Volleyball Probability

It is possible to express the sequence of points won as a chain of As and Bs, e.g. AABBA. We're only interested in there being 3 As to signify that A has won the game, and that means then we are only interested in the chains ending in A.

We can realise that the longest possible chain has a length of 7, because only a maximum of 4 Bs can be used or else B will have won. Since we know the last letter must be an A, then we only need to work out the number of different arrangements for the first letters.

We can also note that the shortest chain is of length 3 (3 As in a row) and with that we can start to count all the possible chains.

For a length 7 chain, there are $\frac {6!}{4!*2!} =15$ different chains (Divided by $2!$ and $4!$ account for the repeat of the As and Bs.

For chains of length 6, there are $\frac {5!}{3!*2!} = 10$ different arrangements.

For chains of length 5, there are $\frac {4!}{2!*2!} = 6$ different arrangements.

For chains of length 4, there are $\frac {3!}{2!} = 3$ different arrangements .

And for chains of length 3, there are $\frac {2!}{2!} = 1$ different arrangements.

If we let $k_{n} =$ the number of arrangements for a chain of length $n$ then the probability of getting that chain is $k_{n} (\frac {1}{2})^{n}$

This means that the probability of Team A winning the game, which is equal to the sum of the probability of the chains of lengths 3 to 7 occurring, is $15 (\frac {1}{2})^{7} + 10 (\frac {1}{2})^{6} + 6 (\frac {1}{2})^{5} + 3(\frac {1}{2})^{4} + (\frac {1}{2})^{3} = \frac {99}{128}$

From this we see that $a = 99$ and $b =128$ Therefore, the answer to $a+b = 99+128 = 227$

Team A needs three more points won, and they could lose at most 4 points, so, for A to win there are 5 cases (according to the number of team A points won and total of disputed points (after the currently score)).

Let “W” be “won point by team A” and “L” be “lose point by team A” It´s clear that the last disputed point in each case has to be won by team A, else we would have less disputed points (contradiction).

The number of ways Team A would win the game in each case (given that the last point is won by team A) is given by the Permutation of (2)W´s and (k-3)L´s =(k-1)!/((k-3)!(2)!) ways, where k is the number of disputed points

• To win 3 points in 3 disputed points.

There is only 2!/2!=1 way in this case. Probability for this case 1(1/8) = 1/8

• To win 3 points in 4 disputed points.

There are 3!/(1!*2!)=3 ways in this case. Probability for this case 3(1/16) = 3/16

• To win 3 points in 5 disputed points.

There are 4!/(2!*2!)=6 ways in this case. Probability for this case 6(1/32) = 3/16

• To win 3 points in 6 disputed points.

There are 5!/(3!*2!)=10 ways in this case. Probability for this case 10(1/64) = 5/32

• To win 3 points in 7 disputed points.

There are 6!/(4!*2!)=15 ways in this case. Probability for this case (1/128) = 15/128

Finally we have to sum the probabilities obtained for each case.

It give us: 1/8 + 3/16 + 3/16 + 5/32 + 15/128 = 99/128 = a/b

So a+b = 227

A needs exactly 3 more points to win while B(other team )can have 1,2,3 or 4 points (that means other team's score can take value of 17,18,19 or 20) but the third point of A should be the last point. so 4 cases are possible-

CASE1. B got 0 point A should get 3 points so it is possible in 2!/2! ways [ fixing last point of A and permuting the rest of 2 points out of which both are of same type so in 2!/2! which is obviously 1] but probability of point going to A is 1/2 so for 3 points it is 1/8 since it is simultaneously happening so probability that A wins by score 21-16 is 1/8 similarly

CASE 2: A-3 points B-1 point probability of wining by 21-17 is (3!/2!)(1/16) [ fixing last point of A and permuting 2 points of A and 1 point of B which can be done in 3!/2! and probability of each point is 1/2 so for 4 points it is 1/16]

CASE 3: A-3 points B-2 points A wining by 21-18 probability that A will win is 4!/(2!2!) multiplied by 1/32

CASE 4 A-3 points B-3 points A wining by 21-19 probability that A will win is 5!/(2!3!) multiplied by 1/64

CASE 4: A-3 points B-4 points A wining by 21-20 probability that A will win is 6!/(2!4!) multiplied by 1/128

Since events are mutually exclusive adding up gives 99/128 and 99+128=227

In the end, Team A has to win 3 more points and Team B has to win 0,1,2,3, or 4 more points.

Additionally, Team A has to win the last point to win the game.

Let's split this into 5 cases, depending on what number of points Team B won. We will denote a score by Team A by "A", and a score by Team B by "B".

1. B wins 0 points: The only way this can happen is AAA. This means there is only one configuration out of a possible $2^3=8$ configurations (because there are two possible choices for a point, and there are three points), so the probability of this is $(\frac{1}{2})^3$ , or $\frac{1}{8}$ .

2. B wins 1 point: A must win the last point, and the first three points must be of the form BAA, in any order. There are ${3 \choose 1}$ ways to order BAA, so there are three configurations out of a possible $2^4=16$ . $\frac{3}{2^4}=\frac{3}{16}$ .

3. B wins 2 points: A must win the last point, and the first three points must be of the form BBAA, in any order. There are ${4\choose2}=6$ ways to order BBAA, so $\frac{6}{2^5}=\frac{3}{16}$ .

