Question concerning probablity

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Discussion Overview

The discussion revolves around calculating the probability of identical numbers appearing consecutively in a sequence of randomly chosen numbers from the set {1, 2, 3, 4, 5, 6}. The scope includes theoretical probability calculations and the exploration of methods to determine occurrences of multiple pairs of identical numbers in a line.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant asks how to calculate the probability of two identical numbers being adjacent in a sequence of 100 random numbers.
  • Another participant suggests that the probability of having at least two identical numbers in a row is nearly certain, proposing a method to calculate it based on smaller cases with fewer numbers.
  • A follow-up question seeks to generalize the calculation for 'n' pairs of identical numbers being adjacent.
  • Participants discuss examples to clarify what constitutes a "set" of identical numbers, with differing interpretations of sequences like 1112345 and 1111234.
  • There is a request for a method to calculate the probability of multiple sets of identical numbers appearing in longer sequences, specifically for sequences of 50 numbers.

Areas of Agreement / Disagreement

Participants express differing views on what constitutes a "set" of identical numbers, and there is no consensus on the method for calculating the probability of multiple sets appearing in a sequence.

Contextual Notes

Participants have not fully defined the parameters for what constitutes a "set" or how to approach the calculation for multiple sets, leading to uncertainty in the proposed methods.

Mins
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How can I calculate the probability of 2 same numbers being right next to each other, when 100 random numbers chosen from '1,2,3... 6' form a line?
 
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That's pretty well certain. To five decimal places, S = 100%.


To see why, and to get a more precise answer:
Let S denote a success, that is at least two in a row are the same. Consider instead the smaller problem with only two such dice in a row:

(1-6)(1-6)

S is just the chance that the two are the same, which is 6/36 = 1/6.

For three dice:

(1-6)(1-6)(1-6)

The middle die has a 1/6 chance of being the same as the one before it, and the last die has a 1/6 chance of being like the die before it. That's 1/6 + 1/6, except that now you're double counting when all three are the same, so it's 1/6 + 1/6 - 1/36.

This is easier if you calculate 1-S, which is 5/6 with two dice and (5/6)^2 for three dice.
 
Thanks. Another question

CRGreathouse said:
This is easier if you calculate 1-S, which is 5/6 with two dice and (5/6)^2 for three dice.

How can you calculate the probability of 'n' pairs of same numbers being right next to each other?
 
Mins said:
How can you calculate the probability of 'n' pairs of same numbers being right next to each other?

I'm not sure what you mean, give an example.
 
CRGreathouse said:
I'm not sure what you mean, give an example.

10 numbers will be chosen from 1,2,3...6, and they will form a line, like this

1442345662

There are 2 pairs of sets, as underlined above.

What I want to find out is the probability of not just only one, but 2 or more sets(2 same numbers being right next to each other) appearing in a line.
 
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Mins said:
What I want to find out is the probability of not just only one, but 2 or more sets(2 same numbers being right next to each other) appearing in a line.

Does 1112345 count as having two "sets"? Does 1111234?
 
Does 1112345 count as having two "sets"? Does 1111234?
First one has one "set"(I defined the word, think you should know).
And second one has two "sets".
But if ruling those out simiplify your calculation, think it's OK.
 
What I want is a method to calculate those :

two sets appearing in a line of 50 numbers
three sets appearing in a line of 50 numbers
four sets appearing in a line of 50 numbers
etc.

But a solution that can be used in similar situations will help me.
 
Last edited:

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