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September 19th, 2017, 10:27 AM   #1
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Complicated Probability With Dice

Dice are rolled with 9 trials. On each trial, 2 dice are rolled, and one is re-rolled if they are the same number until there are different numbers, with re-rolls 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?
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September 19th, 2017, 11:25 AM   #2
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Quote:
Originally Posted by EvanJ View Post
Dice are rolled with 9 trials. On each trial, 2 dice are rolled, and one is re-rolled if they are the same number until there are different numbers, with re-rolls 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?
What does this mean? Of course at least one number will be rolled on every trial.
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September 20th, 2017, 10:19 AM   #3
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What does this mean? Of course at least one number will be rolled on every trial.
I mean that at least one of these will happen:

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
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September 20th, 2017, 11:37 PM   #4
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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 $9-k+1=10-k$ possible consecutive runs among the 9 trials.

Thus

$P[\text{given digit appears in at least 4 consecutive runs}]= \displaystyle \sum_{k=4}^9~(10-k)p_d^k (1-p_d)^{9-k} = \dfrac{107}{6561}$
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September 21st, 2017, 12:04 AM   #5
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(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

$P[\text{a given pair occurs on at least 3 trials}]=P[k\geq 3]=1-P[k<3]$

$P[k<3] = (1-p)^9 + 9 p (1-p)^8 + 36p^2 (1-p)^7= \dfrac{37738034432}{38443359375}$

$1-P[k<3] = 1 - \dfrac{37738034432}{38443359375} = \dfrac{705324943}{38443359375}$
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September 21st, 2017, 04:53 PM   #6
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Quote:
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$P[\text{given digit appears in at least 4 consecutive runs}]= \displaystyle \sum_{k=4}^9~(10-k)p_d^k (1-p_d)^{9-k} = \dfrac{107}{6561}$
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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