Exercise 1.7

Question

Two chess players, A and B, are going to play 7 games. Each game has three possible
outcomes: a win for A (which is a loss for B), a draw (tie), and a loss for A (which is
a win for B). A win is worth 1 point, a draw is worth 0.5 points, and a loss is worth 0
points.
\begin{enumerate}[label=(\alph*)]
\item How many possible outcomes for the individual games are there, such that overall
player A ends up with 3 wins, 2 draws, and 2 losses?
\item How many possible outcomes for the individual games are there, such that A ends
up with 4 points and B ends up with 3 points?
\item Now assume that they are playing a best-of-7 match, where the match will end
when either player has 4 points or when 7 games have been played, whichever is first.
For example, if after 6 games the score is 4 to 2 in favor of A, then A wins the match
and they don’t play a 7th game. How many possible outcomes for the individual games
are there, such that the match lasts for 7 games and A wins by a score of 4 to 3?
\end{enumerate}

fifthist x · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item There are ${7 \choose 3}$ ways to assign three wins to player $A$. For a 
specific combination of three games won by $A$, there are ${4 \choose 2}$ ways 
to assign two draws to $A$. There is only one way to assign two losses to $A$ 
from the remaining two games, namely, $A$ losses both games.

$$ {7 \choose 3} 	imes {4 \choose 2} 	imes {2 \choose 2} $$

\item   If $A$ were to draw every game, there would need to be at least $8$ 
games for $A$ to obtain $4$ points, so $A$ has to win at least $1$ game. 
Similarly, if $A$ wins more than $4$ games, they will have more than $4$ points.
  
  Case 1: $A$ wins $1$ game and draws $6$.
  
  This case amounts to selecting $1$ out of $7$ for $A$ to win and assigning a 
  draw for the other $6$ games. Hence, there are $7$ possibilities.
    
  Case 2: $A$ wins $2$ games and draws $4$.
  
  There are ${7 \choose 2}$ ways to assign $2$ wins to $A$. For each of them, 
  there are ${5 \choose 4}$ ways to assign four draws to $A$ out of the 
  remaining $5$ games. Player $B$ wins the remaining game. The total number 
  of possibilities for this case is ${7 \choose 2} 	imes {5 \choose 4}$.
  
  Case 3: $A$ wins $3$ games and draws $2$.
  
  There are ${7 \choose 3}$ ways to assign $3$ wins to $A$. For each of them, 
  there are ${4 \choose 2}$ ways to assign two draws to $A$ out of the remaining 
  $4$ games. $B$ wins the remaining $2$ games. The total number of possibilities 
  for this case is ${7 \choose 3} 	imes {4 \choose 2}$.
  
  Case 4: $A$ wins $4$ games and loses $3$.
  
  There are ${7 \choose 4}$ ways to assign $4$ wins to $A$. $B$ wins the 
  remaining $3$ games.
  The total number of possibilities for this case is ${7 \choose 4}$.
  
  Summing up the number of possibilities in each of the cases we get
  
  $$ {7 \choose 1} + {7 \choose 2} 	imes {5 \choose 4} + {7 \choose 3} 	imes 
  {4 \choose 2} + {7 \choose 4} $$

\item If $B$ were to win the last game, that would mean that $A$ had already 
obtained $4$ points prior to the last game, so the last game would not be 
played at all. Hence, $B$ could not have won the last game.
  
  Case 1: $A$ wins $3$ out of the first $6$ games and wins the last game.
  
  There are ${6 \choose 3}$ ways to assign $3$ wins to $A$ out of the first $6$ 
  games. The other $3$ games end in a draw. The number of possibilities then 
  is ${6 \choose 3}$.
  
  Case 2: $A$ wins $2$ and draws $2$ out of the first $6$ games and wins the 
  last game.
  
  There are ${6 \choose 2}$ ways to assign $2$ wins to $A$ out of the first 
  $6$ games. From the $4$ remaining games, there are ${4 \choose 2}$ ways to 
  assign $2$ draws. The remaining $2$ games are won by $B$. The number of 
  possibilities is ${6 \choose 2} 	imes {4 \choose 2}$.
  
  Case 3: The last game ends in a draw.
  
  This case implies that $A$ had $3.5$ and $B$ had $2.5$ points by the end of 
  game $6$.
  
  Case 3.1: $A$ wins $3$ and draws $1$ out of the first $6$ games.
  
  There are ${6 \choose 3}$ ways to assign $3$ wins to $A$ out of the first 
  $6$ games. There are ${3 \choose 1}$ ways to assign a draw out of the 
  remaining $3$ games. $B$ wins the other $2$ games. The number of 
  possibilities is ${6 \choose 3} 	imes {3 \choose 1}$.
  
  Case 3.2: $A$ wins $2$ and draws $3$ out of the first $6$ games.
   
  There are ${6 \choose 2}$ ways to assign $2$ wins to $A$ out of the 
  first $6$ games. There are ${4 \choose 3}$ ways to assign $3$ draws out of 
  the remaining $4$ games. $B$ wins the remaining game. The number of 
  possibilities is ${6 \choose 2} 	imes {4 \choose 3}$.
  
  Case 3.3: $A$ wins $1$ and draws $5$ of the first $6$ games.
  
  There are ${6 \choose 1}$ ways to assign a win to $A$ out of the first 
  $6$ games.
  
  The total number of possibilities then is
  
  $$ {6 \choose 3} + {6 \choose 2} 	imes {4 \choose 2} + {6 \choose 3} 	imes 
  {3 \choose 1} + {6 \choose 2} 	imes {4 \choose 3} + {6 \choose 1}$$
\end{enumerate}

Exercise 1.7

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Exercise 1.7

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