# Probability that team $A$ has more points than team $B$

Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

I got that since, Team $B$ already has one loss, it doesnt matter how many games team $B$ wins.

We must find the probability that team $A$ wins the rest $5$ games.

Since it says: "outcome of games is independent, I am confused."

My first approach was:

$$P(\text{A Wins 5}) = \frac{1}{32} \implies m + n = 33$$

This was wrong.

Second approach.

Suppose $A$ has a match with Team $C$.

$$P(\text{Team A wins, Team C loses}) = \frac{1}{2}\frac{1}{2} = \frac{1}{4}$$

But then overall: $\frac{1}{4^5} > 1000$ too big of an answer ($m + n < 1000$ requirement).

EDIT: I did some casework and the work is very messy and I don't think I got the right answer anyway. I have to find:

$$\binom{5}{1} = 5, \binom{5}{2} = 10, \binom{5}{3} = 10, \binom{5}{4} = 5, \binom{5}{5} = 1.$$

Let $A = x$

$$P(B=1, x \ge 1) + P(B = 2, x\ge 2) + P(B=3, x\ge 3) + ... + P(B=5, x = 5)$$

$$P(B=1, x \ge 1) = \binom{5}{1}(0.5)^{5} \cdot \bigg(\binom{5}{1} (0.5)^5 + \binom{5}{2} (0.5)^5 + ... + (0.5)^5\binom{5}{1} \bigg) = \frac{5}{1024} \cdot \bigg(31\bigg) = \frac{155}{1024}$$

$$P_2 = \frac{10}{1024} \bigg(\binom{5}{2} + ... + \binom{5}{5}\bigg) = \frac{260}{1024}$$

$$P_3 = \frac{10}{1024} \bigg(\binom{5}{3} + ... + \binom{5}{5} \bigg) = \frac{160}{1024}$$

$$P_4 = \frac{5}{1024} \bigg( \binom{5}{4} + \binom{5}{5}\bigg) = \frac{60}{1024}$$

$$P_5 = \frac{1}{1024} \bigg( 1\bigg) = \frac{1}{1024}$$

$$P(\text{Total}) = \frac{636}{1024} = \frac{318}{512} = \text{wrong}$$

What is wrong with this method?

• Hint: Not sure why you are so quick to dismiss poor B! There are still a lot more games to play and B only has a single loss. If millions of games were yet to be played this one game would have no significance.
– lulu
Jul 12, 2015 at 21:00
• @lulu, but each team only plays another team once. Jul 12, 2015 at 21:01
• Sure. But even so, B might win all her other matches while A might never win again.
– lulu
Jul 12, 2015 at 21:02
• @lulu, yes, so that is why I was finding the probability that $A$ wins ALL other games, so $B$ has no chance. Jul 12, 2015 at 21:04
• @Amad27 but that is just one possible outcome. You need to consider all the outcomes. There are many outcomes in which B outperforms A, for example.
– lulu
Jul 12, 2015 at 21:06

You can actually get away with computing only one binomial coefficient; I'll leave out one detail so this qualifies as a hint.

As @GrahamKemp observes, both $X$ and $Y$ are distributed as $\mathcal{Bin}(5,1/2)$; we want the probability that $X\ge Y$, i.e. that $Z:=X-Y\ge 0$. But $Z+5$ is distributed as $\mathcal{Bin}(10,1/2)$ (why?) From this you can compute $p=\textrm{Pr}(X=Y)=\textrm{Pr}(Z=0)$. Finally, by symmetry, we have $\textrm{Pr}(X<Y)=\textrm{Pr}(X>Y)$, so both of these are equal to $(1-p)/2$. It follows that $$\textrm{Pr}(X\ge Y)=(1+p)/2.$$

Incidentally, I got $m+n=831.$

$A$ and $B$ have played one game against each pther, and $B$ lpst.

Bpth teams have each to play against the remaining five teams (independently).   Let $X$ be the count of those games $A$ wins (ie: the points), and let $Y$ be the count of games $B$ wins.   Since $A$ has at least one point from the game between the two teams, you wish to find: $\;\mathsf P(X+1>Y)\;$.

Independence means that how many and which teams $A$ wins against has no influence on the probabilities of $B$ winning against any team.

Hint: The random variables $X$, $Y$ have iid binomial distributions, parameters $p=0.5, n=5$. $$X,Y\mathop{\sim}^{iid}\mathcal{Bin}(5, 0.5)$$

\begin{align} m/n & = \mathsf P(X+1> Y) \\[1ex] & = \sum_{x=0}^5 \mathsf P(X=x \cap Y< x+1) \\[1ex] & = \sum_{x=0}^5 \mathsf P(X=x) \sum_{y=0}^{x}\mathsf P(Y=y) \end{align}

The rest is left to you.

• (+1) great answer although I have no clue what the notation is. But I'll try to understand: (1) What does $P(X = x \cap Y < x + 1)$ mean written out? Jul 13, 2015 at 11:35
• Probability that X has x wins and Y has less than x+1 wins, P() is probabilty, ∩ is "and", X and Y are explained in the beginning of this answer Jul 13, 2015 at 15:22
• Exactly as Inuyaki said, @Amad27 Jul 13, 2015 at 22:43
• @GrahamKemp, how are the probabilities symmetric though? I think because I have no clue of notation Im having this problem. Jul 14, 2015 at 12:17

Let us compute some figures so that you can confirm answer !

Firstly, notice that when p = q = 1/2, the binomial distribution formula simplifies to P(X) = $\dfrac{n\choose X}{2^n}$

To simply computations, we can leave the division by $2^n$ till the end.

n(A wins) = $n[A = X]\cdot n[B\le X] =1\cdot1 + 5\cdot6 + 10\cdot16 + 10\cdot26 + 5\cdot31 + 1\cdot32$= 638

and P(A wins) = 638/1024 = 319/512

Finally, m+n = 831

Hints:

1. Team A has $5$ more games to play, and so does team B.
2. The outcome of games is independent. Thus, for team A or B, the probability that the team wins exactly $k$ more games (after their game with each other) has binomial distribution, namely

$$P(\text{exactly k more games won})={5 \choose k}\left(\frac 12\right)^5$$

1. There are really four main ways that B can finish with more points than A: (a) A wins $0$ more games and B wins $2$ through $5$ more games; (b) A wins $1$ more game and B wins $3$ through $5$ more games; (c) A wins $2$ more games and B wins $4$ through $5$ more games; (d) A wins $3$ more games and B wins $5$ more games.
2. The probability of (a) happening is

$$P(\text{A wins 0 more games}) \times [P(\text{B wins 2 more games}) + P(\text{B wins 3 more games}) + P(\text{B wins 4 more games}) + P(\text{B wins 5 more games})]$$

1. Possibilities (a) through (d) are mutually exclusive, so you can just add their probabilities.

I'm sure you get the idea. The final calculation is made easier by the fact that $\left(\frac 12\right)^5$ factors out nicely. Calculating everything else has $6$ additions, $4$ multiplications, $3$ more additions, and $1$ more multiplication. This may be more calculating than you would like, but it does get the job done in a reasonable amount of time. Some calculators, like the TI-84 and TI-Nspire, can add ranges of binomial probabilities together all at once, which makes the calculations easier.

Of course, when you find the final probability, finding $m+n$ is easy.