# Probability that the tournament will produce neither an undefeated team nor a winless team?

In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integers. Find $m+n$.

The probability that one team wins all games is $5\cdot (\frac{1}{2})^4=\frac{5}{16}$.

Similarity, the probability that one team loses all games is $\frac{5}{16}$.

I did this much, but after that what should I do to reach the final answer ? I'm confused. Can someone explain?

• Is this an integer type question? Just curious... – SchrodingersCat Nov 28 '15 at 16:55
• yes....@Aniket.. – user220382 Nov 28 '15 at 16:56

Define events:

\begin{align} W &= \text{Some player wins $4$ times} \\ L &= \text{Some player loses $4$ times}. \end{align}

\begin{align} P(W) &= 5\left(\dfrac{1}{2}\right)^4 = \dfrac{5}{16} \qquad\text{Since any of the $5$ players can win $4$ games} \\ P(L) &= 5\left(\dfrac{1}{2}\right)^4 = \dfrac{5}{16} \qquad\text{Since any of the $5$ players can lose $4$ games} \\ P(W\cap L) &= 5\cdot 4\cdot\left(\dfrac{1}{2}\right)^7 = \dfrac{5}{32} \qquad\text{Since there are $5$ ways to pick the 4-winner,} \\ & \qquad\text{$4$ ways to pick the 4-loser, and $7$ games needing a particular result.} \\ \end{align}

We need probability:

\begin{align} P(W^c\cap L^c) &= 1 - P(W\cup L) \\ &= 1 - P(W) - P(L) + P(W\cap L) \\ &= 1 - \dfrac{5}{16} - \dfrac{5}{16} + \dfrac{5}{32} \\ &= \dfrac{17}{32}. \\ \therefore\quad m+n &= 49. \end{align}

• Why not 5 ways to pick the 4-loser?There are 5 teams right? – user220382 Nov 29 '15 at 5:10
• @SanchayanDutta Once you've picked the 4-winner there are only 4 players left. – Mick A Nov 29 '15 at 5:13
• Oh sorry i didnt notice.Thanks a lot!! – user220382 Nov 29 '15 at 5:14

This is a round robin situation, so there will be $\binom52 = 10$ games in total, so $10$ wins in total !

Permissible patterns of win distributions between teams are:

$3-3-2-1-1,\; 3-2-2-2-1,\;2-2-2-2-2\;$ and considering permutations,

$\dfrac{5!}{2!2!} + \dfrac{5!}{3!} + 1 = 51$ ways

Total possible results $= 2^{10} = 1024$

$Pr = \dfrac{51}{1024}$

$m+n = 1075$

• There will be more than 10 games...as each team plays with every other team once. – user220382 Nov 29 '15 at 5:05
• That's not possible, if each team plays only once with each other team. $\binom52 = 10.$ – true blue anil Nov 29 '15 at 7:28
• $2-2-2-2-2$ should count more than once and I suspect $24$ times (Which two does the first player beat? Which of those beats the other? And of the two the first player loses to, which beats the other?) Similarly with the others – Henry Oct 2 '18 at 21:57