(adsbygoogle = window.adsbygoogle || []).push({}); 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=(Z_{1,..., 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(X_{i}=1), if {Z_{i}!= Z_{i+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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# Expected number of runs

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