
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
September 19th, 2017, 11:27 AM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 573 Thanks: 79  Complicated Probability With Dice
Dice are rolled with 9 trials. On each trial, 2 dice are rolled, and one is rerolled if they are the same number until there are different numbers, with rerolls counting as part of the same trial. Within each trial, it doesn't matter which die lands on which number. 1. What is the probability that at least one number will be rolled on at least 4 consecutive trials? 2. What is the probability that at least one of the 15 possible pairs of numbers (1 and 2, 1 and 3,... 5 and 6) will be rolled on at least 3 trials, without caring if they are consecutive? 
September 19th, 2017, 12:25 PM  #2  
Senior Member Joined: Sep 2015 From: USA Posts: 1,653 Thanks: 840  Quote:
 
September 20th, 2017, 11:19 AM  #3  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 573 Thanks: 79  Quote:
1 is rolled on at least 4 consecutive trials 2 is rolled on at least 4 consecutive trials 3 is rolled on at least 4 consecutive trials 4 is rolled on at least 4 consecutive trials 5 is rolled on at least 4 consecutive trials 6 is rolled on at least 4 consecutive trials  
September 21st, 2017, 12:37 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 1,653 Thanks: 840 
Essentially what you've got on each trial is a uniform distribution on the 15 possible unordered pairs of unlike numbers. Each pair has a probability of $p=\dfrac {1}{15}$ Any given digit is contained in 5 pairs. So there is a probability of $p_d=\dfrac 1 3$ of that digit being in the pair produced by the roll scheme for a given trial. For a given consecutive run of $k$ trials there are $9k+1=10k$ possible consecutive runs among the 9 trials. Thus $P[\text{given digit appears in at least 4 consecutive runs}]= \displaystyle \sum_{k=4}^9~(10k)p_d^k (1p_d)^{9k} = \dfrac{107}{6561}$ 
September 21st, 2017, 01:04 AM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 1,653 Thanks: 840 
(2) is a direct application of the binomial distribution Here we have the probability of a given pair is $p=\dfrac {1}{15}$ so, letting a roll of a given pair as a success $P[\text{a given pair occurs on at least 3 trials}]=P[k\geq 3]=1P[k<3]$ $P[k<3] = (1p)^9 + 9 p (1p)^8 + 36p^2 (1p)^7= \dfrac{37738034432}{38443359375}$ $1P[k<3] = 1  \dfrac{37738034432}{38443359375} = \dfrac{705324943}{38443359375}$ 
September 21st, 2017, 05:53 PM  #6 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 573 Thanks: 79  Is that the probability of "1 is rolled on at least 4 consecutive trials" or the probability that at least one number will be rolled on at least 4 consecutive trials (like the 6 lines I wrote out)? The first part is easier. The second part is hard because if 1 is not rolled on at least 4 consecutive trials, the other numbers are more likely to be rolled on at least 4 consecutive trials. To make an analogy, if somebody tells you that a die didn't land on 1, it's more likely (1/5 compared to 1/6) that the die landed on 2 (and the same for 3 through 6).


Tags 
complicated, dice, probability 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A bit more complicated probability  david0925  Probability and Statistics  7  February 6th, 2015 07:32 AM 
Complicated dice issue!  Theguru  Probability and Statistics  1  April 10th, 2014 06:52 AM 
Complicated Probability  cosmic  Advanced Statistics  4  November 23rd, 2011 09:40 AM 
Dice probability with multicolored dice pips  Brimstone  Algebra  10  May 30th, 2008 08:36 PM 
Very complicated dice problem  Infinity  Advanced Statistics  6  January 1st, 2007 02:17 PM 