- #1
TheSodesa
- 224
- 7
Homework Statement
4 Players (A, B, C, and D) enter a chess tournament. The probabilities of them winning each other have been calculated based on their previous games. The results are in the below table:
\begin{array}{|c|cccc|}
\hline\\ \\
& \text{A wins} & \text{B wins} & \text{C wins} & \text{D wins} \\
\hline\\
A & - & 0.4 & 0.6 & 0.7\\
B & 0.6 & - & 0.2 & 0.3\\
C & 0.4 & 0.8 & - & 0.4\\
D & 0.3 & 0.7 & 0.6 & -\\
\hline
\end{array}
What is the probability of A winning the tournament, assuming no games will end in a tie?
The correct answer is 0.196
Homework Equations
Combinations:
\begin{equation}
{n \choose k} = \frac{n!}{k!(n-k)!}
\end{equation}
Permutations:
\begin{equation}
p(k) = k!
\end{equation}
Independent probability:
\begin{equation}
P(A \cap B) = P(A)P(B)
\end{equation}
The additive rule
\begin{equation}
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\end{equation}
The multiplicative rule:
\begin{equation}
P(A \cap B) = P(B)P(A|B)
\end{equation}
The Attempt at a Solution
[/B]
Since the rules were not specified in the problem statement, I'm assuming everybody will play against everybody, meaning each player has 3 opponents. Therefore the total number of games, ##N##, will be
[tex]N = 3! = 6[/tex]
For A to win, he has to win the most games. If A wins all of his games, he is sure to win the tournament. If he loses 1 game, he still has a chance of winning, assuming the other players don't reach 2 victories. He is certainly not going to win if he loses more than 1 game.
Now let's denote ##A## = "Player A wins", and let ##X## denote the number of victories player A has.
The number of ways ##X = 3## is ##{3 \choose 3} = 1##.For ##X = 2## the number is ##{3 \choose 2} = 3##
Then
\begin{align*}
P(X = 3) &= (0.6)(0.4)(0.3) = 0.072\\
P(X = 2) &= \stackrel{\text{Loses against B}}{(0.4)(0.4)(0.3)}
+ \stackrel{\text{Loses against C}}{(0.6)(0.6)(0.3)}
+ \stackrel{\text{Loses against D}}{(0.6)(0.4)(0.7)}\\
&= 0.324
\end{align*}
Now the total probability (there's another formula for this, but I don't think it applies here since, I don't know the probabilities of the partitions of the sample space ##\Omega##):
\begin{align*}
P(A)
&= P((X = 3) \cup (X = 2))\\
&= P(X=3) + P(X=2) - P((X =3) \cap (X=2))
\end{align*}
This is where I'm stuck (And I'm not sure my approach is "the" correct one in the first place.) I can't seem to get rid of that last probability of the intersections by using ##(5)##, because none of the events seem to be dependent on each other.
What to do?