# What is the probability each team wins the tournament?

$A, B, C, D$ are four teams in a tournament where $A$ faces $B$ and $C$ faces $D$ in the first round and the winners of those rounds face each other. $A$ has a $0.7$ probability of beating any other team, while $B, C, D$ have a $0.5$ probability of beating each other. If the games are independent, what is the probability that each team wins the tournament?

My idea is $P(\text{$A$wins})=P[A \text{ beats } B$ and ($A \text{ beats } C$ or $A \text{ beats D}$)$]$ $=P[(A \text{ beats B and A beats C) or (A beats B and A beats D)}]$ $=0.7*0.7 + 0.7*0.7 =98$%, and repeating this process for each $B,C,D$ but this seems incorrect.

• $A$ wins with probability $0.7\cdot 0.5=0.35$. How you got $0.98$ I have no idea. Similarly, $B$ will win by first beating $A$ and then beating whichever of the remaining opponents are left, which occurs with probability... Finally, we can get the probability for $C$ and $D$ by appealing to symmetry and that it must be half of the remaining available probability. – JMoravitz Feb 20 '18 at 3:52
• @JMoravitz He said $A$ wins with probability $0.7$ against all teams. Should it not be $0.49$? – bames Feb 20 '18 at 3:55
• @JMoravitz I edited my question to show specifically how I got $0.98$, what have I incorrectly assumed to get that number? – Oliver G Feb 20 '18 at 3:58
• @bames sure would, good catch. And now that he has edited, we can see where his mistake lies. When $A$ beats $B$, he moves on to play against the other remaining player, but this is conditioned on that player having won his own game!. If you insist on keeping track of who the second opponent is for $A$, then you need a factor of $\frac{1}{2}$ in front of each. Note, it doesn't make sense for the probability that $A$ wins the tournament as a whole to be higher than the probability that $A$ wins his first game. – JMoravitz Feb 20 '18 at 3:58
• So assuming $P(A \text{ wins })=0.7 * 0.7 = 0.49$, $P(B) = 0.3 * 0.5 = 0.15$, $P(C) = 0.5 * 0.5=0.25$ and $P(D)= 0.5 * 0.5 = 0.25$ then we have the probability of the sample space is greater than $1$. What mistake did I make here? – Oliver G Feb 20 '18 at 4:06

Since $A$ has probability $0.7$ of beating all teams, it doesn't matter who wins in the $C$ vs. $D$ match. We have $$P(A\text{ wins})= P(A\text{ wins twice})=0.7\cdot0.7=0.49$$ Similarly, since $B$ has probability $0.5$ of beating both $C$ and $D$, we can write: $$P(B\text{ wins}) = P(B\text{ beats }A)P(B\text{ wins next game}) = 0.3 \cdot 0.5 = 0.15$$ By symmetry, the probability that $C$ wins is the same as the probability that $D$ wins. We have: \begin{align} P(C\text{ wins}) &= P(C\text{ beats }D)\left[P(A\text{ beats }B)P(C\text{ beats }A) + P(B\text{ beats }A)P(C\text{ beats }B)\right]\\ &=0.5(0.7\cdot0.3+0.3\cdot0.5) \\ &=0.5\cdot 0.36 \\ &= 0.18\\ &=P(D\text{ wins}) \end{align} An easier way to get $P(C\text{ wins})$ is to divide the remaining probability $1 - (0.49 + 0.15) = 0.36$ by $2$, since $C$ and $D$ have the same chance of winning the tournament.
• What is the sample space you're using here? I don't understand what you mean by "$A$ wins twice". I was assuming the sample space had two ways to win for each team as in $A$ wins by beating $B$ and $C$ or $A$ wins by beating $B$ and $D$. – Oliver G Feb 20 '18 at 4:18
• @OliverG $A$ wins the first game with probability $0.7$, and $A$ wins the second game with probability $0.7$ as well, regardless of whether $C$ won or $D$ won in the $C$ vs $D$ game because $A$ wins with probability $0.7$ against everyone. "$A$ wins twice" means that $A$ wins in both the first and second round of the tournament. – bames Feb 20 '18 at 4:19
• @OliverG It is the same sample space as yours. You simply calculated the probabilities incorrectly, as JMoravitz pointed out. The probability that $C$ wins the second game is not $0.5$, since the probability of $C$ winning against $A$ ($0.3$) is different from the probability that $C$ wins against $B$ ($0.5$)... – bames Feb 20 '18 at 4:23
• @OliverG nothing is wrong with the samplespace either person chose, but you are ignoring the chance that $A$ gets the chance to play against $C$ in the first place in your calculations. If you want, you could write it like barnes did, but it gets tedious to parse: $A$ wins tournament $=$($A$ beats $B$ in rd1) and ((($A$ beats $C$ in rd2) and ($C$ beats $D$ in rd1)) or (($A$ beats $D$ in rd2) and ($D$ beats $C$ in rd1))), yielding the calculation $0.7\cdot((0.7\cdot 0.5)+(0.7\cdot 0.5))=0.49$ – JMoravitz Feb 20 '18 at 4:26