Question

Two chess players, A and B, are going to play 7 games. Each game has three possibleoutcomes: a win for A (which is a loss for B), a draw (tie), and a loss for A (which isa win for B). A win is worth 1 point, a draw is worth 0.5 points, and a loss is worth 0points.(A) How many possible outcomes for the individual games are there, such that overallplayer A ends up with 3 wins, 2 draws, and 2 losses?(B) How many possible outcomes for the individual games are there, such that A endsup with 4 points and B ends up with 3 points?(C) Now assume that they are playing a best-of-7 match, where the match will end assoon as either player has 4 points. For example, if after 6 games the score is 4 to 2 infavor of A, then A wins the match and they don’t play a 7th game. How many possibleoutcomes for the individual games are there, such that the match lasts for 7 games andA wins by a score of 4 to 3?

Answer 1

Answer:

A) the possible outcomes for individual game is 210 games.

B) The possible outcome for this is 357 games.

C) Then the Possible Games are 267 games.

Step-by-step explanation:

A) Total number of individual games are 7 in which A ends up with 3 wins which give 4 remaining games. then there are two draw games from them. and in the end the remaining games are losses so,we useformula for combination:

                       $(7C3)*(4C2)=210 \ games$

B) Now Player A has 4 point that gives us 5 possibilities until we reach 7 games. Similarly B have 3 points which is same as player A gives us 5 possibilities until we reach 7 games. thus those two cases for player A and Player B can be stated as:

           $(7C3)+(7C3)*(4C2)+(7C1)*(6C6)+(7C2)*(5C4)=357\ games$

C) Lets say that player A ahs 4 points and player B has 3 points and all seven games have been played so.

c1) If player A has 4 wins and 3 losses then last win have to be in 7th match thus:the answer is $(6C3)$.

c2) If player A has 3 win, 2 draws and 3 losses thus it means that final match cannot be a loss thus the answer is $(6C2)*(5C3)$.

c3) Now lastly if player A has 1 win and 5 draws we can arrange them arbitrarily thus the answer here is $(7C1)$.

c4) now If player A has 2 wins, 4 draws and 2 losses thus answer is $(6C1)*(6C2)$

Sum of all the cases is

               $(6C3)+(6C2)*(5C3)+(7C1)+(6C1)*(6C2)=267\ games$