Spin a dial that has a pointer to n regions

[lizarton]

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)<br /> (1,2,n-3)<br /> (1,3,n-4)<br /> \ldots<br /> (1,n-2,1)<br /> (1,n-3,2)<br /> \ldots<br />

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!

 

[SW VandeCarr]

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?

 

[lizarton]

Thanks for replying!
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)}?

 

[SW VandeCarr]

Thanks for replying!

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?

 

[lizarton]

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?

 

[SW VandeCarr]

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?

 

[lizarton]

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.

 

[SW VandeCarr]

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.

 

[lizarton]

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?

 

[SW VandeCarr]

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.

 

[lizarton]

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.