Boston Celtics VS. LA Lakers- Expectation of series of games? 
Boston celtics & LA Lakers play a series of games. the first team who win 4 games, win the whole series.

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*The probability of win or lose game is equal (1/2)

a. what is the expectation of the number of games in the series?
    


So i defined an indicator: $x_i=1$ if the game $i$ was played.
It is clear that $E[x_1]=E[x_2]=E[x_3]=E[x_4]=1$.
for the 5th game, for each team we have 5 different scenarios (W=win, L=lose) and the probability to win is:


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*W W W L   $((\frac12)^3=\frac18)$

*W W L L   $((\frac12)^2=\frac14)$

*W L L L   $((\frac12)^1=\frac12)$


$\frac12+\frac14+\frac18=\frac78$
We can use the complementary event: if only one game is L in a series of 4 games (so the 5th game will be played): $1-\frac18=\frac78$
The same calculation is for the 6th and 7th game:
6th:  W W W L L or W W L L L    => $(\frac18+\frac14=\frac38)$
7th:  W W W L L L =>  $(\frac18)$
And the expectation is $E[x]=1+1+1+1+\frac78+\frac38+\frac18=\frac{43}8$
What am i missing here, and how can I fix it?
 A: We find the probability the series lasts $4$ games, $5$, $6$, $7$. Then we can find the expectation using a basic calculation. 
Let $X$ be the number of games. 
It is clear that $\Pr(X=4)=\frac{2}{2^4}$. For the probability the Celtics win in $4$ is $\frac{1}{2^4}$. Double that to take into account the probability the Lakers win in $4$.
To find $\Pr(X=5)$, again there are two cases, Celtics win series in $5$ games and Lakers win series in $5$. Let's find the probability Celtics win in $5$. They must win exactly $3$ of the first $4$ games, and win the fifth. 
The probability of this is $\binom{4}{1}(1/2)^3(1/2)(1/2)$. Multiply by $2$ for the probability the series lasts exactly $5$ games. 
Similarly, the probability the Celtics win in $6$ is $\binom{5}{3}(1/2)^3(1/2)^2(1/2)$. 
Again, multiply by $2$ to find $\Pr(X=6)$.
Similar reasoning gives us the probability the series lasts $7$ games. Or else we can find that by subtracting the sum of the other $3$ probabilities from $1$.
As mentioned at the beginning, now that you have the distribution of $X$, calculation of $E(X)$ is straightforward.
Remarks: $1.$ Note that the probability the series lasts exactly $6$ games is equal to the probability that the series lasts $7$. For suppose that after the $5$-th game, the series has not been decided. Then one team is leading, $3$ games to $2$. With probability $\frac{1}{2}$ it will win the $6$-th game, and the series will be over in $6$. And with probability $\frac{1}{2}$ it will lose the $6$-th game, and the series will require $7$ games. Thus $\Pr(X=6)=\Pr(X=7)$. 
Precisely the same is true of a series where the first team to win $n+1$ games, where $n\ge 1$, wins the series. The probability the series lasts $2n$ games 
is equal to the probability the series lasts $2n+1$ games. (We are again assuming independence, and that each team has probability $\frac{1}{2}$ of winning any particular game.)
$2.$ During our calculations, we implicitly assumed independence. We need to know not only that each team has probability $1/2$ of winning any game, we must also assume that game results are independent. 
A: If we have the family of all length-7 sequences composed of W and L, we see that each of these sequences represent one of $2^7$ outcomes to our task at hand with equal probability. Then, we see that the number of games played is pre-decided for each such given sequence (e.x.: WWWLWLL and WWWLWWW both result in five games played, while WLWLWLW result in seven games). So, we can find the probability of each event (number of games played) by counting how many sequences fall into each category.
Note: O indicates that this can be either W or L; this occurs when the outcome of the series is already decided and additional games become irrelevant to the total number of games played. Also, let us assume that the W team wins the seven-game series without loss of generality.
Four games: WWWWOOO
$2^3=8$ sequences
Five games: (combination of WWWL)WOO
$2^2{{4}\choose{3}} = 16$ sequences
Six games: (combination of WWWLL)WO
$2{{5}\choose{3}} = 20$ sequences
Seven games: (combination of WWWLLL)W
${{6}\choose{3}} = 20$ sequences
So, the expected value for number of games played is:
$4(\frac{8}{64})+5(\frac{16}{64})+6(\frac{20}{64})+7(\frac{20}{64})=\frac{93}{16}=5.8125$ games
