How Many Four-Game Series?

A best out of five game means that one team must win three times. If we want this to occur in exactly four games, the team that wins 3 of the 4 games cannot win the first 3 games in a row (this would cause the series to end in only 3 games). Labeling the two teams p and q, we consider all options for team p winning the series. The first situation is that team p wins the first two games, loses the third, and wins the first. The probability of the p team winning the first game is simply $\frac{1}{2}$ . Since they won the previous game, the second game has a probability of $\frac {7}{10}$ . The probability of the third game being won by the q team is only $\frac {3}{10}$ . The probability that the p team comes back to win the fourth game is $\frac {3}{10}$ . Multiplying these together (because it takes into account all possible outcomes in the denominator and the specific outcome of the top) yields $\frac {63}{2000}$ . Continuing in this manner with team p winning games 1, 3, and 4 yields the probability $\frac{1}{2} \cdot \frac {3}{10} \cdot \frac {3}{10} \cdot \frac {7}{10}$ . The third, and final, possible way for team p to win in 4 games is to win games 2, 3, and 4. This yields the probability $\frac {1}{2} \cdot \frac {3}{10} \cdot \frac {7}{10} \cdot \frac {7}{10}$ . Once these values are added together, it can be noticed that the exact same probabilities are true for team q winning the series in 4 games. As both teams have the same conditions imposed upon them, they are interchangeable. We can then simply multiply the sum of the first three probabilities we determined by two. This results in the probability that the series lasts exactly 4 games equaling $\frac {273}{1000}$ . To get the final answer, we multiply this by 1000. This results in the greatest integer function of 273, which is just 273.

The series of 5 matches can be won by A or B(the two teams). 'A' can win the series in 4 matches in the following three ways. Case 1(E1) :'A' wins the first match,wins the second match, loses the third match and wins the fourth match. Case 2(E2):'A' wins the first match,loses the second match and wins the third and fourth matches. Case 3(E3):'A' loses the first match,wins the second ,third and fourth matches. According to the data given; Probability of E1=1/2 * 7/10 * 3/10 * 3/10 =(7 * 9)/2000=63/2000. Probability of E2=1/2 3/10 *3/10 * 7/10 =(9 * 7)/2000=63/2000. Probability of E3=1/2 * 3/10 * 7/10 * 7/10 =(49 * 3)/2000=147/2000. As the three events are not related to each other,the probability of A winning the series in 4 matches =P(E1) union P(E2) union P(E3) =P(E1)+P(E2)+P(E3)=(63+63+147)/2000=273/2000. Similarly, B can also win the series in 4 matches in the above stated ways. So, the probability of B winning the series in 4 matches is also 273/2000. So, the probability of the series being decided in 4 matches = P(A) union P(B) =P(A)+P(B)=(273+273)/2000=546/2000=273/1000;since the event A and B are not related i.e. their intersection is a null set. If 'p' is the probability that the series is decided in 4 matches then p=273/1000. So,the value of 1000p = (273/1000) 1000=273.

Consider all cases for 4 games for a team to win, namely WWLW, WLWW and LWWW. (win means a win for a team, L a loss).

The probability of the first happening is 0.5 0.7 0.3 0.3=0.0315. The 2nd case happening has probability 0.5 0.3 0.3 0.7=0.0315. The last case has 0.5 0.3 0.7 0.7=0.0735 probability happening. This gives a total probability of 0.1365. However, this can happen for each of both teams, thus the required answer is 0.1365 2*1000 = 273.

Let the teams be A and B. Then the possible results are: AABA,ABAA,BAAA and the analogous results when team B wins (which, by the way, just multiplies the probability by 2).

$1000p = 2000(\frac12\frac7{10}\frac3{10}\frac3{10}+\frac12\frac3{10}\frac3{10}\frac7{10}+\frac12\frac3{10}\frac7{10}\frac7{10})=273$ .

Let the two teams be A and B respectively.

Note that for there to be exactly 4 games, the victorious team has to have won thrice and lost once. The one game that they lost also cannot be the last. Otherwise, the team would have won the finals in three games.

So, the three possibilities (with team A winning) are as follows:

$1. \text{AABA} \;$

$2. \text{ABAA} \;$

$3. \text{BAAA} \;$

The probabilities are as follows:

$1. \frac{1} {2} \times \frac{7} {10} \times \frac{3} {10} \times \frac{3} {10} = \frac{63} {2000}.$ This is because for the first game, A has a $\frac{1} {2}$ chance of winning. For the second game, since A won the first game , it has a $\frac{7} {10}$ chance of winning. For the third game, since B lost the previous game, it has a $\frac{3} {10}$ chance of winning. Lastly, for the last game, since A lost the previous game, it also has a $\frac{3} {10}$ chance of winning. This results in a $\frac{63} {2000}$ chance of this case happening.

