# Spin a dial that has a pointer to n regions

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

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

The total number of outcomes is $n^3$ (I think)

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

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

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

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

My problem is identifying $n$ and $k$ in these problems. Any help would be greatly appreciated!

## Answers and Replies

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 $2 ^ {n-1}$. 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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Are compositions the same as selections in this context? If so, here's my attempt, provided I understand you correctly:

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

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

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

So then, to answer the question in general for arbitrary $n$, the probability is

$\frac{(n-1)!}{2^{n}(n-3)!} = \frac{1}{2^{n}(n-3)(n-2)}$?

So then, to answer the question in general for arbitrary $n$, the probability is

$\frac{(n-1)!}{2^{n}(n-3)!} = \frac{1}{2^{n}(n-3)(n-2)}$?

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?

For the denominator, we would have $n=k$, right? The formula I wrote for arbitrary $n$ has been simplified; $(3-1)!=2$, and I divided the entire quantity by $2^{n-1}$. I'm not sure that I understand the relationship we draw here between $2^{n-1}$ and $\frac{(n-1)!}{(k-1)!(n-k)!}$. Was my estimate for $n=9, k=3$ correct?

For the denominator, we would have $n=k$, right? The formula I wrote for arbitrary $n$ has been simplified; $(3-1)!=2$, and I divided the entire quantity by $2^{n-1}$. I'm not sure that I understand the relationship we draw here between $2^{n-1}$ and $\frac{(n-1)!}{(k-1)!(n-k)!}$. Was my estimate for $n=9, k=3$ 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?

The dial can stop at any arbitrary natural number $1..n$. The number of regions isn't specified further than that, but there are $k=3$ trials that we consider. The problem asks to compute the probability that the sum of the three spins will equal $n$. However, your version of the problem is just a simpler case of the general problem I need to solve, I believe.

The dial can stop at any arbitrary natural number $1..n$. The number of regions isn't specified further than that, but there are $k=3$ trials that we consider. The problem asks to compute the probability that the sum of the three spins will equal $n$. 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.

Yes, the largest possible sum would be $3n$.

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

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

Then the total number of outcomes is $n^{k}$ by the product rule.

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

Does this look correct to you?

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Yes, the largest possible sum would be $3n$.

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

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

Then the total number of outcomes is $n^{k}$ by the product rule.

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

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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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 $k$ unordered outcomes from $n$ 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 $n=3,4,5,6,7$, then the coefficient that will give me the number of ways this can happen for arbitrary $n$ is $\left(\stackrel{n-1}{2}\right) = \frac{1}{2} (n-2)(n-1)$.

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

Thanks for your help.