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September 19th, 2017, 10:27 AM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 561 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, 11:25 AM  #2  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757  Quote:
 
September 20th, 2017, 10:19 AM  #3  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 561 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 20th, 2017, 11:37 PM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757 
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, 12:04 AM  #5 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757 
(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, 04:53 PM  #6 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 561 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).


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