Probability of ending with 2 blue balls in a row after n drawsThe 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?

Question

Probability of ending with 2 blue balls in a row after n drawsThe scenario is, we&#039;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?

Terry Avery · Accepted Answer

Step 1Here is an idea how to solve the problem using the Markov&#039;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 &quot;active&quot;   3  ×  3 block of the matrix onto the   3  ×  1 vector of &quot;active&quot; states [highlighted in (1)] be considered. In what follows these reduced versions of the transition matrix and the state vector are assumed.Step 2The 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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