$2. \frac{1} {2} \times \frac{3} {10} \times \frac{3} {10} \times \frac{7} {10} = \frac{63} {2000}.$ The reasoning here is also similar to the above case.

$3. \frac{1} {2} \times \frac{3} {10} \times \frac{7} {10} \times \frac{7} {10} = \frac{147} {2000}.$ In this case, the chance of B winning the first game is $\frac{1} {2}$ . The chance of A winning the next game is thus $\frac{3} {10}$ , since it lost the previous game. For the next game, A has a $\frac{7} {10}$ chance of winning, as it has won the previous game. It is the same for the last game. Multiplying these probabilities together, the probability of this case will be calculated as $\frac{147} {2000}.$

When these 3 probabilities are added up, $\frac{273} {2000}$ is the result. However, we also have to take into account the cases where team B wins. Thus, we have to take $p = \frac{273} {2000} \times 2 = \frac{273} {1000}.$

Therefore, $1000p = \frac{273} {1000} \times 1000 = 273.$

probability of occurence independent events simultaneously is product of individual probablities either of the two can win and possible wins are WWLW-------(1/2)(7/10)(3/10)(3/10) WLWW-------(1/2)(3/10)(3/10)(7/10) LWWW-------(1/2)(3/10)(7/10)(7/10) ADDING AND MULTIPLYING BY 2 FOR EACH TO WIN p=1000p=273 hence the result

First of all, only focus on when teams when. Secondly, let a 1 represent a win for the first team and 2 represent a win for the second team.

Then, compile a list of all the possible combinations of the two digits as 4-digit numbers and you get:

1111

1112

1121

1122

1211

1212

1222

2222

2111

2112

2121

2211

2212

Next, break the list down to the possible outcomes of the first to five match that last EXACTLY 4 games. This means that numbers like 1112 can't be counted since the first team got three wins before the fourth game, meaning they wouldn't have played the fourth game.

The new list is:

1121

1211

1222

2111

2122

2212

With these six events, you can work out the individual probabilities of the happening. Remember that the probability of winning changes depending on who won the previous game.

You can also note that all the denominators will be 2000, so you can focus on the numerator's only for now.

1121 $\rightarrow$ $1\times7\times3\times3 = 63$

1211 $\rightarrow$ $1\times3\times3\times7 = 63$

1222 $\rightarrow$ $1\times3\times7\times7 = 147$

2111 $\rightarrow$ $1\times3\times7\times7 = 147$

2122 $\rightarrow$ $1\times3\times3\times7 = 63$

2212 $\rightarrow$ $1\times7\times3\times3 = 63$

Now add up the numerators and you get:

$147\times2+4\times63 = 546$

$\frac {546}{2000} \times 1000$

= $\frac{546}{2}$

= 273

$\left \lfloor {273} \right \rfloor = 273$ , and that's your answer.

If the game is a best-of-five, and we know it ends in four games, there has to be a team that won three games out of the four (not the first three though, otherwise, the series would have ended in three games.)

In how many ways could this have happened? Let us consider the possibilities through the perspective of one team, representing wins as Ws and losses as Ls.

1. LWWW
2. WLWW

3. WWLW

4. WLLL

5. LWLL
6. LLWL

Let us calculate the probability case by case:

1. The first loss has a probability of $\frac{1}{2}$ . The next win (given a previous loss) has a probability of $\frac{3}{10}$ . The next win (given a previous win) has a probability of $\frac{7}{10}$ . The final win (given a previous win) has a probability of $\frac{7}{10}$ .

$P_{1} = \dfrac{1}{2} \times \dfrac{3}{10} \times \dfrac{7}{10} \times \dfrac{7}{10} = \dfrac{147}{2000}$

Using the same logic:

$P_{2} = \dfrac{1}{2} \times \dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10} = \dfrac{63}{2000}$

$P_{3} = \dfrac{1}{2} \times \dfrac{7}{10} \times \dfrac{3}{10} \times \dfrac{3}{10} = \dfrac{63}{2000}$

It is important to note the symmetry between the winning cases and the losing cases for our chosen team; they have identical probabilities. Hence:

$P_{1+2+3+4+5+6} = 2\left(\dfrac{147}{2000}+\dfrac{63}{2000}+\dfrac{63}{2000}\right) = \dfrac{273}{1000} \Longrightarrow \dfrac{273}{1000} \times 1000 = \boxed{273}$

Let's assume the two teams as $A$ and $B$ .

Ok Since it says "Exactly 4 games" so there must be 3 wins and 1 loss for team $A$ .The same case goes for Team $B$ .

Now There are 4 cases for Each possibility (Assuming team $A$ wins):

1.WWWL

2.WWLW

3.WLWW

4.LWWW

Case 1) not possible, since team $A$ has won 3 times first and there is no need for the 4th game.

