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Three players $A,B,C$ play tennis matches. There is always one player waiting to face the winner of the match between the other two. For a given match both players have the same probability of winning. The tournament ends whenever a player wins two consecutive matches. What is the probability of the tournament never ending? Do all players have same chance of winning?

Intuitively I would say that the probability of the match going on forever is $0$. How can I show this formally. For the tournament to keep on going we need that player that wins first match loses the second one, the player that wins second match loses the third and so on. How can I show that the probability of the intersection of these events is $0$.

For the second question the players that play the first game clearly have the same probability of winning the tournament, however I don't know whether the one that doesn't play the first match has the same probability of winning the tournament than the other two.

Peter Sampodiras
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  • Assume P-1 plays P-2 initially, with P-3 watching. Let $~A~$ denote the probability of P-1 winning the tournament, assuming that P-1 wins the first match. Let $~B~$ denote the probability of P-1 winning the tournament, assuming that P-1 loses the first match. Then P-1's overall chances are $~\dfrac{A+B}{2}.~$ Assume that P-1 wins the first match and now plays P-3 in the 2nd match. If P-3 does not immediately win the tournament is over. If he does win, then he is in the exact same position that P-1 was in just after P-1 won the first match. – user2661923 Dec 13 '23 at 23:22
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    after the first game, the series ends with probability $\frac 12$ on each successive round, so the odds of the series running $n$ games (not counting the first) is $\frac 1{2^n}$. Now let $n\to \infty$. – lulu Dec 13 '23 at 23:33
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    This is equivalent to flipping a fair coin and never getting tails. – Daniel Mathias Dec 13 '23 at 23:35
  • @lulu You mean the probability of the tournament ending in round $n$ is $\frac{1}{2^n}$ after the first one? – Peter Sampodiras Dec 14 '23 at 02:12
  • No...I meant what I wrote. But since each game is a $50-50$ business, the answers would be the same. – lulu Dec 14 '23 at 05:29
  • TLDR; ... BUT ... if victory involves an $x : (x + 1)$ format, an even number of wins for both sides means the game will ... go on ... forever. This takes the shape of winning alternate games, which from $1$ player's drishti looks like WLWLWLWL .... If the probability of winning for both players is 0.5, the chances of ananata is $0.5 \times 0.5 \times 0.5 \times 0.5 \times ...$ and of the game ending on the third match is $0.5 \times 0.5 \times 0.5$, basicially $0.5^{n}$, where $n$ is odd. – Agent Smith Dec 14 '23 at 13:21
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    I read your name as "pete sampras", probably because of the tennis games theme – Rei Henigman Dec 14 '23 at 21:36

1 Answers1

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The probability that the tournament runs past the round $n$ is $p_n=\frac1{2^{n-1}}$. This is seen from the fact that the game ends if the player who won the previous round wins again (with probability $\frac12$). Since $\lim_{n\to\infty}p_n=0$, the game will certainly end.

The probability that the third player wins the tournament is less than that of his opponents, since they can win the tournament even if they loose their first match. Contrary, the third player has to win his first match to remain in the game.

user
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