Spin a dial that has a pointer to n regions

  • Thread starter lizarton
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Homework Statement



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]?

Homework Equations



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!

The Attempt at a Solution



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]
 

Answers and Replies

  • #2
Ibix
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Simplify to two spins and draw a table. Then think how that generalises to three spins.

Alternatively look into how to determine the PDF of a sum of independent random variables from their PDFs. This is the more powerful way of doing it because it works for any PDF, discreet or continuous. But the first way will solve the problem at hand.
 
  • #3
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I'm not even sure that I know how to form the simpler problem, but here's a shot:

For two spins, there are [itex]n-1[/itex] combinations that will sum to [itex]n.
(1,n-1)
(2,n-2)
...
(n-2,2)
(n-1,1)[/itex]

The total number of outcomes is n^2 (I think?).

Where do I go from here?
 

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