What Is the Probability of Rolling a Six 70 Times in 360 Dice Tosses?

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Discussion Overview

The discussion revolves around calculating the probability of rolling a six at least 70 times in 360 tosses of a fair die. Participants explore various statistical methods, including the normal approximation to the binomial distribution and direct binomial calculations, while addressing potential errors in their approaches.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the normal approximation to the binomial distribution, calculating the expected number of sixes and standard deviation, and applying a continuity correction.
  • Another participant proposes a series approach to calculate the probability, although they later express concern about the validity of their method due to an assumption of specific order.
  • A different participant mentions using the binomial distribution directly, indicating that summing the first 290 terms of a specific expression could yield an exact answer.
  • Some participants express confusion about their calculations, noting that their results seem implausibly low or near certainty, suggesting potential errors in their methods.
  • There are repeated references to modifying the original equations to correct for perceived mistakes, indicating uncertainty in the approaches taken.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to calculate the probability, with multiple competing views and approaches presented, and ongoing uncertainty about the correctness of their calculations.

Contextual Notes

Participants acknowledge limitations in their approaches, including assumptions made in calculations and the need for corrections in their mathematical expressions.

lhuyvn
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Hi All,

With the given problem below, any suggestion?

A fair die is tossed 360 times. The probability that a six comes up on 70 or more of the tosses is

A. greater than 0.50
B.between 0.16 and 0.50
C.between 0.02 and 0.16
D.between 0.01 and 0.02
E. less than 0.01
 
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With 360 tosses it probably best to use the normal approximation to the binomial distribution. The probability of a 6 on one roll is 1/6 so the expected number of 6s in 360 rolls is (1/6)(360)= 60. The standard deviation is sqrt((1/6)(5/6)(360= sqrt(50)= 5sqrt(2). Since the number of 6s must be an integer while the normal variable is continuous, use the "half-integer" correction: instead of "more than 70" use "more than 69.5". The standard variable would be (69.5- 60)/(5sqrt(2)). Use a table of the standard normal distribution to find the probability that z is larger than that.
 
Surely you can use the series:

[tex]\frac{1}{6^{360}}\sum^{290}_{n=0} 5^n = \frac{1-5^{291}}{6^{360}(1-5)}[/tex]
 
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my answer seems ridculously low though and it doesn't equal unity where it should :confused:

edited to add I see the mistake now I've assumed a specific order so the series doesn't work.
 
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Well, you could take the binomial distribution approach. If you take the sum of the first 290 terms of:
[tex](\frac{1}{6} + \frac{1}{6})^{360})[/tex]
Then you get an exact answer.
 
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jcsd said:
Surely you can use the series:

[tex]\frac{1}{6^{360}}\sum^{290}_{n=0} 5^n = \frac{1-5^{291}}{6^{360}(1-5)}[/tex]

I think what you mean is :

[tex]1- \frac {1} {6^{360}}\sum^{70}_{n=0} 5^{360-n}[/tex]

Edited to add : this makes it look like a near certainty (~ 1 - 10^-29), something's wrong !
 
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Gokul43201 said:
I think what you mean is :

[tex]1- \frac {1} {6^{360}}\sum^{70}_{n=0} 5^{360-n}[/tex]

Edited to add : this makes it look like a near certainty (~ 1 - 10^-29), something's wrong !

I'm pretty sure the original equation was sound, but it needs the following modification to hold true:

[tex]\frac{1}{6^{360}}\sum^{290}_{n=0} ^{360}P_{n}5^n[/tex]
 
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