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

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SUMMARY

The probability of rolling a six 70 times in 360 tosses of a fair die can be approximated using the normal distribution due to the large number of trials. The expected number of sixes is calculated as 60, with a standard deviation of approximately 5√2. By applying the half-integer correction, the probability of obtaining more than 69.5 sixes is determined using the standard normal distribution. The discussion also highlights the binomial distribution approach as an alternative method for calculating the exact probability.

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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:

\frac{1}{6^{360}}\sum^{290}_{n=0} 5^n = \frac{1-5^{291}}{6^{360}(1-5)}
 
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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:
(\frac{1}{6} + \frac{1}{6})^{360})
Then you get an exact answer.
 
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jcsd said:
Surely you can use the series:

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

I think what you mean is :

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

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 :

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

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:

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