Spin a dial that has a pointer to n regions

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Main Question or Discussion Point

Consider a dial having a pointer that is equally likely to point to each of [itex]n[/itex] region numbered [itex]1,2,...,n.[/itex] When we spin the dial three times, what is the probability that the sum of the selected numbers is [itex]n[/itex]?

I have to use summations, and i'm sure binomial coefficients. I believe that this is a selection; it seems to imitate rolling an [itex]n[/itex]-sided die three times, but I even have trouble computing that problem.

The total number of outcomes is [itex]n^3[/itex] (I think)

I started counting ordered triples of some n terms...

[itex]n=3:[/itex] There is only one way, [itex]{(1,1,1)}[/itex]
[itex]n=4:[/itex] There are 3 ways, [itex]{(1,2,1),(1,1,2),(2,1,1)}[/itex]
[itex]n=5:[/itex] There are 6 ways, [itex]{(1,1,3),(1,2,2),(1,3,1),(2,1,2,),(2,2,1),(3,1,1)}[/itex]
[itex]n=6:[/itex] There are 10 ways...
[itex]n=7:[/itex] 15 ways...

For arbitrary n, you can start making ordered triples...
[itex](1,1,n-2)
(1,2,n-3)
(1,3,n-4)
\ldots
(1,n-2,1)
(1,n-3,2)
\ldots
[/itex]

A Theorem I believe is relevant:
With repetition allowed, there are [itex]\left(\stackrel{n+k-1}{k - 1}\right)[/itex] ways to select [itex]n[/itex] objects from [itex]k[/itex] types. This also equals the number of nonnegative integer solutions to [itex]x_{1} + \ldots + x_{k} = n.[/itex]

My problem is identifying [itex]n[/itex] and [itex]k[/itex] in these problems. Any help would be greatly appreciated!
 

Answers and Replies

  • #2
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If I understand you, you want to know the probability of getting a sum = n for in k tries where n is a positive integer and the regions are the natural numbers: 1, 2, 3, ..........n. This involves compositions of integers which you demonstrated. The formula for the number of compositions for any natural number n is [itex] 2 ^ {n-1} [/itex]. Since you are only interested in the compositions of three integers (k=3) and are not including zero, the formula is (n-1)!/(k-1)!(n-k)!

How would you use this information to find the probability for n=9, k=3?
 
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  • #3
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Thanks for replying!
Are compositions the same as selections in this context? If so, here's my attempt, provided I understand you correctly:

For [itex]n = 9[/itex] and [itex]k = 3[/itex], the total number of compositions would be [itex]2^{8}[/itex].

Then the number of favorable outcomes/compositions would be [itex]\frac{8!}{2!6!}[/itex].

So the probability that the sum of the regions equals 9 is [itex]\frac{8*7}{2^{9}}[/itex].

So then, to answer the question in general for arbitrary [itex]n[/itex], the probability is

[itex]\frac{(n-1)!}{2^{n}(n-3)!} = \frac{1}{2^{n}(n-3)(n-2)}[/itex]?
 
  • #4
2,123
79
Thanks for replying!

So then, to answer the question in general for arbitrary [itex]n[/itex], the probability is

[itex]\frac{(n-1)!}{2^{n}(n-3)!} = \frac{1}{2^{n}(n-3)(n-2)}[/itex]?
Think about it. The probability is a fraction. Both the numerator and denominator is (n-1)!/(k-1)!(n-k)!. What are n and k for the numerator and denominator respectively?
 
  • #5
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For the denominator, we would have [itex]n=k[/itex], right? The formula I wrote for arbitrary [itex]n[/itex] has been simplified; [itex](3-1)!=2[/itex], and I divided the entire quantity by [itex]2^{n-1}[/itex]. I'm not sure that I understand the relationship we draw here between [itex]2^{n-1}[/itex] and [itex]\frac{(n-1)!}{(k-1)!(n-k)!}[/itex]. Was my estimate for [itex]n=9, k=3[/itex] correct?
 
