# Probability of winning a tournament by winning all matches in tournament

Ques: Two players are competing in a tournament which consists of three matches. The probability of player1 winning the first match is 0.2, winning the second match is 0.5 and winning the third match is 0.6. The probability of player2 winning the first match is 0.8, winning the second match is 0.5 and winning the third match is 0.4.

A player wins the tournament is he wins all the matches. Otherwise, the tournament is played again. The tournament is played again and again until a player wins all matches and hence wins the tournament.

What is the probability that player1 wins the tournament?

My approach: Since the matches are independent of each other, the probability of player1 winning all the matches is 0.2*0.5*0.6. There are 7 other possible outcomes of tournaments:-

0.2*0.5*(1-0.6) i.e 0.2*0.5*0.4

0.2*(1-0.5)*0.6 i.e 0.2*0.5*0.4

... and so on.

I'm am doubtful as to how to progress from here. Any help would be appreciated.

• Your terminology is very confusing: You are using "tournament" for two different things, a combination of three matches, and a series of tournaments in the first sense. In particular, your players can not win the tournament, and yet win the tournament … because the "tournaments" in both parts of the sentence are different things. – celtschk Jul 12 '15 at 7:12

The probability Player 1 wins all three matches is $a=0.06$. The probability for Player 2 is $b=0.16$. The probability neither wins all three matches is $t=0.78$.

Player 1 wins the tournament if she wins in the first round (probability $a$) or if neither wins in the first round and Player 1 wins in the second round (probability $ta$), or if neither wins in the first two rounds but 1 wins in the third round (probability $t^2a$) and so on. So the probability Player 1 ultimately wins is $a+ta+t^2a+t^3a+\cdots$. The sum of this infinite geometric series is $\frac{a}{1-t}$. Similarly, the probability Player 2 ultimately wins is $\frac{b}{1-t}$.

I believe the simplest way is to ignore altogether rounds where neither win.

P(A wins in a round) = $0.2\cdot0.5\cdot0.6 = 0.06$

P(B wins in a round) = $0.8\cdot0.5\cdot0.4 = 0.16$

Odds in favour of A = 6/16 = 3/8, so

P(A wins) = 3/11

P(B wins) = 8/11

Be $p_k$ the probability that player $k$ wins a single tournament. Be $P$ the probability that player 1 wins the tournament series (the probability that player 2 wins the tournament series is, of course, $1-P$).

Now player 1 has two ways to win the tournament series: Either he wins the first tournament, or the first tournament is a draw, but he wins a later tournament. Those two cases are mutually exclusive, therefore the probabilities add up.

The probability of player winning the first tournament, we know, it's $p_1$. The probability of the first tournament being a draw is $1-p_1-p_2$. But after a draw, the situation is exactly the same as at the start of the tournament series (past tournaments don't influence the winning rules of future tournaments), therefore the probability of the first player to win the tournament is exactly the same as it was before the tournament series started. So we have $$P = p_1 + (1-p_1-p_2)P$$ This equation is easily solved for $P$, giving $$P = \frac{p_1}{p_1+p_2}$$ This is actually a quite intuitive result: The relative probability for each player to win is unchanged, only the total probability that one of the players wins is scaled up to $1$.

Now all that's left is to calculate $p_1$ and $p_2$, the probabilities for each player to win a single tournament. But you already know how to do that.

There's a much simpler way of doing this problem. The key observation in this problem is that there are essentially seven distinct states:

1. The tournament just started or restarted; i.e. both players have won 0 games in a row. Call this state $S0$.
2. Player 1 has won the first game. Call this state $L1$. Notice now that if he loses the second game, the state reverts to $S0$.
3. Player 1 has won the first two games. Call this state $L2$.
4. Player 1 has won first three games. Call this $L3$. Note that this means he's won the tournament.
5. Player 2 won the first game. $R1$.

6. Player 2 won the first two games. $R2$.

7. Player 2 won the first three games. $R3$.

Now, each state can change to another adjacent state based on some probability; e.g $S0$ goes to $L1$, with probability $0.2$, and goes $R1$ is $0.8$. Define a function $P(s)$, where $s$ is one of the states I've mentioned. $P(s)$ gives the probability that player 1 will win the tournament given that the tournament is in stage $s$. What you're asking for now is the value of $P(S0)$. You already know the value of $P(L3)=1$ and $P(R3)=0$. It's easy now to determine the other probabilities. Refer to this blog post for details on this method.