# Probability Game: Find probability she wins the game on nth point played for n=4,5,6

Two people play a game. They play a series of points, each producing a winner and a loser, until one player has won at least 4 points and has won at least 2 more points than the other. Anne wins each point with probability p. What is the probability that she wins the game on the kth point played for k=4,5,6,...

Ok so I started with k=4 P(win on 4th point) can happen 3 ways 4-0,4-1,4-2 =p^4(1 + (4 choose 1)q +(5 choose 2)q^2), which is a geometric series for q. I'm not sure what to do with the changing binomial coefficients though. Similarly, P(win on 5th point can happen 4 ways) 5-0, 5-1, 5-2, 5-3 =p^5(1 + (5 choose 1)q + (6 choose 2) q^2 +(7 choose 3)q^3)

So it seems like we will have two geometric series one of powers of p, each containing one of powers of q. I'm not sure the exact pattern and how to formalize this. Thank you!

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You can right-click on any formula (e.g. in my answer) and select "Show Source" to see how you can format questions using $\TeX$ to make them more readable. – joriki Oct 13 '11 at 5:13
If the game ended with a score 5-3, it ended on the 8th point played, did it not? – Jyrki Lahtonen Oct 13 '11 at 12:06
To add to Jyrki Lahtonen's comment, "P(win on 5th point) can happen 4 ways) 5-0, 5-1, 5-2, 5-3" (meaning Anne wins when she wins her 5th point) is incorrect. Anne's $5$-th (winning) point is necessarily the last point of the game. Thus the game cannot end with a score of $5$-$x$, $x = 0, 1, 2$ (game would have ended as soon as Anne had won $4$ points earlier in the game). In short, the event "Anne wins when she has scored $5$ points" occurs exactly when the game ends with a score of $5$-$3$, and this has probability $\binom{7}{3}p^5q^3$ as you correctly computed. – Dilip Sarwate Oct 13 '11 at 12:35

I am wondering if the interpretation used by caligirl11 and @joriki is the one that was intended by whomsoever devised the problem.

What is the probability that she wins the game on the kth point played for k=4,5,6,...

I suggest that the $k$-th point played is not the same as the $k$-th point won by Anne.

Ok so I started with k=4 P(win on 4th point) can happen 3 ways 4-0,4-1,4-2 =p^4(1 + (4 choose 1)q +(5 choose 2)q^2)

I would say here that Anne wins the game on the $4$th point played with probability $p^4$; the probability that she wins on the $5$th point played is $\binom{4}{1}p^4q$; and the probability that she wins on the $6$th point played is $\binom{5}{2}p^4q^2$. The corresponding win probabilities for Betty are obtained by interchanging $p$ and $q$ in the previous sentence.

If $6$ points have been played and the game is not over, the score must be $3$-$3$ (deuce) and this event has probability $\binom{6}{3}p^3q^3 = 20p^3q^3$. The game can now end only when an even number of additional points have been played, and the probability of Anne winning on the $k$-th point played (where $k = 6+2n, n>0$) is $\binom{6}{3}p^3q^3\cdot p^2(2pq)^{n-1}$. To get the probability that Anne wins the game, sum the probabilities that she wins in $4$, $5$, $6$, $8$, $10, \ldots$ points; a geometric series is involved.

If $f(p)$ denotes the probability that Anne wins the game, then $f(0.5) = 0.5$ but $\left . \frac{d}{dp}f(p)\right\vert_{p = 0.5} = 2.5$ so that $f(0.5 + \epsilon) = 0.5 + 2.5\epsilon + \cdots$, that is, a small difference $p-q = 2\epsilon$ in the point win probabilities for Anne and Betty is amplified by the rules of tennis into a greater difference in the game win probabilities. See "The Drunken Tennis Player" in Ian Stewart's Game, Set, and Math, Penguin Books, 1981.

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This would have been my interpretation, too. – Jyrki Lahtonen Oct 13 '11 at 12:09
That is also my interpretation. – Michael Lugo Oct 13 '11 at 15:58

I think you've already got the right result for $k=4$. For higher $k$, note that the game has to go through a tie in order to continue. For instance, for someone to win 5:3, the score must have been 3:3 (since it must have been 4:3, and 4:2 would have ended the game). So you can calculate the probability for 3:3, which is $\binom63p^3q^3$, and from then on the probability is $p^2$ for Anne to win and $2pq$ (two ways for one win and one lose) to reach the next tie, so for $k\ge5$,

$$p_k=\binom63p^3q^3p^2\left(2pq\right)^{k-5}=\frac5{8q^2}(2pq)^k\;.$$

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I don't think there is -- I think you'll just have to add the two -- you can simplify the result a bit, though (under "Alternate forms"). – joriki Oct 13 '11 at 5:45
Here's something sort of nice: If you express this symmetrically by setting $p=(1+x)/2$, so $x=0$ at equal probability, you get this, which can be written as $$\frac{(x+1)^4-(x^2-1)^4}{8x(x^2+1)}\;.$$ Don't know whether there's any deeper meaning to that... – joriki Oct 13 '11 at 5:52