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February 26th, 2015, 01:15 AM  #1 
Newbie Joined: Feb 2015 From: Belgrade, Serbia Posts: 1 Thanks: 0  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 132 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. 
February 26th, 2015, 12:53 PM  #2 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
For the first part I created the program  the probability is: $\displaystyle \frac{1443}{46656}$ 
February 26th, 2015, 01:44 PM  #3 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
Yes, for the first, $\displaystyle \sum_{i=1}^{5} \left[ i^3 \cdot ((6i)^3  (5i)^3) \right]/46656 = 1443 / 46656$

February 26th, 2015, 05:14 PM  #4 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
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. 
February 26th, 2015, 05:35 PM  #5  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Code: small=vector(32,n,v[n]=binomial(32n,2)) large=vector(32,n,v[n]=binomial(n1,2)) sum(m=1,31,sum(n=m+1,32,small[n])*large[m])/binomial(32,3)^2 %*1.0  
February 27th, 2015, 03:17 AM  #6 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
Maybe we should declare v=vector(32)? I get a typeerror when running that code. Equivalently, we have Code: sum(i=3,29,binomial(i1,2)*binomial(32i,3))/binomial(32,3)^2+0. 

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cards, die, persons, probability, rolling, times 
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