3
$\begingroup$

The winner of a tennis tie break is the first to get to 7 points and lead by 2.

Let $p$ be the probability of player 1 winning when serving, and let $m$ be the probabiliity of player 1 winning when receiving the serve.

Every time the score sums to an odd number, the server changes.

What is the probability that player 1 will win the entire tie-break?

$\endgroup$
2
  • $\begingroup$ Player 1 serves first? $\endgroup$ Jan 31 '15 at 9:56
  • $\begingroup$ @barakmanos, it shouldn't matter who serves first since the winner must lead by 2. $\endgroup$
    – Kenshin
    Jan 31 '15 at 10:51
5
$\begingroup$

The ways for Player $A$ to win the tiebreaker are with scores: $7-0,\;\;$ $7-1,\;\;$ $7-2,\;\;$ $7-3,\;\;$ $7-4,\;\;$ $7-5,\;$ and then after reaching $6-6$ gaining an advantage of $2$ points.

We assume Player $A$ serves first. Then the server of each ball will be:

\begin{matrix} 1&2&3&4&5&6&7&8&9&10&11&12&\cdots \\ A&B&B&A&A&B&B&A&A&B&B&A&\cdots \\ \end{matrix}

We'll look at the probability of each scoreline separately. For example, $7-4$. This has $11$ points, the last of which Player $B$ serves and Player $A$ wins. Player $A$ wins $6$ of the other $10$ points and this can be done in the following ways:

\begin{matrix} \text{$A$ wins $5$ of $A$'s serves and $1$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{5}\binom{5}{1}p^5m^2(1-m)^4 \\ \text{$A$ wins $4$ of $A$'s serves and $2$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{4}\binom{5}{2}p^4(1-p)m^3(1-m)^3 \\ \text{$A$ wins $3$ of $A$'s serves and $3$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{3}\binom{5}{3}p^3(1-p)^2m^4(1-m)^2 \\ \text{$A$ wins $2$ of $A$'s serves and $4$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{2}\binom{5}{4}p^2(1-p)^3m^5(1-m) \\ \text{$A$ wins $1$ of $A$'s serves and $5$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{1}\binom{5}{5}p(1-p)^4m^6. \end{matrix}

The probability of $A$ to win $7-4$ is the sum of those $5$ values.

A similar method is used to obtain the probability for other scorelines:

$$P(7-0) = \binom{3}{3}\binom{4}{4}p^3m^4 $$ $$\\$$ $$P(7-1) = \binom{3}{3}\binom{4}{3}p^4m^3(1-m) + \binom{3}{2}\binom{4}{4}p^3(1-p)m^4$$ $$\\$$ $$P(7-2) = \binom{4}{4}\binom{4}{2}p^5m^2(1-m)^2 + \binom{4}{3}\binom{4}{3}p^4(1-p)m^3(1-m) + \binom{4}{2}\binom{4}{4}p^3(1-p)^2m^4$$ $$\\$$ $$P(7-3) = \binom{5}{5}\binom{4}{1}p^5m^2(1-m)^3 + \binom{5}{4}\binom{4}{2}p^4(1-p)m^3(1-m)^2 + \binom{5}{3}\binom{4}{3}p^3(1-p)^2m^4(1-m) + \binom{5}{1}\binom{4}{4}p^2(1-p)^3m^5$$ $$\\$$ $$P(7-4) = \binom{5}{5}\binom{5}{1}p^5m^2(1-m)^4 + \binom{5}{4}\binom{5}{2}p^4(1-p)m^3(1-m)^3 + \binom{5}{3}\binom{5}{3}p^3(1-p)^2m^4(1-m)^2 + \binom{5}{2}\binom{5}{4}p^2(1-p)^3m^5(1-m) + \binom{5}{1}\binom{5}{5}p(1-p)^4m^6$$ $$\\$$ $$P(7-5) = \binom{5}{5}\binom{6}{1}p^5m^2(1-m)^5 + \binom{5}{4}\binom{6}{2}p^4(1-p)m^3(1-m)^4 + \binom{5}{3}\binom{6}{3}p^3(1-p)^2m^4(1-m)^3 + \binom{5}{2}\binom{6}{4}p^2(1-p)^3m^5(1-m)^2 + \binom{5}{1}\binom{6}{5}p(1-p)^4m^6(1-m) + \binom{5}{0}\binom{6}{6}(1-p)^5m^7$$ $$\\$$

To reach $6-6$, there is no restriction on who wins the $12^{th}$ point. So,

$$P(6-6) = \binom{6}{6}\binom{6}{0}p^6(1-m)^6 + \binom{6}{5}\binom{6}{1}p^5(1-p)m(1-m)^5 + \binom{6}{4}\binom{6}{2}p^4(1-p)^2m^2(1-m)^4 + \binom{6}{3}\binom{6}{3}p^3(1-p)^3m^3(1-m)^3 + \binom{6}{2}\binom{6}{4}p^2(1-p)^4m^4(1-m)^2 + \binom{6}{1}\binom{6}{5}p(1-p)^5m^5(1-m) + \binom{6}{0}\binom{6}{6}(1-p)^6m^6$$ $$\\$$

Let $P_D$ be the probability of Player $A$ winning from $6-6$. Then we have,

\begin{eqnarray*} P_D &=& pm + (p(1-m) + (1-p)m)P_D \\ && \\ \therefore P_D &=& \dfrac{pm}{1 - p(1-m) + (1-p)m} \\ && \\ &=& \dfrac{pm}{1 - p - m + 2pm} \\ \end{eqnarray*}

Therefore, the probability of $A$ winning the tiebreaker is obtained by gathering all the components above:

$$P(\text{$A$ wins}) = P(7-0) + P(7-1) + P(7-2) + P(7-3) + P(7-4) + P(7-5) + P(6-6)P_D.$$

$\endgroup$
2
  • 1
    $\begingroup$ It would seem that you double-counted the scenario "7-5 after 5-5". This can be mitigated by passing to 6-6. $\endgroup$
    – Lord_Farin
    Jan 31 '15 at 9:58
  • $\begingroup$ @Lord_Farin yes, right, thanks. I'll fix. $\endgroup$
    – Mick A
    Jan 31 '15 at 10:02
0
$\begingroup$

Maybe one should try first to simplify the problem a little bit. First one could assume, that the probability of winning the point is $p$. So first assume $m=p$, so we don't distinguish between serving and returning. Than the probability of winning the tiebreak ($P_w$)can be expressed by Bernoulli.

So we get $P_w\leq {7 \choose 5}p^7(1-p)^5+{7 \choose 4}p^7(1-p)^4 +...+ {7 \choose 0}p^7(1-p)^0+ \sum\limits_{k=6}^{\infty} {n+2 \choose n}p^{n+2}(1-p)^n$. All these summands are a little too high, because it is important, that the game is finished if 7 points are reached. One could subtract these terms.

This is an first step of an ansatz. This would be good task to simulate. Maybe try to implement it to get an idea of how the behavior of the probability is.

$\endgroup$
1
  • $\begingroup$ Intuitively the probability is greater than one half if $p+m>1$ $\endgroup$
    – epsilon
    Jan 30 '15 at 22:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.