^2(4!)^1}.

    \binom{4}{2}\cdot2 ways to get {1,3} resulting in \frac{(12)(10!)}{(3!)^3(1!)^1}.

    \binom{4}{2} ways to get {2,2} resulting in \frac{(6)(10!)}{(3!)^2(2!)^2}.

    \binom{4}{3}\cdot3 ways to get {1,1,2} resulting in \frac{(12)(10!)}{(3!)^2(1!)^2(2!)^1}.

    1 way to get {1,1,1,1} resulting in \frac{(1)(10!)}{(3!)^2(1!)^4}.

    Which can be summarized as

    P=\left(\frac{10!}{(3!)^2 6^{10}}\right)\left(\frac{4}{(4!)^1}+\frac{12}{(1!  )^1(3!)^1} + \frac{6}{(2!)^2} + \frac{12}{(1!)^2(2!)^1} + \frac{1}{(1!)^4}\right)

    We get P=\frac{350}{19683} \approx 0.01778.

    By the way, I also programmed a Monte Carlo to verify the result. (Because initially there was a discrepancy between the two answers, due to my writing a 12 instead of a 6 in one spot.)

    Code:
    import java.util.Random;
    
    public class diceRollsProblem() {
    	Random g = new Random();
    	public void monteCarlo() {
    		int successes = 0;
    		for (int i = 0; i < Integer.MAX_VALUE; i++) {
    			if (i % 500 == 0)
    				System.out.println(i + " " + successes/(double)i);
    			int[] rolls = new int[10];
    			for (int j = 0; j < 10; j++)
    				rolls[j] = g.nextInt(6)+1;
    			int num1s = 0, num2s = 0;
    			for (int j = 0; j < 10; j++) {
    				if (rolls[j] == 1) num1s++;
    				if (rolls[j] == 2) num2s++;
    			}
    			if (num1s == 3 && num2s == 3) successes++; 
    		}
    	}
    }
    Edit: This program's a bit better.

    Code:
    import java.util.Random;
    
    public class diceRollsProblem() {
    	Random g = new Random();
    	public void monteCarlo() {
    		int successes = 0;
    		for (int i = 0; i < Integer.MAX_VALUE; i++) {
    			if (i % 100000 == 0)
    				System.out.println(i + " " + successes/(double)i);
    			int[] distr = new int[6];
    			for (int j = 0; j < 10; j++)
    				distr[g.nextInt(6)]++;
    			if (distr[0] == 3 && distr[1] == 3) successes++; 
    		}
    	}
    }
    Last edited by undefined; May 8th 2010 at 05:41 PM. Reason: new code, several minor refinements
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  9. #9
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    Great explanation, undefined! You're right, when you put it this way it's much easier to comprehend.
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