The scenario is, we're playing a game where we have in a box 9 balls: 4 red, 3 green, 2 blue. We draw one ball at a time, and note down its color and then return it for the next draw, the game stops only when we have drawn the same color twice in a row. What is the probability of the game ending after n draws? P(X=n)? What is the probability of the last two balls being blue?

Paige Vasquez

Paige Vasquez

Open question

2022-08-21

Probability of ending with 2 blue balls in a row after n draws
The scenario is, we're playing a game where we have in a box 9 balls: 4 red, 3 green, 2 blue.
- We draw one ball at a time, and note down its color and then return it for the next draw, the game stops only when we have drawn the same color twice in a row.
- What is the probability of the game ending after n draws? P ( X = n )?
- What is the probability of the last two balls being blue?

Answer & Explanation

Terry Avery

Terry Avery

Beginner2022-08-22Added 10 answers

Step 1
Here is an idea how to solve the problem using the Markov's chain.
Let the probabilities to draw the balls be p 1 , p 2 , p 3 , respectively and let the state of the sequence be characterized by the color (type) of the last ball. Then the transition matrix and the initial state read:
(1) T = ( 0 p 1 p 1 0 p 2 0 p 2 0 p 3 p 3 0 0 p 1 p 2 p 3 1 ) and P 1 = ( p 1 p 2 p 3 0 ) , respectively, and the state of the sequence after n draws is (2) P n = T n 1 P 1 .
To make the things simpler one can note that there is no back coupling from the final state, so that it can be omitted and only the action of the "active" 3 × 3 block of the matrix onto the 3 × 1 vector of "active" states [highlighted in (1)] be considered. In what follows these reduced versions of the transition matrix and the state vector are assumed.
Step 2
The characteristic polynomial of the active block is:
(3) Q ( x ) = x 3 ( p 1 p 2 + p 2 p 3 + p 3 p 1 ) x 2 p 1 p 2 p 3 ,
so that the probability vector P n can be computed by the recurrence relation:
(4) P n = ( p 1 p 2 + p 2 p 3 + p 3 p 1 ) P n 2 + 2 p 1 p 2 p 3 P n 3
with the elements of the three initial state vectors being:
(5) P 0 , i = 1 2 , P 1 , i = p i , P 2 , i = ( 1 p i ) p i .
Finally the probability of ending with two balls of i-th color after n-th draw is: (6) P n 1 , i p i .

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