Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question A and B play until one has 2 more points than the other. Assuming that each point is independently won by A with probability p, what is the probability they will play a total of 2n points? What is the probability that A will win?

My attempt for the solution:

what is the probability they will play a total of 2n points?

For the first question, A and B will be "exchanging wins" until $|A-B| = 2$, if one wins twice in a row then its terminal state. Also, winning guarantees even total points,

$A = B + 2\\ A+B=B+B+2=2B+2$

A win always means even points then, making the job simpler. Probability prior to double winning streak shown in the following,

Let $E_{i}$ be a set of game from start till achieving $2*i$ points.

$P(E_{1}, E_{2}, \cdots, E_{n-1}) = P(E_{1}) + P(E_{2}) + \cdots + P(E_{n-1})\\ P(E_{1}, E_{2}, \cdots, E_{n-1}) = 1 + p*(p-1) + \cdots + (p*(p-1))^{2i-2}\\ P(E_{1}, E_{2}, \cdots, E_{n-1}) = \frac{1}{1-(p*(p-1))^{2}} - (p*(p-1))^{2i-1} - (p*(p-1))^{2i}$

*Note that $P(E_{1}) = P(\emptyset)$ since this is not the two winning streak.

The winning streaks can be a union of either $A$ wins twice in a row or $B$ wins twice in a row:

Let $S$ be the probability of 2n final points.

$P(S) = P(E_{1}, E_{2}, \cdots, E_{n-1})*p^{2} + P(E_{1}, E_{2}, \cdots, E_{n-1})*(p-1)^{2}$


What is the probability that A will win? I didn't really understand this question, does it mean from a tie points to a double winning streak of A? Anyway, in that case, I just use the result from first question,

$P(A_{win}) = P(E_{1}, E_{2}, \cdots, E_{n-1})*p^{2}$

share|improve this question

2 Answers 2

up vote 3 down vote accepted

The probability $a$ that A (ultimately) wins is easy to compute. She wins if she wins the first two games, or if the players are tied after $2$ games, but A ultimately wins. Thus, conditioning on the outcome of the first two games, we have $$a=p^2+2p(1-p)a.$$ Solve this linear equation for $a$.

For the probability the game lasts for a total of $2n$ points, we need that there is a tie at $2n-2$ points, and no one has won by then, and then one of the two players gets $2$ in a row.

For tie at $2n-2$ points, with no one having won, we need $n-1$ occurrences of the pattern AB or BA (we can "mix" these). This sort of pattern has probability $2p(1-p)$, so the required probability is $$[2p(1-p)]^{n-1}(p^2+(1-p)^2).$$

Remark: We can alternately use the formula for winning in $2n$ games to find A's probability of winning. Her probability of winning in $2$ is $p^2$. Her probability of winning in $4$ is $(2p-2p^2)p^2$. Her probability of winning in $6$ is $(2p-2p^2)^2p^2$. And so on. So her probability of winning is $$p^2+p^2(2p-2p^2)+p^2(2p-2p^2)^2+p^2(2p-2p^2)^3+\cdots.$$ This is an infinite geometric series, first term $p^2$, common ratio $2p-2p^2$. Now use the formula for the sum of such a series.

More complicated for sure than the simple linear equation approach we gave in the answer!

share|improve this answer
    
I'll digest this for a minute or 2, I just hope your still online then so I can clarify stuff. –  JoeyAndres Jun 28 at 5:21
    
I will be around for a while, I have made the argument somewhat compact. –  André Nicolas Jun 28 at 5:23
    
I got your answer for the first question, may I ask for clarification on the answer on the second question? What is "She wins if she wins first two games, or if the players are tied after 2 games". –  JoeyAndres Jun 28 at 5:29
    
Player A wins if she wins the first two games. Clear. She also wins if the first two games end AB or BA, but A ultimately manages to win. Given they are tied after the first two games, A's probability of winning is exactly the same as her initial probability of winning, hence the result. I can make an explicit conditional probability calculation if you wish. I will also add a remark giving a different way of solving the A wins problem. –  André Nicolas Jun 28 at 5:37
    
I have difficulty with the 2nd question, I get that $p_{2}$ if the player wins the first two games. I get that $2*p*(1-p)$ is to account for the pattern AB or BA, but why $2*p*(1-p)*a$? How do you literally translate $2*p*(1-p)*a$ in words? –  JoeyAndres Jun 28 at 5:37

If the players are currently tied, and then one wins twice in a row, that player wins and the game ends. If one player is currently leading, then the other player has to win the next game to reach a tie. Therefore, in order for the result of the first $2n - 2$ games to be a tie, the wins have to be in the form of $n - 1$ pairs, each of the form $AB$ or $BA$ (for example: $ABBABAAB$). Thus, the wins are in the pattern $\underbrace{(P_1)(P_2)\ldots(P_n)}_{n-1}$, where each pair $P_i$ is either $AB$ or $BA$ (independently of the other pairs). If $q = 1 - p$, each of the pairs $AB$ and $BA$ has probability $pq$ (for $qp = pq$). As there are two possibilities for each of the $n - 1$ pairs $P_i$, there are $2^{n-1}$ possible configurations. Thus the probability of a tie after $2n - 2$ games is $2^{n-1}(pq)^{n-1}$.

Now, the last two wins can be either $AA$ with probability $p^2$, or $BB$ with probability $q^2$. So the probability of there being exactly $n$ games is $\boxed{2^{n-1}(pq)^{n-1}(p^2 + q^2)}$.

If $A$ wins, then the only possibility for the last two wins is $AA$, so the probability that $A$ wins in $2n$ games is $2^{n-1}(pq)^{n-1}p^2$.

The probability that $A$ wins after an indefinite number of games can be obtained by summing the probabilities for $n = 1, 2, \ldots$

$\displaystyle\sum\limits_{n=1}^{\infty}(2pq)^{n-1}p^2 = \boxed{\dfrac{p^2}{1 - 2pq}}$.

share|improve this answer
    
@AndréNicolas Now we agree! Thanks for pointing out the mistake. –  M. Vinay Jun 28 at 5:55
1  
@JoeyAndres Note that the probability that $B$ wins is, by symmetry, $\dfrac{q^2}{1 - 2pq}$. And either $A$ or $B$ must ultimately win (the probability of a tie approaches $0$ as the number of games approaches infinity - note that $2pq < 1$). So the total $\dfrac{p^2 + q^2}{1 - 2pq}$ should be $1$. Check whether it is so. –  M. Vinay Jun 28 at 6:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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