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February 26th, 2015, 01:15 AM   #1
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Probability of rolling one die by two persons 3 times each

So suppose you have one die. Person A rolls one die 3 times. Person B does the same after A is finished. What is the probability that person A's smallest number (of his/hers 3 rolls) is greater than person B's biggest number?

Extra addition: Now problem is the same but instead of 1 die we have 32 cards with numbers 1-32 on each of the cards. I want to know the probability that person A's smallest number taken from 32 cards is greater than person B's biggest number. Each person takes 3 cards. When A is finished he puts back his cards. Than it's person B's turn and he also takes 3 cards out of 32.
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February 26th, 2015, 12:53 PM   #2
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For the first part I created the program - the probability is:
$\displaystyle \frac{1443}{46656}$
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February 26th, 2015, 01:44 PM   #3
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Yes, for the first, $\displaystyle \sum_{i=1}^{5} \left[ i^3 \cdot ((6-i)^3 - (5-i)^3) \right]/46656 = 1443 / 46656$
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February 26th, 2015, 05:14 PM   #4
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For the second, there are $6{i - 1 \choose 2}$ ways to have a maximum of i and there are $6{32 - l \choose 2}$ ways to have a minimum of l with three distinct cards, so there are $6{33 - l \choose 3}$ ways to have a minimum of at least l. (We could also omit the number of permutations of the 3 cards all the time, they cancel with the total possibilities.

So we have $$36\sum_{i=3}^{29} {i - 1 \choose 2}\cdot {33 - (i+1) \choose 3}$$, which is 32622912. There are possibilities of $\displaystyle 6{32 \choose 3} = 29760$ for each player so a total of 29760^2 = 885657600

Gives a final answer of 32622912/885657600 = 1827/49600 ~= 0.03683.
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February 26th, 2015, 05:35 PM   #5
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Quote:
Originally Posted by anicicn View Post
Now problem is the same but instead of 1 die we have 32 cards with numbers 1-32 on each of the cards. I want to know the probability that person A's smallest number taken from 32 cards is greater than person B's biggest number. Each person takes 3 cards. When A is finished he puts back his cards. Than it's person B's turn and he also takes 3 cards out of 32.
This is how I found the answer, with PARI/GP (see my .sig below).

Code:
small=vector(32,n,v[n]=binomial(32-n,2))
large=vector(32,n,v[n]=binomial(n-1,2))
sum(m=1,31,sum(n=m+1,32,small[n])*large[m])/binomial(32,3)^2
%*1.0
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February 27th, 2015, 03:17 AM   #6
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Maybe we should declare v=vector(32)? I get a type-error when running that code.

Equivalently, we have
Code:
sum(i=3,29,binomial(i-1,2)*binomial(32-i,3))/binomial(32,3)^2+0.
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