  • #6
2,123
79
For the denominator, we would have [itex]n=k[/itex], right? The formula I wrote for arbitrary [itex]n[/itex] has been simplified; [itex](3-1)!=2[/itex], and I divided the entire quantity by [itex]2^{n-1}[/itex]. I'm not sure that I understand the relationship we draw here between [itex]2^{n-1}[/itex] and [itex]\frac{(n-1)!}{(k-1)!(n-k)!}[/itex]. Was my estimate for [itex]n=9, k=3[/itex] correct?
If I understand your problem correctly, you spin the dial just 3 times and it can stop at any number 1-9. Your question is the probability that the sum of three spins will be exactly 9. However, it's possible the dial can stop, for example, at 9 three times, no?
 
  • #7
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The dial can stop at any arbitrary natural number [itex]1..n[/itex]. The number of regions isn't specified further than that, but there are [itex]k=3[/itex] trials that we consider. The problem asks to compute the probability that the sum of the three spins will equal [itex]n[/itex]. However, your version of the problem is just a simpler case of the general problem I need to solve, I believe.
 
  • #8
2,123
79
The dial can stop at any arbitrary natural number [itex]1..n[/itex]. The number of regions isn't specified further than that, but there are [itex]k=3[/itex] trials that we consider. The problem asks to compute the probability that the sum of the three spins will equal [itex]n[/itex]. However, your version of the problem is just a simpler case of the general problem I need to solve, I believe.
Well then, I'm not sure what you're trying to do. Just say n instead of 9. The dial can still stop at n three times, no? How do you account for sums larger than n? If I understand you, you can have sums up to n*k.
 
  • #9
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Yes, the largest possible sum would be [itex]3n[/itex].

Using my formula above, the number of ways that the 3 trials can add up to n is [itex]\left(\stackrel{n+3-1}{3 - 1}\right) = \left(\stackrel{n + 2}{2}\right)[/itex]

Does that seem right to you? Since I'm looking for the number of nonnegative integer solutions to [itex]x_{1} + x_{2} + ... + x_{k} = n[/itex], where the number of trials is [itex]k = 3[/itex].

Then the total number of outcomes is [itex]n^{k}[/itex] by the product rule.

This gives the probability as
[itex]\frac{\left(\stackrel{n + 2}{2}\right)}{n^{3}}[/itex].

Does this look correct to you?
 
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  • #10
2,123
79
Yes, the largest possible sum would be [itex]3n[/itex].

Using my formula above, the number of ways that the 3 trials can add up to n is [itex]\left(\stackrel{n+3-1}{3 - 1}\right) = \left(\stackrel{n + 2}{2}\right)[/itex]

Does that seem right to you? Since I'm looking for the number of nonnegative integer solutions to [itex]x_{1} + x_{2} + ... + x_{k} = n[/itex], where the number of trials is [itex]k = 3[/itex].

Then the total number of outcomes is [itex]n^{k}[/itex] by the product rule.

This gives the probability as
[itex]\frac{\left(\stackrel{n + 2}{2}\right)}{n^{k}}[/itex].

Does this look correct to you?
No. Your problem is to find the number of compositions of length k for n and for kn. The probability then becomes the fraction of the former over the latter. The proper formula for the number of compositions of length k for n is the factorial one I gave you. I suggest you look up compositions and verify this formula for yourself before you respond.
 
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  • #11
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I looked up compositions, and after some research I think you may have meant to say that we seek combinations, the number of ways of picking [itex]k[/itex] unordered outcomes from [itex]n[/itex] possibilities. I have not dealt with either of these terms in my class, so I apologize for not recognizing them off-hand.

If my reckoning of the number of ways this can happen (all the way at the top) is correct with [itex]n=3,4,5,6,7[/itex], then the coefficient that will give me the number of ways this can happen for arbitrary [itex]n[/itex] is [itex]\left(\stackrel{n-1}{2}\right) = \frac{1}{2} (n-2)(n-1)[/itex].

The probability is this quantity divided by the total number of outcomes.

Thanks for your help.
 

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