Sarah and Rachel play a series of games. The probability that Sarah wins a game
ID: 3049539 • Letter: S
Question
Sarah and Rachel play a series of games. The probability that Sarah wins a game is p, which is constant from game to game; the probability that Rachel wins the game is 1-p. the outcome of each game is independent of the other games. Play stops when the total number of games won by one player is two greater than that of the other player and the player with the greater number of total wins is the series winner.
A. Find the probability that a total of four games are played.
B. Find the Probabilty that Sarah wins the series.
Explanation / Answer
Answer to the question)
Win of Sarah = p
Win of Rachel = (1-p)
Since it is stated that the games in the series are independent of each other, this implies that the multiplication rule of probability applies here
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Answer to part a)
When total four games are played following are the possible scenarios
All four games are won by Sarah
But this is not possible because after two consecutive wins of sarah , the series would be oover as per the rule that one player wins two times more than the other player,that marks the end of the series
Likewise even if Rachel wins all four games , it would not be possible to go up to four games.
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This implies for four games to be played, the first condition is that first two games have to be won by different players: either Sarah and then Rachel , or Rache first and then Sarah. Only then the series proceeds to the third level and then the fourth level too.
So the possible sequence of winnings that would assure thate xact 4 games are played in the series are as follows:
Case I : First win : Sarah , Second win: Rachel, thrid win: Sarah , Fourth win; Sarah
The probability of this case is: p*(1-p)*p*p = (1-p)*p^3
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Case II: First win: Rachel, Second win: Sarah , Thrid win: Rachel , Fourth win: Rachel
The probability of case II = (1-p)*p*(1-p)*(1-p) = p*(1-p)^3
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Thus total probability fo exactly 4 games played = (1-p)*p^3 + p*(1-p)^3
Total probability = p*(1-p)*(p^2 + (1-p)^2)
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Answer to part b)
Sarah can win the series in two games only: p*p = p^2
We move to three games only when the first tow games are not won by the same player , so when three games are player , there would be 2 wins of one player , and one win of another player, ultimately no one clearly wins here to end the series, so the fourth game comes up:
Probability Sarah wins when four games are played = (1-p)*p^3
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If no clear winner comes it in the fourth game as well , fifth game is played and no clear winner can be there in fifth one , so the sixth game appears , and for Sarah to win in the sixth game , the sequene of winnings is as follows:
Sarah wins,Rachel wins, Sarah wins , Rachel wins, Sarach wins , Sarach wins.... like this we can observe that Sarach wins four games and Rachel wins 2 games , thus Sarach is 2 wins ahead of Rachel , and thats what ends the series
The Probability of win of Sarach in 6 games = (1-p)^2*p^4
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Like this the total probability of Sarah's win depends on the number of games that happens in the series:
Total probability of Sarah's win = P(winning in 2 games) + P(winning in four games) + P9winning in six games) and so on....it keeps on adding
Probability of Sarah wins = p^2 + (1-p)*p^3 + (1-p)^2*p^4 .....
Probability of Sarah wins = p^2 [1 + (1-p)*p + (1-p)^2*P^2 ..........]
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