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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 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. 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: $\displaystyle \frac{1443}{46656}$ Thanks from Hoempa February 26th, 2015, 01:44 PM #3 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Yes, for the first, $\displaystyle \sum_{i=1}^{5} \left[ i^3 \cdot ((6-i)^3 - (5-i)^3) \right]/46656 = 1443 / 46656$ Thanks from anicicn 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. Thanks from anicicn February 26th, 2015, 05:35 PM   #5
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Quote:
 Originally Posted by anicicn 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 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 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. Tags cards, die, persons, probability, rolling, times Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Mr Davis 97 Probability and Statistics 1 January 26th, 2015 02:58 PM Bogauss Advanced Statistics 4 February 12th, 2012 09:05 AM pksinghal Probability and Statistics 3 October 23rd, 2010 03:03 PM robb Advanced Statistics 1 September 7th, 2009 08:37 AM Dungeoneer Algebra 1 December 13th, 2008 11:39 AM

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