## Expected number of runs

1. The problem statement, all variables and given/known data

Hi,

Suppose we have a die with 3 colors on it.

4 sides are blue => P(Z=Blue) = 2/3
1 side is green => P(Z=Green) = 1/6
1 side is red => P(Z=Red) = 1/6

I throw it 20 times and have Z=(Z1,..., Z20). Now what is the expected number of "runs"?

Run is defined as the number of times the color changes, or equivalently, as the number of consistent blocks of a color.

For example: string "bbgrg" has 4 runs ( |bb|, |g|, |r|, |g| )

2. Relevant equations

3. The attempt at a solution

Attempt #1:
Change the representation of the sequence from "bbgrg" into a sequence of 1 and 0. One being a new color block (a success), 0 being just another ball of the previous color.
"bbgrg" becomes 10111.

In other words, P(Xi=1), if {Zi != Zi+1}.

This is, however, only a restatement of the problem and doesn't solve the initial problem: how many "1" do I have in 20 throws?

Attempt #2:

The number of throws before a given color occurs is geometrically distributed (Geo(p)). Thus:

E(number of throws until blue occurs) = 1/P(Z=Blue) = 3/2
E(number of throws until green occurs) = 1/P(Z=Green) = 6
E(number of throws until red occurs) = 1/P(Z=Red) = 6

I also know E(# Blue) = n * P(Z=Blue) = 20*2/3 = 40/3
E(# Green) = E(# Red) = 20/6

I could maybe use those 2 pieces of information but I can't see how. Any comments are welcomed.

Thank you for help.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
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Homework Help
 Quote by binjip 1. The problem statement, all variables and given/known data Hi, Suppose we have a die with 3 colors on it. 4 sides are blue => P(Z=Blue) = 2/3 1 side is green => P(Z=Green) = 1/6 1 side is red => P(Z=Red) = 1/6 I throw it 20 times and have Z=(Z1,..., Z20). Now what is the expected number of "runs"? Run is defined as the number of times the color changes, or equivalently, as the number of consistent blocks of a color. For example: string "bbgrg" has 4 runs ( |bb|, |g|, |r|, |g| ) 2. Relevant equations 3. The attempt at a solution Attempt #1: Change the representation of the sequence from "bbgrg" into a sequence of 1 and 0. One being a new color block (a success), 0 being just another ball of the previous color. "bbgrg" becomes 10111. In other words, P(Xi=1), if {Zi != Zi+1}. This is, however, only a restatement of the problem and doesn't solve the initial problem: how many "1" do I have in 20 throws? Attempt #2: The number of throws before a given color occurs is geometrically distributed (Geo(p)). Thus: E(number of throws until blue occurs) = 1/P(Z=Blue) = 3/2 E(number of throws until green occurs) = 1/P(Z=Green) = 6 E(number of throws until red occurs) = 1/P(Z=Red) = 6 I also know E(# Blue) = n * P(Z=Blue) = 20*2/3 = 40/3 E(# Green) = E(# Red) = 20/6 I could maybe use those 2 pieces of information but I can't see how. Any comments are welcomed. Thank you for help. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
I would do it by an iterative method. If B(n) = expected number of runs in n tosses, given the first toss is Blue, and G(n), R(n) are defined similarly, I would get the answer in terms of B(20), G(20) and R(20). Then I would get recursions for B(n), G(n) and R(n) by noting how B(n) is related to B(n-1), G(n-1) and R(n-1) by looking at the next colour, etc.

RGV

 Tags die, probability, run