4. B wins 3 points: ${5\choose3}=10$ , $\frac{10}{2^6}=\frac{5}{32}$ .

5. B wins 4 points: ${6\choose4}=15$ , $\frac{15}{2^7}=\frac{15}{128}$ .

Adding all of these together, we get $\frac{1}{8}+\frac{3}{16}+\frac{3}{16}+\frac{5}{32}+\frac{15}{128}=\frac{99}{128}$ , which is in simplest form. So the answer is $99+128=\boxed{227}$ .

In order for A to win the game, they must win 3 games before B wins 5. The possible cases are: (1) A wins 3, B wins 0 (2) A wins 3, B wins 1 (3) A wins 3, B wins 2 (4) A wins 3, B wins 3, or (5) A wins 3, B wins 4

In each case, assume A wins the last game in order to make the cases distinct.

In any case in which A wins 3 games, B wins k games, with A winning the last game - there are $\binom{k+2}{2}$ ways for A to win 2 of the first k+2 games. Therefore the probability of A winning the series will be $\sum_{k=0}^4 \binom{k+2}{2} \left( \frac{1}{2} \right)^{3+k}$ .

This equals 99/128. So a + b = 227.

Notice that Team A needs 3 wins, and Team B needs 5. This means that for A to win, the total number of games cannot exceed 7. Once Team A has 3 wins, we ignore the "remaining" games as they would not be played.

Divide the problem into the probability that the number of games played is k for 3<=k<=7. Then the number of successes for Team B is k-3. To determine the probability of a given outcome, take (1/2)^k and multiply by (k-1)choose(k-3). To get the overall probability, add these together since they are disjoint outcomes. The sum is: (1/8)+(3/16)+(6/32)+(10/64)+(15/128)=(99/128)=a/b

Then a+b is simply 99+128=227.

The problem is equivalent to winning of 3 points. by A before B wins 5 points. This can happen if i)A wins consecutively 3 points= 1/2^3 ii)A wins 3 points and B wins 1 point= 3C1 (1/2)^4 iii)A wins 3 points & B wins 2 points= 4C2 (1/2)^5 iv)A wins 3 points & B wins 3 points=5C3 (1/2)^6 v)A wins 3 points & B wins 4 points=6C4 (1/2)^7 Thus, total probability= 99/128=a/b a+b=227

Solution 1: We will determine this result indirectly. If we assume that the teams play exactly 7 more points, then after these points, exactly one team will have at least 21 points. Team A will win if they score 3, 4, 5, 6, or 7 points, and team B will win if A scores 0, 1, or 2, points. The number of ways team B can win is $\binom{7}{0} + \binom{7}{1} + \binom{7}{2}$ , and the total number of possible games is $2^{7} = 128$ , so the probability that B win is $\frac{1 + 7 + 21}{128} = \frac{29}{128}$ , and the probability that A wins is $\frac{99}{128}$ . So $a + b = 99 + 128 = 227$ .

Solution 2: We determine the probability that team B wins the game. For team B to win, they must score 5 more points before team A scores 3. We consider 3 cases, either A scores 0, 1, or 2 points before B scores 5. There is 1 way for B to score 5 points and A to score 0, and there is a $\left(\frac{1}{2}\right)^{5} = \frac{1}{32}$ chance of this occurring. There are $\binom{5}{1} = 5$ ways for A to score 1 point before B scores 5 (it is not $\binom{6}{1}$ , since A needs to get its point before B gets its last point). The probability of each of these occurring is $\left(\frac{1}{2}\right)^{6}$ . There are $\binom{6}{2}$ ways for A to score 2 points before B scores 5 points, and the probability of each of these occurring is $\left(\frac{1}{2}\right)^{7}$ . So the probability that B wins is $\frac{1}{32} + \frac{5}{64} + \frac{15}{128} = \frac{29}{128}$ . So the probability that A wins is $1 - \frac{29}{128} = \frac{99}{128}$ . So $a + b = 99 + 128 = 227$ .

Solution 3: For team A to win, they must score 3 points before team B scores 5 points. We make a chart for the probability that team A wins and the game lasts exactly $k$ more points. The first column is the value of $k$ , the second column is the number of ways that this can occur, the third column is the probability that each of these happens, and the fourth column is the overall probability for all these ways (the second column multiplied by the third column).

$\begin{array}{|r|ccc|} \hline k & \mbox{ways} & \mbox{individual prob.} & \mbox{overall prob.}\\ \hline 3 & 1 & \frac{1}{8} & \frac{1}{8}\\ 4 & \binom{3}{1} & \frac{1}{16} & \frac{3}{16}\\ 5 & \binom{4}{2} & \frac{1}{32} & \frac{6}{32}\\ 6 & \binom{5}{3} & \frac{1}{64} & \frac{10}{64}\\ 7 & \binom{6}{4} & \frac{1}{128} & \frac{15}{128}\\ \hline \end{array}$

Thus the probability that A wins is $\frac{1}{8} + \frac{3}{16} + \frac{6}{32} + \frac{10}{64} + \frac{15}{128} = \frac{99}{128}$ . So $a + b = 99 + 128 = 227$ .

1/8 + 3/16 + 4!/2!*2!/32..... + = 99/128