Die tossing problem

  • Thread starter Mogarrr
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  • #1
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Homework Statement


A fair die is cast until a 6 appears. What is the probability that it must be cast more than five times.


Homework Equations


The die is fair, hence like most of the problems I can assume equally likely outcomes.
[itex]P(A^c)=1-P(A)[/itex] for any event A

The Attempt at a Solution


Theoretically, a 6 may never come up. It should be better to calculate the complement of the event.
The complement, I think, is the event in at least 5 tosses, a 6 occurs. So a 6 may occur in the 1st toss or the 2nd toss or... or the 5th toss.

So I have [itex] \frac 16 + \frac 56 \cdot \frac 16 + .... + (\frac 56)^4 \cdot \frac 16 [/itex]
Then factoring out [itex]\frac 16[/itex] and writing the probabilities as a summation, I have
[itex] \frac 16 \cdot \sum_{k=1}^5 (\frac 56)^{5-i} [/itex]

Is this correct? I don't have the answer. I suspect this can be derived from a probability distribution. If my suspicions are correct, which probability distribution?
 

Answers and Replies

  • #2
Fredrik
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A fair die is cast until a 6 appears. What is the probability that it must be cast more than five times.
Isn't this just the probability that none of the first five is a 6?
 
  • #3
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Isn't this just the probability that none of the first five is a 6?

Right you are. In my originial post, I should have wrote [itex] 1 - (\frac 16 \cdot \sum_{i=1}^5(\frac 56)^{5-i})[/itex].

Which is approximately 0.4.

What you suggested (and I agree with) is [itex]P(Event)= (\frac 56)^5[/itex]. Which is the same number.

This seems to much like a coincidence. Any thoughts?
 
  • #4
Fredrik
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I didn't really look at the LaTeX in your post. I just looked at the problem statement and immediately thought that the problem should be equivalent to finding the probability of five consecutive results in the 1-5 range. Then I read enough of your post to see that you appeared to be solving a more complicated problem. For a moment I thought that I must have misunderstood the problem, so I thought about it some more, but I couldn't find anything wrong with my first thought.

My interpretation of the problem is this: Suppose that we roll the die over and over until we get a six. Then we write down the sequence we got. Repeat this indefinitely. The problem is asking what fraction of the sequences we're writing down will not have a 6 among the first five numbers. If there's less than five numbers in the sequence, we can just pad it with more sixes so that we have a 5-digit sequence. The frequency of 5-digit initial sequences without sixes has to be equal to the probability that an arbitrary 5-digit sequence doesn't contain any sixes.
 
  • #5
AlephZero
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Your sum is the first few terms of a geometric series. It's not hard to find the sum as a fraction.
 
  • #6
Mentallic
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I just want to add that your summation is unnecessarily more complicated than it should be. Given
[tex] \frac 16 + \frac 56 \cdot \frac 16 + .... + (\frac 56)^4 \cdot \frac 16 [/tex]

This summation expression that you gave
[tex]\frac 16\cdot \sum_{k=1}^5 (\frac 56)^{5-i} [/tex]
(by the way, you used k and i as your dummy variable accidentally)

is equivalent to
[tex]\frac 16\cdot \sum_{k=0}^4 (\frac 56)^{k} [/tex]

which I'm sure you would agree is easier to read.
 
  • #7
Ray Vickson
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Right you are. In my originial post, I should have wrote [itex] 1 - (\frac 16 \cdot \sum_{i=1}^5(\frac 56)^{5-i})[/itex].

Which is approximately 0.4.

What you suggested (and I agree with) is [itex]P(Event)= (\frac 56)^5[/itex]. Which is the same number.

This seems to much like a coincidence. Any thoughts?

There is no coincidence here; you are just computing the right-tail of the geometric distribution. Google is your friend. See, eg., http://www.math.uah.edu/stat/bernoulli/Geometric.html .
 
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  • #8
HallsofIvy
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The probability you do NOT get a 6 on any one throw is 5/6. What is the probability you will NOT get a 6 on any of 5 consecutive throws?
 

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