# Probability tree question

3 tennis players A,B and C play in a tennis tournament. The first match is between A and B. The winner plays with C and so on until one player wins 2 matches (overall, not necessarily in a row). All 3 players are equally strong.

What is the probability that A, B or C wins the tournament?

My problem is that the solution in my book says that these probabilites are equal, but with my calculations it came out, that C has a smaller chance to win the competion.

• What is this book? What are your calculations?
– Did
Sep 22, 2014 at 11:07
• It is written in my native language. Sep 22, 2014 at 15:21

Clearly $A$ and $B$ have the same chance of winning, and $C$ has whatever probability remains. So let's look at $A$.

There are unique sequences for $A$ to win in rounds $2$, $4$, $5$, $7$, ... (any nonmultiple of $3$, except for $1$). These sequences (of match winners) are:

$AA$

$BCAA$

$ACBAA$

$BCABCAA$

$ACBACBAA$

etc.

So the probability that $A$ wins in round $2,5,8,...$ is $\frac{1}{4}+\frac{1}{32}+\frac{1}{256}+\dots=\frac{\frac14}{\frac78}=\frac{2}{7}$

Similarly the probability that $A$ wins in round $4,7, 10,...$ is $\frac{1}{14}$.

Hence $A$'s overall chance of winning is $\frac{2}{7}+\frac{1}{14}=\frac{5}{14}$. $B$'s chance is the same, leaving $C$ with a $\frac{4}{14}$ chance of winning.

Define $P$ the chance a player will win the tournament when he gets a new opportunity to, playing against last game's victor. Now what can happen?

• If he loses, two games in a row are won by the same player, and the tournament is over , ie. $P_{L}=0$
• If he wins twice, he wins the tournament: $P_{WW} = \frac1{2}\frac1{2} = \frac1{4}$
• If he wins first, and loses his next game, he will have to hope for a new opportunity, ie. that the player he lost to will not win the next game. The chances that this happens are $P_{WLX} = \frac1{2}\frac1{2}\frac1{2} = \frac1{8}$. In this case he will get an other opportunity to win twice in a row.

$P$ thus becomes $\frac1{4} + \frac1{4}\frac1{8} + \frac1{4}\frac1{8}^2 + \ldots = \sum_{n=0}^{\infty} \frac1{4}\left(\frac1{8}\right)^n = \sum_{n=0}^{\infty} \frac1{4}\frac1{1-\frac1{8}} = \frac{2}{7}$ where the sum is that of a geometric serie. Since player C's first game is already against another game's victor, $P$ is also the chance player C will win the tournament. For player A and B, both equally likely, what remains is $\frac{1 - P}{2} = \frac{5}{14}$.

To doublecheck this result, one can check the chance for player A (or B) to win directly. This is the chance to win the first to games in row $\left(\frac1{4}\right)$ + the chance to win first, then lose, and then get another shot at winning $\left(\frac1{8}P\right)$ + the chance to lose first and get another opportunity to win $\left(\frac1{4}P\right)$, or in total $\frac1{4} + \frac{3}{8}P = \frac1{4} + \frac{3}{8}\frac{2}{7} = \frac{5}{14}$, as predicted.

NOTE The above derivation is done for the interpretation that a player has to win two times in a row. To win just 2 times overall, one can point out that player C's only hope is to win his first two games outright; winning only the first game will result in a game between A and B where both players have already one win (which you can check by drawing the probability tree). In this case, the chance for player C is $\frac1{2}\frac1{2}=\frac1{4}$, where A and B both share the remainder, ie. $\frac{3}{8}$.

That the chances are not equal was already pointed out, using the simple argument that in the second game (player C's first), there will be a player with already one win while C has none.

• Thanks, yes a player doesn't have to win 2 times in a row, I got $3/8$, $3/8$ and $1/4$ as well, so I guess that is an error in my book. Sep 22, 2014 at 15:15

It seems pretty obvious that C has a lesser chance, since one of his two opponents will already have a win under his belt before C even gets to play.

Does the book imply that the three players draw lots for order of play first - in which case the three players do have equal chance before the draw happens? In that case "A", "B" and "C" are not arbitrary labels but are assigned according to the draw (C being the one that gets the worst draw).

What's the chance that C wins?

In his first game he plays someone who has already won a game, either A or B. C must win that game to have any chance (50%). He then plays the player who lost that first game (either B or A). If C loses, then both A and B have one win, and they play each other in the next round. One of them will have two wins and win the tournament. So to have any chance, C must win his second match as well (chance = 25%). At that point C is the tournament winner, so his chance to win the tournament is exactly 25%.

(Obviously that's answering the question as it was asked; "win two games in a row" would likely have a different answer).