Two people play a game. They play a series of points, each producing a winner and a loser, until one player has won at least 4 points and has won at least 2 more points than the other. Anne wins each point with probability p. What is the probability that she wins the game on the kth point played for k = 4 , 5 , 6 , . . .

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Two people play a game. They play a series of points, each producing a winner and a loser, until one player has won at least 4 points and has won at least 2 more points than the other. Anne wins each point with probability p. What is the probability that she wins the game on the kth point played for   k  =  4  ,  5  ,  6  ,  .  .  .

ri1men4dp · Accepted Answer

Step 1For higher k, note that the game has to go through a tie in order to continue. For instance, for someone to win 5:3, the score must have been 3:3 (since it must have been 4:3, and 4:2 would have ended the game).Step 2So you can calculate the probability for 3:3, which is                     (                    6      3                      )                  p    3        q    3  , and from then on the probability is       p    2   for Anne to win and 2pq (two ways for one win and one lose) to reach the next tie, so for   k  ≥  5,       p    k    =                    (                    6      3                      )                  p    3        q    3        p    2              (      2      p      q      )              k      −      5        =      5          8              q        2              (  2  p  q      )    k      .

Kenya Leonard · Accepted Answer

Step 1What is the probability that she wins the game on the kth point played for   k  =  4  ,  5  ,  6  ,  .  .  .I suggest that the k-th point played is not the same as the k-th point won by Anne.So I started with   k  =  4 P(win on 4th point) can happen 3 ways   4  −  0  ,  4  −  1  ,  4  −  2  =      p    4    (  1  +  (  4     choose     1  )  q  +  (  5     choose     2  )      q    2    )I would say here that Anne wins the game on the 4th point played with probability       p    4  ; the probability that she wins on the 5th point played is                     (                    4      1                      )                  p    4    q; and the probability that she wins on the 6th point played is                     (                    5      2                      )                  p    4        q    2  . The corresponding win probabilities for Betty are obtained by interchanging p and q in the previous sentence.Step 2If 6 points have been played and the game is not over, the score must be 3-3 (deuce) and this event has probability                     (                    6      3                      )                  p    3        q    3    =  20      p    3        q    3  . The game can now end only when an even number of additional points have been played, and the probability of Anne winning on the k-th point played (where   k  =  6  +  2  n  ,  n  &amp;gt;  0) is                     (                    6      3                      )                  p    3        q    3    ⋅      p    2    (  2  p  q      )          n      −      1      . To get the probability that Anne wins the game, sum the probabilities that she wins in 4, 5, 6, 8, 10,… points; a geometric series is involved.If f(p) denotes the probability that Anne wins the game, then   f  (  0.5  )  =  0.5 but                           d                  d          p                    f      (      p      )      |              p      =      0.5        =  2.5 so that   f  (  0.5  +  ϵ  )  =  0.5  +  2.5  ϵ  +  ⋯, that is, a small difference   p  −  q  =  2  ϵ in the point win probabilities for Anne and Betty is amplified by the rules of tennis into a greater difference in the game win probabilities.

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