Case 2) team $A$ wins the first game by $frac{1}{2}$ chance. It wins the second game by a chance of $frac{7}{10}$ . Since team $A$ won the second game the chance of team $B$ winning is $frac{3}{10}$ . At the last game since team $B$ won the previous game the chance of team $A$ winning is $frac{3}{10}$ .

So in total: $frac{1}{2}\timesfrac{7}{10}\timesfrac{3}{10}\timesfrac{3}{10}=frac{63}{2000}$

Case 3) team $A$ wins the first game by $frac{1}{2}$ chance. Since team $A$ won the first game, the chance of team $B$ winning is frac{3}{10}\. At the third game since team \(B won the previous game, the chance of team $A$ winning is $frac{3}{10}$ . Team $A$ wins the last game by a probability of $frac{7}{10}$

So in total: $frac{1}{2}\timesfrac{3}{10}\timesfrac{3}{10}\timesfrac{7}{10}=frac{63}{2000}$

Case 4) team $B$ wins the first game by $frac{1}{2}$ chance. Since team $B$ won the first game, the chance of team $A$ winning is frac{3}{10}\. At the third game since team \(A won the previous game, the chance of it winning is $frac{7}{10}$ . Team $A$ wins the last game by a probability of $frac{7}{10}$

So in total: $frac{1}{2}\timesfrac{3}{10}\timesfrac{7}{10}\timesfrac{7}{10}=frac{147}{2000}$

Adding The probabilities we get: $0+frac{63}{2000}+frac{63}{2000}+frac{147}{2000}=frac{237}{2000}$

Since there were 2 ways from the beginning (team $A$ winning or team $B$ ): $p=2\timesfrac{237}{2000}=frac{237}{1000}$

so 1000p is: $p\times1000=frac{237}{1000}\times1000=237$

so the answer is $\boxed{237}$

Let O denote the first team and T denote the second team. If a series lasts exactly 4 games, the following scenarios can occur (when the O team ends up winning the series)

• OOTO
• OTOO
• TOOO

The probabilities for these events respectively are

• OOTO $\dfrac{1}{2} \times \dfrac{7}{10} \times \dfrac{3}{10} \times \dfrac{3}{10}$

• OTOO $\dfrac{1}{2} \times \dfrac{3}{10} \times \dfrac{3}{10} \times \dfrac{7}{10}$

• TOOO $\dfrac{1}{2} \times \dfrac{3}{10} \times \dfrac{7}{10} \times \dfrac{7}{10}$

Calculating the sum of these probabilities, we get $0.1365$

But remember, we only calculated the chances when the O team wins. The same probabilities apply when the T team wins. (TTOT,TOTT,OTTT)

Therefore, we multiply the answer we got above by 2. We get $0.1365 \times 2 = 0.273$

Thus, $p = 0.273$ and $\lfloor{1000p} \rfloor= \boxed{273}$

Let teams be $A$ and $B$

If it is best out of $5$ , the first team to win $3$ times is the winner. So a configuration of $4$ games with $3$ wins for one team and $1$ win for the other team must be found. Without loss of generality, let $A$ win.

The configurations are:

$1)$ $AABA$

$2)$ $ABAA$

$3)$ $BAAA$

$4)$ $AAAB$ but this last one is not possible since A would have won after $3$ games.

$P(1)$ = $\frac{1}{2}*\frac{7}{10}*\frac{3}{10}*\frac{3}{10}=\frac{63}{2000}$

$P(2)$ = $\frac{1}{2}*\frac{3}{10}*\frac{3}{10}*\frac{7}{10}=\frac{63}{2000}$

$P(3)$ = $\frac{1}{2}*\frac{3}{10}*\frac{7}{10}*\frac{7}{10}=\frac{147}{2000}$

so $P(1)$ + $P(2)$ + $P(3)$ $=\frac{63}{2000}+\frac{63}{2000}+\frac{147}{2000}=\frac{273}{2000}$

But we must double this to get $\frac{273}{1000}$ since we could have $B$ win instead and repeat those same configurations with $A$ replaced by $B$

so $p=\frac{273}{1000}$ and so $\lfloor 1000p \rfloor= \lfloor 273\rfloor= \boxed{273}$

a probability distribution can be made as follows.the numbers outside are proportional to probability brackets show score and star indicates who won.

game 1: 1(1*,0).

game 2: 7(2* 0) 3(1 1*).

game 3: 49(3* 0) 21(2 1 *) 30(2 *1){this is because (1 1 * )is symmetrical for both the teams).

since the outcome 3,0 doesn't require game 4 its outcome doesn't matter.

game 4 490(_) 63(3 * 1) 147(2 2 *) 210(3 * 1) 90(2 2 *).

thus p= 63+210/490+63+147+210+90.

=273/1000